Evaluate $int_{-pi/4}^{pi/4}frac{x}{sin x}mathrm{d}x$
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I am working on the integral
$$I=int_{-pi/4}^{pi/4}frac{x}{sin x}mathrm{d}x=2int_0^{pi/4}frac{x}{sin x}mathrm{d}x$$
Which I am fairly confident has a closed form, as $$int_{0}^{pi/2}frac{x}{sin x}mathrm{d}x=2G$$
Where $G$ is Catalan's constant.
Preforming a tangent half angle substitution, we have that
$$I=4int_0^{sqrt{2}-1}frac{arctan x}{x}mathrm{d}x$$
Then using $$arctan x=sum_{ngeq0}(-1)^nfrac{x^{2n+1}}{2n+1}$$
We have
$$I=4sum_{ngeq0}frac{(-1)^n}{(2n+1)^2}(sqrt{2}-1)^{2n+1}$$
Which is painfully similar to $G$. I do not know how to deal with that extra $(sqrt{2}-1)^{2n+1}$ bit though...
In another post of mine I showed that
$$I=pisum_{ngeq1} nlogbigg(frac{4n+1}{4n-1}bigg)prod_{kgeq1\kneq n}frac{k^2}{k^2-n^2}$$
And similarly I showed that
$$sum_{ngeq1}nlogbigg(frac{2n+1}{2n-1}bigg)prod_{kgeq1\kneq n}frac{k^2}{k^2-n^2}=frac{4G}pi$$
So I have to questions. How do I find an exact value for $I$? And are the last two series representations correct? Thanks.
Major Edit:
Okay so I found a closed form for the integral. Wolfy gave me
$$intfrac{x}{sin x}mathrm{d}x=ibigg(text{Li}_2(-e^{ix})-text{Li}_2(e^{ix})bigg)+xlogfrac{1-e^{ix}}{1+e^{ix}}$$
I guess that Wolfy didn't want to do the algebra, so I did it by hand. It took me like $10$ minutes, but I am pretty sure that
$$I=-frac34G+frac{pi^2}4bigg(frac{13}{24}-ibigg)-frac{ipi}4log(1+sqrt{2})+frac{i-1}{32sqrt{2}}bigg[psi^{(1)}bigg(frac{5}{8}bigg)-psi^{(1)}bigg(frac{1}{8}bigg)bigg]+frac{i+1}{32sqrt{2}}bigg[psi^{(1)}bigg(frac{3}{8}bigg)-psi^{(1)}bigg(frac{7}{8}bigg)bigg]$$
Where $psi^{(1)}$ is the first derivative of the di-gamma function.
integration sequences-and-series definite-integrals closed-form
asked Dec 21 '18 at 2:43
clathratusclathratus
5,1141439
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show 2 more comments
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I am working on the integral
$$I=int_{-pi/4}^{pi/4}frac{x}{sin x}mathrm{d}x=2int_0^{pi/4}frac{x}{sin x}mathrm{d}x$$
Which I am fairly confident has a closed form, as $$int_{0}^{pi/2}frac{x}{sin x}mathrm{d}x=2G$$
Where $G$ is Catalan's constant.
Preforming a tangent half angle substitution, we have that
$$I=4int_0^{sqrt{2}-1}frac{arctan x}{x}mathrm{d}x$$
Then using $$arctan x=sum_{ngeq0}(-1)^nfrac{x^{2n+1}}{2n+1}$$
We have
$$I=4sum_{ngeq0}frac{(-1)^n}{(2n+1)^2}(sqrt{2}-1)^{2n+1}$$
Which is painfully similar to $G$. I do not know how to deal with that extra $(sqrt{2}-1)^{2n+1}$ bit though...
In another post of mine I showed that
$$I=pisum_{ngeq1} nlogbigg(frac{4n+1}{4n-1}bigg)prod_{kgeq1\kneq n}frac{k^2}{k^2-n^2}$$
And similarly I showed that
$$sum_{ngeq1}nlogbigg(frac{2n+1}{2n-1}bigg)prod_{kgeq1\kneq n}frac{k^2}{k^2-n^2}=frac{4G}pi$$
So I have to questions. How do I find an exact value for $I$? And are the last two series representations correct? Thanks.
Major Edit:
Okay so I found a closed form for the integral. Wolfy gave me
$$intfrac{x}{sin x}mathrm{d}x=ibigg(text{Li}_2(-e^{ix})-text{Li}_2(e^{ix})bigg)+xlogfrac{1-e^{ix}}{1+e^{ix}}$$
I guess that Wolfy didn't want to do the algebra, so I did it by hand. It took me like $10$ minutes, but I am pretty sure that
$$I=-frac34G+frac{pi^2}4bigg(frac{13}{24}-ibigg)-frac{ipi}4log(1+sqrt{2})+frac{i-1}{32sqrt{2}}bigg[psi^{(1)}bigg(frac{5}{8}bigg)-psi^{(1)}bigg(frac{1}{8}bigg)bigg]+frac{i+1}{32sqrt{2}}bigg[psi^{(1)}bigg(frac{3}{8}bigg)-psi^{(1)}bigg(frac{7}{8}bigg)bigg]$$
Where $psi^{(1)}$ is the first derivative of the di-gamma function.
integration sequences-and-series definite-integrals closed-form
asked Dec 21 '18 at 2:43
clathratusclathratus
5,1141439
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1
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I'm not sure why you are confident that your integral has a closed form. Not that Wolfram Alpha is the last word on integration, but it is usually pretty good at recognizing constants, and in this case it does not.
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– DeficientMathDude
Dec 21 '18 at 2:49
1
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@DeficientMathDude - You just contradicted yourself.
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– user150203
Dec 21 '18 at 3:31
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Wolframalpha numerical value for integral = 1.62706 and 2*G = 1.83193118
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– user150203
Dec 21 '18 at 3:39
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You can write $frac{x}{sin(x)}$ as $xcsc(x)$. See where I'm going with this?
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– Aniruddh Venkatesan
Dec 21 '18 at 3:42
1
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@DeficientMathDude see the edit I just made
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– clathratus
Dec 21 '18 at 3:50
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show 2 more comments
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I am working on the integral
$$I=int_{-pi/4}^{pi/4}frac{x}{sin x}mathrm{d}x=2int_0^{pi/4}frac{x}{sin x}mathrm{d}x$$
Which I am fairly confident has a closed form, as $$int_{0}^{pi/2}frac{x}{sin x}mathrm{d}x=2G$$
Where $G$ is Catalan's constant.
Preforming a tangent half angle substitution, we have that
$$I=4int_0^{sqrt{2}-1}frac{arctan x}{x}mathrm{d}x$$
Then using $$arctan x=sum_{ngeq0}(-1)^nfrac{x^{2n+1}}{2n+1}$$
We have
$$I=4sum_{ngeq0}frac{(-1)^n}{(2n+1)^2}(sqrt{2}-1)^{2n+1}$$
Which is painfully similar to $G$. I do not know how to deal with that extra $(sqrt{2}-1)^{2n+1}$ bit though...
In another post of mine I showed that
$$I=pisum_{ngeq1} nlogbigg(frac{4n+1}{4n-1}bigg)prod_{kgeq1\kneq n}frac{k^2}{k^2-n^2}$$
And similarly I showed that
$$sum_{ngeq1}nlogbigg(frac{2n+1}{2n-1}bigg)prod_{kgeq1\kneq n}frac{k^2}{k^2-n^2}=frac{4G}pi$$
So I have to questions. How do I find an exact value for $I$? And are the last two series representations correct? Thanks.
Major Edit:
Okay so I found a closed form for the integral. Wolfy gave me
$$intfrac{x}{sin x}mathrm{d}x=ibigg(text{Li}_2(-e^{ix})-text{Li}_2(e^{ix})bigg)+xlogfrac{1-e^{ix}}{1+e^{ix}}$$
I guess that Wolfy didn't want to do the algebra, so I did it by hand. It took me like $10$ minutes, but I am pretty sure that
$$I=-frac34G+frac{pi^2}4bigg(frac{13}{24}-ibigg)-frac{ipi}4log(1+sqrt{2})+frac{i-1}{32sqrt{2}}bigg[psi^{(1)}bigg(frac{5}{8}bigg)-psi^{(1)}bigg(frac{1}{8}bigg)bigg]+frac{i+1}{32sqrt{2}}bigg[psi^{(1)}bigg(frac{3}{8}bigg)-psi^{(1)}bigg(frac{7}{8}bigg)bigg]$$
Where $psi^{(1)}$ is the first derivative of the di-gamma function.
integration sequences-and-series definite-integrals closed-form
asked Dec 21 '18 at 2:43
clathratusclathratus
5,1141439
$endgroup$
I am working on the integral
$$I=int_{-pi/4}^{pi/4}frac{x}{sin x}mathrm{d}x=2int_0^{pi/4}frac{x}{sin x}mathrm{d}x$$
Which I am fairly confident has a closed form, as $$int_{0}^{pi/2}frac{x}{sin x}mathrm{d}x=2G$$
Where $G$ is Catalan's constant.
Preforming a tangent half angle substitution, we have that
$$I=4int_0^{sqrt{2}-1}frac{arctan x}{x}mathrm{d}x$$
Then using $$arctan x=sum_{ngeq0}(-1)^nfrac{x^{2n+1}}{2n+1}$$
We have
$$I=4sum_{ngeq0}frac{(-1)^n}{(2n+1)^2}(sqrt{2}-1)^{2n+1}$$
Which is painfully similar to $G$. I do not know how to deal with that extra $(sqrt{2}-1)^{2n+1}$ bit though...
In another post of mine I showed that
$$I=pisum_{ngeq1} nlogbigg(frac{4n+1}{4n-1}bigg)prod_{kgeq1\kneq n}frac{k^2}{k^2-n^2}$$
And similarly I showed that
$$sum_{ngeq1}nlogbigg(frac{2n+1}{2n-1}bigg)prod_{kgeq1\kneq n}frac{k^2}{k^2-n^2}=frac{4G}pi$$
So I have to questions. How do I find an exact value for $I$? And are the last two series representations correct? Thanks.
Major Edit:
Okay so I found a closed form for the integral. Wolfy gave me
$$intfrac{x}{sin x}mathrm{d}x=ibigg(text{Li}_2(-e^{ix})-text{Li}_2(e^{ix})bigg)+xlogfrac{1-e^{ix}}{1+e^{ix}}$$
I guess that Wolfy didn't want to do the algebra, so I did it by hand. It took me like $10$ minutes, but I am pretty sure that
$$I=-frac34G+frac{pi^2}4bigg(frac{13}{24}-ibigg)-frac{ipi}4log(1+sqrt{2})+frac{i-1}{32sqrt{2}}bigg[psi^{(1)}bigg(frac{5}{8}bigg)-psi^{(1)}bigg(frac{1}{8}bigg)bigg]+frac{i+1}{32sqrt{2}}bigg[psi^{(1)}bigg(frac{3}{8}bigg)-psi^{(1)}bigg(frac{7}{8}bigg)bigg]$$
Where $psi^{(1)}$ is the first derivative of the di-gamma function.
integration sequences-and-series definite-integrals closed-form
integration sequences-and-series definite-integrals closed-form
asked Dec 21 '18 at 2:43
clathratusclathratus
5,1141439
asked Dec 21 '18 at 2:43
clathratusclathratus
5,1141439
edited Dec 21 '18 at 5:12
clathratus
asked Dec 21 '18 at 2:43
clathratusclathratus
5,1141439
asked Dec 21 '18 at 2:43
clathratusclathratus
5,1141439
asked Dec 21 '18 at 2:43
clathratusclathratus
5,1141439
5,1141439
1
$begingroup$
I'm not sure why you are confident that your integral has a closed form. Not that Wolfram Alpha is the last word on integration, but it is usually pretty good at recognizing constants, and in this case it does not.
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– DeficientMathDude
Dec 21 '18 at 2:49
1
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@DeficientMathDude - You just contradicted yourself.
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– user150203
Dec 21 '18 at 3:31
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Wolframalpha numerical value for integral = 1.62706 and 2*G = 1.83193118
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– user150203
Dec 21 '18 at 3:39
$begingroup$
You can write $frac{x}{sin(x)}$ as $xcsc(x)$. See where I'm going with this?
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– Aniruddh Venkatesan
Dec 21 '18 at 3:42
1
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@DeficientMathDude see the edit I just made
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– clathratus
Dec 21 '18 at 3:50
|
show 2 more comments
1
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I'm not sure why you are confident that your integral has a closed form. Not that Wolfram Alpha is the last word on integration, but it is usually pretty good at recognizing constants, and in this case it does not.
$endgroup$
– DeficientMathDude
Dec 21 '18 at 2:49
1
$begingroup$
@DeficientMathDude - You just contradicted yourself.
$endgroup$
– user150203
Dec 21 '18 at 3:31
$begingroup$
Wolframalpha numerical value for integral = 1.62706 and 2*G = 1.83193118
$endgroup$
– user150203
Dec 21 '18 at 3:39
$begingroup$
You can write $frac{x}{sin(x)}$ as $xcsc(x)$. See where I'm going with this?
$endgroup$
– Aniruddh Venkatesan
Dec 21 '18 at 3:42
1
$begingroup$
@DeficientMathDude see the edit I just made
$endgroup$
– clathratus
Dec 21 '18 at 3:50
1
1
$begingroup$
I'm not sure why you are confident that your integral has a closed form. Not that Wolfram Alpha is the last word on integration, but it is usually pretty good at recognizing constants, and in this case it does not.
$endgroup$
– DeficientMathDude
Dec 21 '18 at 2:49
$begingroup$
I'm not sure why you are confident that your integral has a closed form. Not that Wolfram Alpha is the last word on integration, but it is usually pretty good at recognizing constants, and in this case it does not.
$endgroup$
– DeficientMathDude
Dec 21 '18 at 2:49
1
1
$begingroup$
@DeficientMathDude - You just contradicted yourself.
$endgroup$
– user150203
Dec 21 '18 at 3:31
$begingroup$
@DeficientMathDude - You just contradicted yourself.
$endgroup$
– user150203
Dec 21 '18 at 3:31
$begingroup$
Wolframalpha numerical value for integral = 1.62706 and 2*G = 1.83193118
$endgroup$
– user150203
Dec 21 '18 at 3:39
$begingroup$
Wolframalpha numerical value for integral = 1.62706 and 2*G = 1.83193118
$endgroup$
– user150203
Dec 21 '18 at 3:39
$begingroup$
You can write $frac{x}{sin(x)}$ as $xcsc(x)$. See where I'm going with this?
$endgroup$
– Aniruddh Venkatesan
Dec 21 '18 at 3:42
$begingroup$
You can write $frac{x}{sin(x)}$ as $xcsc(x)$. See where I'm going with this?
$endgroup$
– Aniruddh Venkatesan
Dec 21 '18 at 3:42
1
1
$begingroup$
@DeficientMathDude see the edit I just made
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– clathratus
Dec 21 '18 at 3:50
$begingroup$
@DeficientMathDude see the edit I just made
$endgroup$
– clathratus
Dec 21 '18 at 3:50
|
show 2 more comments
5 Answers
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Let
$$I = int_{-pi/4}^{pi/4} x , text{cosec} , x , dx = 2 int_0^{pi/4} x , text{cosec} , x , dx.$$
After integrating by parts we have
begin{align}
I &= -frac{pi}{2} ln (1 + sqrt{2}) + 2 int_0^{pi/4} ln (text{cosec} , x + cot x) , dx\
&= -frac{pi}{2} ln (1 + sqrt{2}) + 2 int_0^{pi/4} ln (1 + cos x) , dx - 2 int_0^{pi/4} ln (sin x) , dx.
end{align}
The second of these integrals is perhaps (?) reasonably well known (for an evaluation, see here). The result is:
$$int_0^{pi/4} ln (sin x) , dx = -frac{1}{2} mathbf{G} - frac{pi}{4} ln 2,$$
where $mathbf{G}$ is Catalan's constant. Thus
$$I = -frac{pi}{2} ln (1 + sqrt{2}) + mathbf{G} + frac{pi}{2} ln 2 + 2I_1.$$
For the first of the integrals we will make use of the following formula, a proof of which can be found here
$$ln (1 + cos x) = 2 sum_{n = 1}^infty (-1)^{n + 1} frac{cos (nx)}{n} - ln 2.$$
So
begin{align}
I_1 &= int_0^{pi/4} ln (1 + cos x) , dx\
&= 2 sum_{n = 1}^infty frac{(-1)^{n + 1}}{n} int_0^{pi/4} cos (nx) , dx - ln 2 int_0^{pi/4} dx\
&= -frac{pi}{4} ln 2 + 2 sum_{n = 1}^infty frac{(-1)^{n + 1}}{n^2} sin left (frac{n pi}{4} right )\
&= -frac{pi}{4} ln 2 + 2 sum_{stackrel{{Large{n = 1}}}{n in text{odd}}}^infty frac{(-1)^{n + 1}}{n^2} sin left (frac{n pi}{4} right ) + + 2 sum_{stackrel{{Large{n = 1}}}{n in text{even}}}^infty frac{(-1)^{n + 1}}{n^2} sin left (frac{n pi}{4} right )\
&= -frac{pi}{4} ln 2 + 2 sum_{n = 1}^infty frac{1}{(2n - 1)^2} sin left [frac{pi}{4} (2n - 1) right ] - frac{1}{2} sum_{n = 1}^infty frac{1}{n^2} sin left (frac{n pi}{2} right )\
&= -frac{pi}{4} ln 2 + 2 S_1 - frac{1}{2} S_2.
end{align}
For the second of these sums,
begin{align}
S_2 &= sum_{stackrel{{Large{n = 1}}}{n in text{odd}}}^infty frac{1}{n^2} sin left (frac{pi n}{2} right ) + sum_{stackrel{{Large{n = 1}}}{n in text{even}}} frac{1}{n^2} sin left (frac{pi n}{2} right )\
&= sum_{n = 0}^infty frac{(-1)^n}{(2n + 1)^2}\
&= mathbf{G},
end{align}
after a shift of the index $n mapsto 2n + 1$ in the odd sum has been made while the even sum is identically equal to zero.
For the first of the sums, as it converges absolutely we can split it up as folows:
$$S_1 = sum_{stackrel{{Large{n = 1}}}{n in 1,5,9,ldots}}^infty frac{1}{(2n - 1)^2} sin left [frac{pi}{4} (2k - 1) right ] + sum_{stackrel{{Large{n = 1}}}{n in 2,6,10,ldots}}^infty frac{1}{(2n - 1)^2} sin left [frac{pi}{4} (2k - 1) right ] + sum_{stackrel{{Large{n = 1}}}{n in 3,7,11,ldots}}^infty frac{1}{(2n - 1)^2} sin left [frac{pi}{4} (2k - 1) right ] + sum_{stackrel{{Large{n = 1}}}{n in 4,8,12,ldots}}^infty frac{1}{(2n - 1)^2} sin left [frac{pi}{4} (2k - 1) right ].$$
Shifting the indices as follows: $n mapsto 4n - 3, n mapsto 4n - 2, n mapsto 4n - 1, n mapsto 4n$ leads to
begin{align}
S_1 &= frac{1}{sqrt{2}} left [sum_{n = 1}^infty frac{1}{(8n - 7)^2} + sum_{n = 1}^infty frac{1}{(8n - 5)^2} - sum_{n = 1}^infty frac{1}{(8n - 3)^2} - sum_{n = 1}^infty frac{1}{(8n - 1)^2} right ]\
&= frac{1}{sqrt{2}} left [sum_{n = 0}^infty frac{1}{(8n + 1)^2} + sum_{n = 0}^infty frac{1}{(8n + 3)^2} - sum_{n = 0}^infty frac{1}{(8n + 5)^2} - sum_{n = 0}^infty frac{1}{(8n + 7)^2} right ]\
&= frac{1}{64 sqrt{2}} left [sum_{n = 0}^infty frac{1}{(n + 1/8)^2} + sum_{n = 0}^infty frac{1}{(n + 3/8)^2} - sum_{n = 0}^infty frac{1}{(n + 5/8)^2} - sum_{n = 0}^infty frac{1}{(n + 7/8)^2} right ]\
&= frac{1}{64 sqrt{2}} left [psi^{(1)} left (frac{1}{8} right ) + psi^{(1)} left (frac{3}{8} right ) - psi^{(1)} left (frac{5}{8} right ) - psi^{(1)} left (frac{7}{8} right ) right ],
end{align}
where we have made use of the series representation for the polygamma function of order one (as known as the trigamma function). So the value for $I_1$ is:
$$I_1 = -frac{pi}{4} ln 2 - frac{1}{2} mathbf{G} + frac{1}{32 sqrt{2}} left [psi^{(1)} left (frac{1}{8} right ) + psi^{(1)} left (frac{3}{8} right ) - psi^{(1)} left (frac{5}{8} right ) - psi^{(1)} left (frac{7}{8} right ) right ],$$
leading to a final result of
$$int_{-pi/4}^{pi/4} x , text{cosec} , x , dx = -frac{pi}{2} ln (1 + sqrt{2}) + frac{1}{16 sqrt{2}} left [psi^{(1)} left (frac{1}{8} right ) + psi^{(1)} left (frac{3}{8} right ) - psi^{(1)} left (frac{5}{8} right ) - psi^{(1)} left (frac{7}{8} right ) right ].$$
Converting the trigamma functions to Clausen functions of order two
Inspired by the answer given by Zacky in terms of the Clausen function of order two, here I will show how to convert my answer in terms of the 4 trigamma functions into 2 Clausen functions of order 2.
The relation between the Clausen function of order two and the trigamma function is given by (a proof of this can be found here)
$$text{Cl}_2 left (frac{q pi}{p} right ) = frac{1}{(2p)^{2m} (2m - 1)!} sum_{n = 1}^p sin left (frac{qnpi}{p} right ) left [psi^{(1)} left (frac{n}{2p} right ) + (-1)^q psi^{(1)} left (frac{n + p}{2p} right ) right ].$$
Setting $m = 1, q = 1, p = 4$ gives
begin{align}
text{Cl}_2 left (frac{pi}{4} right ) &= frac{1}{64} left [frac{1}{sqrt{2}} left {psi^{(1)} left (frac{1}{8} right ) - psi^{(1)} left (frac{5}{8} right ) right } + psi^{(1)} left (frac{1}{4} right ) - psi^{(1)} left (frac{3}{4} right ) right.\
& qquad left. + frac{1}{sqrt{2}} left {psi^{(1)} left (frac{3}{8} right ) - psi^{(1)} left (frac{7}{8} right ) right } right ], qquad (*)
end{align}
and setting $m = 1, q = 3, p = 4$ gives
begin{align}
text{Cl}_2 left (frac{3pi}{4} right ) &= frac{1}{64} left [frac{1}{sqrt{2}} left {psi^{(1)} left (frac{1}{8} right ) - psi^{(1)} left (frac{5}{8} right ) right } - psi^{(1)} left (frac{1}{4} right ) + psi^{(1)} left (frac{3}{4} right ) right.\
& qquad left. + frac{1}{sqrt{2}} left {psi^{(1)} left (frac{3}{8} right ) - psi^{(1)} left (frac{7}{8} right ) right } right ]. qquad (**)
end{align}
On adding ($*$) to ($**$) we see that
$$psi^{(1)} left (frac{1}{8} right ) + psi^{(1)} left (frac{3}{8} right ) - psi^{(1)} left (frac{5}{8} right ) - psi^{(1)} left (frac{7}{8} right ) = 32 sqrt{2} left [text{Cl}_2 left (frac{pi}{4} right ) + text{Cl}_2 left (frac{3pi}{4} right ) right ],$$
giving
$$int_{-pi/4}^{pi/4} frac{x}{sin x} , dx = -frac{pi}{2} ln (1 + sqrt{2}) + 2 , text{Cl}_2 left (frac{pi}{4} right ) + 2 , text{Cl}_2 left (frac{3pi}{4} right ).$$
answered Dec 21 '18 at 10:45
omegadotomegadot
6,2692829
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Nice answer! We might be able to reduce those $4$ values of the trigamma function using: $$psi_1left(xright)+psi_1left(x+frac12right)=4psi_1left(2xright)$$ I mean: $$psi_1left(frac18right)+left(frac58right) = 4psi_1left(frac14right)$$ And same with: $$psi_1left(frac38right)+left(frac78right) = 4psi_1left(frac34right)$$ And the values of $psi_1left(frac14right)$ and $psi_1left(frac34right)$ are known from here: mathworld.wolfram.com/TrigammaFunction.html
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– Zacky
Dec 21 '18 at 12:06
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Perfect beautiful wonderful amazing thank you for a really great answer :)
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– clathratus
Dec 21 '18 at 19:04
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Also: does the closed form I found simplify to the answer you give?
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– clathratus
Dec 21 '18 at 19:56
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For the purpose of an alternative method, mostly relying on Clausen function. We have:
$$I=int_{-pi/4}^{pi/4}frac{x}{sin x} dx=2int_0^{pi/4}xleft(lnleft(tanfrac{x}{2}right)right)' dx=2xlnleft(tanfrac{x}{2}right)bigg|_0^frac{pi}{4}-2int_0^frac{pi}{4}lnleft(tanfrac{x}{2}right)dx=$$
$$=frac{pi}{2}ln(sqrt 2-1)-2int_0^frac{pi}{4}left(lnleft(2sinfrac{x}{2}right)-lnleft(2cosfrac{x}{2}right)right)dx=$$$$=frac{pi}{2}ln(sqrt 2-1)+2text{Cl}_2left(frac{pi}{4}right)+2text{Cl}_2left(frac{3pi}{4}right)$$
The last two integrals can be found on the first link. Now is up to the reader if using Clausen function gives any satisfaction, because in desguise is still a series, but same goes with the trigamma function.
An interesting question might be: For what values of $phi$ does the following integral have an elementary answer (Catalan's constant included)?
$ displaystyle{I(phi)=int_0^phi frac{x}{sin x}dx}$. So far I only know about $Ileft(frac{pi}{6}right)$.
answered Dec 21 '18 at 11:46
ZackyZacky
7,84511062
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I would be happy with the 2 order two Clausen functions as an answer. Far more compact compared to the answer I gave in terms of 4 trigamma functions. And besides, are the elementary functions such as $e^x$ and $sin x$ not themselves just series in disguise?
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– omegadot
Dec 21 '18 at 12:01
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You are definetly right. In my mind, by elementary I reffer of something that is learnt in school (which maybe is not quite the best description of something elementary). Maybe Clausen and Trigamma function are studied somewhere, but I only heard of them from the internet.
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– Zacky
Dec 21 '18 at 12:12
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I love special functions. Thanks Zacky
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– clathratus
Dec 21 '18 at 18:58
1
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$Ileft(frac{pi}{2}right)=2G$ , $G$ is the Catalan constant.
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– FDP
Dec 21 '18 at 20:48
add a comment |
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Too long for comments.
$$I=4sum_{ngeq0}frac{(-1)^n}{(2n+1)^2}(sqrt{2}-1)^{2n+1}=4 left(sqrt{2}-1right) ,
_3F_2left(frac{1}{2},frac{1}{2},1;frac{3}{2},frac{3}{2};2
sqrt{2}-3right)$$
On the other hand, working from Wolfram Alpha expression for the antiderivative and trying to simplify as much as I could the integral, I got
$$I=-frac{pi}{4} , log left(3+2 sqrt{2}right)+frac 1 {16 sqrt 2} left(psi ^{(1)}left(frac{1}{8}right)+psi ^{(1)}left(frac{3}{8}right)-psi
^{(1)}left(frac{5}{8}right)-psi ^{(1)}left(frac{7}{8}right) right)$$
What is interesting to mention is that, doing calculations with another CAS, intermediate steps show $G$ appearing a few times.
answered Dec 21 '18 at 6:49
Claude LeiboviciClaude Leibovici
125k1158135
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Yes $mathbf{G}$ does indeed appear twice before canceling out as can be seen in the answer I provide.
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– omegadot
Dec 21 '18 at 10:58
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@omegadot. Using another way, it did appear four times in complex expressions and ran away ! By the way, your answer is real nice ! $to +1$. Cheers.
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– Claude Leibovici
Dec 21 '18 at 11:22
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So does the closed form I found simplify to this?
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– clathratus
Dec 21 '18 at 19:02
add a comment |
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$newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
newcommand{dd}{mathrm{d}}
newcommand{ds}[1]{displaystyle{#1}}
newcommand{expo}[1]{,mathrm{e}^{#1},}
newcommand{ic}{mathrm{i}}
newcommand{mc}[1]{mathcal{#1}}
newcommand{mrm}[1]{mathrm{#1}}
newcommand{pars}[1]{left(,{#1},right)}
newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
newcommand{root}[2]{,sqrt[#1]{,{#2},},}
newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
newcommand{verts}[1]{leftvert,{#1},rightvert}$
begin{align}
I & equiv int_{-pi/4}^{pi/4}{x over sinpars{x}},dd x,
pars{~mbox{OP already shows that}
I = bbox[10px,#ffd]{!!!!! 4!int_{0}^{root{2} - 1}!!{arctanpars{x} over x},dd x}~}
end{align}
Then,
begin{align}
I & = bbox[10px,#ffd]{4int_{0}^{root{2} - 1}{arctanpars{x} over x},dd x} =
4,Imint_{0}^{root{2} - 1}{lnpars{1 + ic x} over x},dd x
\[5mm] &
stackrel{{large x = ic t} atop {large t = -ic x}}{=},,,
4,Imint_{0}^{-pars{root{2} - 1}ic}
{lnpars{1 - t} over t},dd t
\[5mm] & =
-4,Imint_{0}^{-pars{root{2} - 1}ic}
mrm{Li}_{2}'pars{t},dd t =
-4,Immrm{Li}_{2}pars{-bracks{root{2} - 1}ic}
\[5mm] & =
bbx{4,Immrm{Li}_{2}pars{bracks{root{2} - 1}ic}}
approx 1.6271
end{align}
answered Dec 21 '18 at 15:14
Felix MarinFelix Marin
68.9k7110147
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Your final expression $Immrm{Li}_2([sqrt 2 - 1]i)$ is missing a factor of $-4$ infront otherwise it does not approx to $1.6271$.
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– mrtaurho
Dec 21 '18 at 16:41
1
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@mrtaurho Thanks. Indeed it's $displaystylecolor{red}{+}4$ because I already switched the $displaystylemathrm{Li}_{2} $ argument sign. Note that $displaystyleoverline{mathrm{Li}_{2}left(,{z},right)} = mathrm{Li}_{2}left(,{overline{z}},right)$.
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– Felix Marin
Dec 21 '18 at 17:17
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I appreciate the quick and easy approach. Thanks
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– clathratus
Dec 21 '18 at 19:03
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@clathratus You're welcome !!!.
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– Felix Marin
Dec 22 '18 at 16:22
add a comment |
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begin{align}I&=2int_0^{pi/4}frac{x}{sin x}mathrm{d}x\
&=2Big[xlnleft(tanleft(frac{x}{2}right)right)Big]_0^{pi/4}-2int_0^{pi/4}lnleft(tanleft(frac{x}{2}right)right),dx\
&=frac{pi}{2}ln(sqrt{2}-1)-2int_0^{pi/4}lnleft(tanleft(frac{x}{2}right)right),dx\
end{align}
In the latter integral perform the change of variable $y=dfrac{x}{2}$,
begin{align}I&=frac{pi}{2}ln(sqrt{2}-1)-4int_0^{pi/8}lnleft(tanleft(xright)right),dx\
&=frac{pi}{2}ln(sqrt{2}-1)-4Big[frac{1}{2}text{i}big(text{Li}_2(text{i}tan x)-text{Li}_2(-text{i}tan x)big)+xlnleft(tan xright)Big]_0^{pi/8}\
&=boxed{2text{i}left(text{Li}_2left(text{i}left(1-sqrt{2}right)right)-text{Li}_2left(text{i}left(sqrt{2}-1right)right)right)}
end{align}
answered Dec 21 '18 at 21:17
FDPFDP
6,10211929
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add a comment |
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Let
$$I = int_{-pi/4}^{pi/4} x , text{cosec} , x , dx = 2 int_0^{pi/4} x , text{cosec} , x , dx.$$
After integrating by parts we have
begin{align}
I &= -frac{pi}{2} ln (1 + sqrt{2}) + 2 int_0^{pi/4} ln (text{cosec} , x + cot x) , dx\
&= -frac{pi}{2} ln (1 + sqrt{2}) + 2 int_0^{pi/4} ln (1 + cos x) , dx - 2 int_0^{pi/4} ln (sin x) , dx.
end{align}
The second of these integrals is perhaps (?) reasonably well known (for an evaluation, see here). The result is:
$$int_0^{pi/4} ln (sin x) , dx = -frac{1}{2} mathbf{G} - frac{pi}{4} ln 2,$$
where $mathbf{G}$ is Catalan's constant. Thus
$$I = -frac{pi}{2} ln (1 + sqrt{2}) + mathbf{G} + frac{pi}{2} ln 2 + 2I_1.$$
For the first of the integrals we will make use of the following formula, a proof of which can be found here
$$ln (1 + cos x) = 2 sum_{n = 1}^infty (-1)^{n + 1} frac{cos (nx)}{n} - ln 2.$$
So
begin{align}
I_1 &= int_0^{pi/4} ln (1 + cos x) , dx\
&= 2 sum_{n = 1}^infty frac{(-1)^{n + 1}}{n} int_0^{pi/4} cos (nx) , dx - ln 2 int_0^{pi/4} dx\
&= -frac{pi}{4} ln 2 + 2 sum_{n = 1}^infty frac{(-1)^{n + 1}}{n^2} sin left (frac{n pi}{4} right )\
&= -frac{pi}{4} ln 2 + 2 sum_{stackrel{{Large{n = 1}}}{n in text{odd}}}^infty frac{(-1)^{n + 1}}{n^2} sin left (frac{n pi}{4} right ) + + 2 sum_{stackrel{{Large{n = 1}}}{n in text{even}}}^infty frac{(-1)^{n + 1}}{n^2} sin left (frac{n pi}{4} right )\
&= -frac{pi}{4} ln 2 + 2 sum_{n = 1}^infty frac{1}{(2n - 1)^2} sin left [frac{pi}{4} (2n - 1) right ] - frac{1}{2} sum_{n = 1}^infty frac{1}{n^2} sin left (frac{n pi}{2} right )\
&= -frac{pi}{4} ln 2 + 2 S_1 - frac{1}{2} S_2.
end{align}
For the second of these sums,
begin{align}
S_2 &= sum_{stackrel{{Large{n = 1}}}{n in text{odd}}}^infty frac{1}{n^2} sin left (frac{pi n}{2} right ) + sum_{stackrel{{Large{n = 1}}}{n in text{even}}} frac{1}{n^2} sin left (frac{pi n}{2} right )\
&= sum_{n = 0}^infty frac{(-1)^n}{(2n + 1)^2}\
&= mathbf{G},
end{align}
after a shift of the index $n mapsto 2n + 1$ in the odd sum has been made while the even sum is identically equal to zero.
For the first of the sums, as it converges absolutely we can split it up as folows:
$$S_1 = sum_{stackrel{{Large{n = 1}}}{n in 1,5,9,ldots}}^infty frac{1}{(2n - 1)^2} sin left [frac{pi}{4} (2k - 1) right ] + sum_{stackrel{{Large{n = 1}}}{n in 2,6,10,ldots}}^infty frac{1}{(2n - 1)^2} sin left [frac{pi}{4} (2k - 1) right ] + sum_{stackrel{{Large{n = 1}}}{n in 3,7,11,ldots}}^infty frac{1}{(2n - 1)^2} sin left [frac{pi}{4} (2k - 1) right ] + sum_{stackrel{{Large{n = 1}}}{n in 4,8,12,ldots}}^infty frac{1}{(2n - 1)^2} sin left [frac{pi}{4} (2k - 1) right ].$$
Shifting the indices as follows: $n mapsto 4n - 3, n mapsto 4n - 2, n mapsto 4n - 1, n mapsto 4n$ leads to
begin{align}
S_1 &= frac{1}{sqrt{2}} left [sum_{n = 1}^infty frac{1}{(8n - 7)^2} + sum_{n = 1}^infty frac{1}{(8n - 5)^2} - sum_{n = 1}^infty frac{1}{(8n - 3)^2} - sum_{n = 1}^infty frac{1}{(8n - 1)^2} right ]\
&= frac{1}{sqrt{2}} left [sum_{n = 0}^infty frac{1}{(8n + 1)^2} + sum_{n = 0}^infty frac{1}{(8n + 3)^2} - sum_{n = 0}^infty frac{1}{(8n + 5)^2} - sum_{n = 0}^infty frac{1}{(8n + 7)^2} right ]\
&= frac{1}{64 sqrt{2}} left [sum_{n = 0}^infty frac{1}{(n + 1/8)^2} + sum_{n = 0}^infty frac{1}{(n + 3/8)^2} - sum_{n = 0}^infty frac{1}{(n + 5/8)^2} - sum_{n = 0}^infty frac{1}{(n + 7/8)^2} right ]\
&= frac{1}{64 sqrt{2}} left [psi^{(1)} left (frac{1}{8} right ) + psi^{(1)} left (frac{3}{8} right ) - psi^{(1)} left (frac{5}{8} right ) - psi^{(1)} left (frac{7}{8} right ) right ],
end{align}
where we have made use of the series representation for the polygamma function of order one (as known as the trigamma function). So the value for $I_1$ is:
$$I_1 = -frac{pi}{4} ln 2 - frac{1}{2} mathbf{G} + frac{1}{32 sqrt{2}} left [psi^{(1)} left (frac{1}{8} right ) + psi^{(1)} left (frac{3}{8} right ) - psi^{(1)} left (frac{5}{8} right ) - psi^{(1)} left (frac{7}{8} right ) right ],$$
leading to a final result of
$$int_{-pi/4}^{pi/4} x , text{cosec} , x , dx = -frac{pi}{2} ln (1 + sqrt{2}) + frac{1}{16 sqrt{2}} left [psi^{(1)} left (frac{1}{8} right ) + psi^{(1)} left (frac{3}{8} right ) - psi^{(1)} left (frac{5}{8} right ) - psi^{(1)} left (frac{7}{8} right ) right ].$$
Converting the trigamma functions to Clausen functions of order two
Inspired by the answer given by Zacky in terms of the Clausen function of order two, here I will show how to convert my answer in terms of the 4 trigamma functions into 2 Clausen functions of order 2.
The relation between the Clausen function of order two and the trigamma function is given by (a proof of this can be found here)
$$text{Cl}_2 left (frac{q pi}{p} right ) = frac{1}{(2p)^{2m} (2m - 1)!} sum_{n = 1}^p sin left (frac{qnpi}{p} right ) left [psi^{(1)} left (frac{n}{2p} right ) + (-1)^q psi^{(1)} left (frac{n + p}{2p} right ) right ].$$
Setting $m = 1, q = 1, p = 4$ gives
begin{align}
text{Cl}_2 left (frac{pi}{4} right ) &= frac{1}{64} left [frac{1}{sqrt{2}} left {psi^{(1)} left (frac{1}{8} right ) - psi^{(1)} left (frac{5}{8} right ) right } + psi^{(1)} left (frac{1}{4} right ) - psi^{(1)} left (frac{3}{4} right ) right.\
& qquad left. + frac{1}{sqrt{2}} left {psi^{(1)} left (frac{3}{8} right ) - psi^{(1)} left (frac{7}{8} right ) right } right ], qquad (*)
end{align}
and setting $m = 1, q = 3, p = 4$ gives
begin{align}
text{Cl}_2 left (frac{3pi}{4} right ) &= frac{1}{64} left [frac{1}{sqrt{2}} left {psi^{(1)} left (frac{1}{8} right ) - psi^{(1)} left (frac{5}{8} right ) right } - psi^{(1)} left (frac{1}{4} right ) + psi^{(1)} left (frac{3}{4} right ) right.\
& qquad left. + frac{1}{sqrt{2}} left {psi^{(1)} left (frac{3}{8} right ) - psi^{(1)} left (frac{7}{8} right ) right } right ]. qquad (**)
end{align}
On adding ($*$) to ($**$) we see that
$$psi^{(1)} left (frac{1}{8} right ) + psi^{(1)} left (frac{3}{8} right ) - psi^{(1)} left (frac{5}{8} right ) - psi^{(1)} left (frac{7}{8} right ) = 32 sqrt{2} left [text{Cl}_2 left (frac{pi}{4} right ) + text{Cl}_2 left (frac{3pi}{4} right ) right ],$$
giving
$$int_{-pi/4}^{pi/4} frac{x}{sin x} , dx = -frac{pi}{2} ln (1 + sqrt{2}) + 2 , text{Cl}_2 left (frac{pi}{4} right ) + 2 , text{Cl}_2 left (frac{3pi}{4} right ).$$
answered Dec 21 '18 at 10:45
omegadotomegadot
6,2692829
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Nice answer! We might be able to reduce those $4$ values of the trigamma function using: $$psi_1left(xright)+psi_1left(x+frac12right)=4psi_1left(2xright)$$ I mean: $$psi_1left(frac18right)+left(frac58right) = 4psi_1left(frac14right)$$ And same with: $$psi_1left(frac38right)+left(frac78right) = 4psi_1left(frac34right)$$ And the values of $psi_1left(frac14right)$ and $psi_1left(frac34right)$ are known from here: mathworld.wolfram.com/TrigammaFunction.html
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– Zacky
Dec 21 '18 at 12:06
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Perfect beautiful wonderful amazing thank you for a really great answer :)
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– clathratus
Dec 21 '18 at 19:04
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Also: does the closed form I found simplify to the answer you give?
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– clathratus
Dec 21 '18 at 19:56
add a comment |
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Let
$$I = int_{-pi/4}^{pi/4} x , text{cosec} , x , dx = 2 int_0^{pi/4} x , text{cosec} , x , dx.$$
After integrating by parts we have
begin{align}
I &= -frac{pi}{2} ln (1 + sqrt{2}) + 2 int_0^{pi/4} ln (text{cosec} , x + cot x) , dx\
&= -frac{pi}{2} ln (1 + sqrt{2}) + 2 int_0^{pi/4} ln (1 + cos x) , dx - 2 int_0^{pi/4} ln (sin x) , dx.
end{align}
The second of these integrals is perhaps (?) reasonably well known (for an evaluation, see here). The result is:
$$int_0^{pi/4} ln (sin x) , dx = -frac{1}{2} mathbf{G} - frac{pi}{4} ln 2,$$
where $mathbf{G}$ is Catalan's constant. Thus
$$I = -frac{pi}{2} ln (1 + sqrt{2}) + mathbf{G} + frac{pi}{2} ln 2 + 2I_1.$$
For the first of the integrals we will make use of the following formula, a proof of which can be found here
$$ln (1 + cos x) = 2 sum_{n = 1}^infty (-1)^{n + 1} frac{cos (nx)}{n} - ln 2.$$
So
begin{align}
I_1 &= int_0^{pi/4} ln (1 + cos x) , dx\
&= 2 sum_{n = 1}^infty frac{(-1)^{n + 1}}{n} int_0^{pi/4} cos (nx) , dx - ln 2 int_0^{pi/4} dx\
&= -frac{pi}{4} ln 2 + 2 sum_{n = 1}^infty frac{(-1)^{n + 1}}{n^2} sin left (frac{n pi}{4} right )\
&= -frac{pi}{4} ln 2 + 2 sum_{stackrel{{Large{n = 1}}}{n in text{odd}}}^infty frac{(-1)^{n + 1}}{n^2} sin left (frac{n pi}{4} right ) + + 2 sum_{stackrel{{Large{n = 1}}}{n in text{even}}}^infty frac{(-1)^{n + 1}}{n^2} sin left (frac{n pi}{4} right )\
&= -frac{pi}{4} ln 2 + 2 sum_{n = 1}^infty frac{1}{(2n - 1)^2} sin left [frac{pi}{4} (2n - 1) right ] - frac{1}{2} sum_{n = 1}^infty frac{1}{n^2} sin left (frac{n pi}{2} right )\
&= -frac{pi}{4} ln 2 + 2 S_1 - frac{1}{2} S_2.
end{align}
For the second of these sums,
begin{align}
S_2 &= sum_{stackrel{{Large{n = 1}}}{n in text{odd}}}^infty frac{1}{n^2} sin left (frac{pi n}{2} right ) + sum_{stackrel{{Large{n = 1}}}{n in text{even}}} frac{1}{n^2} sin left (frac{pi n}{2} right )\
&= sum_{n = 0}^infty frac{(-1)^n}{(2n + 1)^2}\
&= mathbf{G},
end{align}
after a shift of the index $n mapsto 2n + 1$ in the odd sum has been made while the even sum is identically equal to zero.
For the first of the sums, as it converges absolutely we can split it up as folows:
$$S_1 = sum_{stackrel{{Large{n = 1}}}{n in 1,5,9,ldots}}^infty frac{1}{(2n - 1)^2} sin left [frac{pi}{4} (2k - 1) right ] + sum_{stackrel{{Large{n = 1}}}{n in 2,6,10,ldots}}^infty frac{1}{(2n - 1)^2} sin left [frac{pi}{4} (2k - 1) right ] + sum_{stackrel{{Large{n = 1}}}{n in 3,7,11,ldots}}^infty frac{1}{(2n - 1)^2} sin left [frac{pi}{4} (2k - 1) right ] + sum_{stackrel{{Large{n = 1}}}{n in 4,8,12,ldots}}^infty frac{1}{(2n - 1)^2} sin left [frac{pi}{4} (2k - 1) right ].$$
Shifting the indices as follows: $n mapsto 4n - 3, n mapsto 4n - 2, n mapsto 4n - 1, n mapsto 4n$ leads to
begin{align}
S_1 &= frac{1}{sqrt{2}} left [sum_{n = 1}^infty frac{1}{(8n - 7)^2} + sum_{n = 1}^infty frac{1}{(8n - 5)^2} - sum_{n = 1}^infty frac{1}{(8n - 3)^2} - sum_{n = 1}^infty frac{1}{(8n - 1)^2} right ]\
&= frac{1}{sqrt{2}} left [sum_{n = 0}^infty frac{1}{(8n + 1)^2} + sum_{n = 0}^infty frac{1}{(8n + 3)^2} - sum_{n = 0}^infty frac{1}{(8n + 5)^2} - sum_{n = 0}^infty frac{1}{(8n + 7)^2} right ]\
&= frac{1}{64 sqrt{2}} left [sum_{n = 0}^infty frac{1}{(n + 1/8)^2} + sum_{n = 0}^infty frac{1}{(n + 3/8)^2} - sum_{n = 0}^infty frac{1}{(n + 5/8)^2} - sum_{n = 0}^infty frac{1}{(n + 7/8)^2} right ]\
&= frac{1}{64 sqrt{2}} left [psi^{(1)} left (frac{1}{8} right ) + psi^{(1)} left (frac{3}{8} right ) - psi^{(1)} left (frac{5}{8} right ) - psi^{(1)} left (frac{7}{8} right ) right ],
end{align}
where we have made use of the series representation for the polygamma function of order one (as known as the trigamma function). So the value for $I_1$ is:
$$I_1 = -frac{pi}{4} ln 2 - frac{1}{2} mathbf{G} + frac{1}{32 sqrt{2}} left [psi^{(1)} left (frac{1}{8} right ) + psi^{(1)} left (frac{3}{8} right ) - psi^{(1)} left (frac{5}{8} right ) - psi^{(1)} left (frac{7}{8} right ) right ],$$
leading to a final result of
$$int_{-pi/4}^{pi/4} x , text{cosec} , x , dx = -frac{pi}{2} ln (1 + sqrt{2}) + frac{1}{16 sqrt{2}} left [psi^{(1)} left (frac{1}{8} right ) + psi^{(1)} left (frac{3}{8} right ) - psi^{(1)} left (frac{5}{8} right ) - psi^{(1)} left (frac{7}{8} right ) right ].$$
Converting the trigamma functions to Clausen functions of order two
Inspired by the answer given by Zacky in terms of the Clausen function of order two, here I will show how to convert my answer in terms of the 4 trigamma functions into 2 Clausen functions of order 2.
The relation between the Clausen function of order two and the trigamma function is given by (a proof of this can be found here)
$$text{Cl}_2 left (frac{q pi}{p} right ) = frac{1}{(2p)^{2m} (2m - 1)!} sum_{n = 1}^p sin left (frac{qnpi}{p} right ) left [psi^{(1)} left (frac{n}{2p} right ) + (-1)^q psi^{(1)} left (frac{n + p}{2p} right ) right ].$$
Setting $m = 1, q = 1, p = 4$ gives
begin{align}
text{Cl}_2 left (frac{pi}{4} right ) &= frac{1}{64} left [frac{1}{sqrt{2}} left {psi^{(1)} left (frac{1}{8} right ) - psi^{(1)} left (frac{5}{8} right ) right } + psi^{(1)} left (frac{1}{4} right ) - psi^{(1)} left (frac{3}{4} right ) right.\
& qquad left. + frac{1}{sqrt{2}} left {psi^{(1)} left (frac{3}{8} right ) - psi^{(1)} left (frac{7}{8} right ) right } right ], qquad (*)
end{align}
and setting $m = 1, q = 3, p = 4$ gives
begin{align}
text{Cl}_2 left (frac{3pi}{4} right ) &= frac{1}{64} left [frac{1}{sqrt{2}} left {psi^{(1)} left (frac{1}{8} right ) - psi^{(1)} left (frac{5}{8} right ) right } - psi^{(1)} left (frac{1}{4} right ) + psi^{(1)} left (frac{3}{4} right ) right.\
& qquad left. + frac{1}{sqrt{2}} left {psi^{(1)} left (frac{3}{8} right ) - psi^{(1)} left (frac{7}{8} right ) right } right ]. qquad (**)
end{align}
On adding ($*$) to ($**$) we see that
$$psi^{(1)} left (frac{1}{8} right ) + psi^{(1)} left (frac{3}{8} right ) - psi^{(1)} left (frac{5}{8} right ) - psi^{(1)} left (frac{7}{8} right ) = 32 sqrt{2} left [text{Cl}_2 left (frac{pi}{4} right ) + text{Cl}_2 left (frac{3pi}{4} right ) right ],$$
giving
$$int_{-pi/4}^{pi/4} frac{x}{sin x} , dx = -frac{pi}{2} ln (1 + sqrt{2}) + 2 , text{Cl}_2 left (frac{pi}{4} right ) + 2 , text{Cl}_2 left (frac{3pi}{4} right ).$$
answered Dec 21 '18 at 10:45
omegadotomegadot
6,2692829
$endgroup$
$begingroup$
Nice answer! We might be able to reduce those $4$ values of the trigamma function using: $$psi_1left(xright)+psi_1left(x+frac12right)=4psi_1left(2xright)$$ I mean: $$psi_1left(frac18right)+left(frac58right) = 4psi_1left(frac14right)$$ And same with: $$psi_1left(frac38right)+left(frac78right) = 4psi_1left(frac34right)$$ And the values of $psi_1left(frac14right)$ and $psi_1left(frac34right)$ are known from here: mathworld.wolfram.com/TrigammaFunction.html
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– Zacky
Dec 21 '18 at 12:06
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Perfect beautiful wonderful amazing thank you for a really great answer :)
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– clathratus
Dec 21 '18 at 19:04
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Also: does the closed form I found simplify to the answer you give?
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– clathratus
Dec 21 '18 at 19:56
add a comment |
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Let
$$I = int_{-pi/4}^{pi/4} x , text{cosec} , x , dx = 2 int_0^{pi/4} x , text{cosec} , x , dx.$$
After integrating by parts we have
begin{align}
I &= -frac{pi}{2} ln (1 + sqrt{2}) + 2 int_0^{pi/4} ln (text{cosec} , x + cot x) , dx\
&= -frac{pi}{2} ln (1 + sqrt{2}) + 2 int_0^{pi/4} ln (1 + cos x) , dx - 2 int_0^{pi/4} ln (sin x) , dx.
end{align}
The second of these integrals is perhaps (?) reasonably well known (for an evaluation, see here). The result is:
$$int_0^{pi/4} ln (sin x) , dx = -frac{1}{2} mathbf{G} - frac{pi}{4} ln 2,$$
where $mathbf{G}$ is Catalan's constant. Thus
$$I = -frac{pi}{2} ln (1 + sqrt{2}) + mathbf{G} + frac{pi}{2} ln 2 + 2I_1.$$
For the first of the integrals we will make use of the following formula, a proof of which can be found here
$$ln (1 + cos x) = 2 sum_{n = 1}^infty (-1)^{n + 1} frac{cos (nx)}{n} - ln 2.$$
So
begin{align}
I_1 &= int_0^{pi/4} ln (1 + cos x) , dx\
&= 2 sum_{n = 1}^infty frac{(-1)^{n + 1}}{n} int_0^{pi/4} cos (nx) , dx - ln 2 int_0^{pi/4} dx\
&= -frac{pi}{4} ln 2 + 2 sum_{n = 1}^infty frac{(-1)^{n + 1}}{n^2} sin left (frac{n pi}{4} right )\
&= -frac{pi}{4} ln 2 + 2 sum_{stackrel{{Large{n = 1}}}{n in text{odd}}}^infty frac{(-1)^{n + 1}}{n^2} sin left (frac{n pi}{4} right ) + + 2 sum_{stackrel{{Large{n = 1}}}{n in text{even}}}^infty frac{(-1)^{n + 1}}{n^2} sin left (frac{n pi}{4} right )\
&= -frac{pi}{4} ln 2 + 2 sum_{n = 1}^infty frac{1}{(2n - 1)^2} sin left [frac{pi}{4} (2n - 1) right ] - frac{1}{2} sum_{n = 1}^infty frac{1}{n^2} sin left (frac{n pi}{2} right )\
&= -frac{pi}{4} ln 2 + 2 S_1 - frac{1}{2} S_2.
end{align}
For the second of these sums,
begin{align}
S_2 &= sum_{stackrel{{Large{n = 1}}}{n in text{odd}}}^infty frac{1}{n^2} sin left (frac{pi n}{2} right ) + sum_{stackrel{{Large{n = 1}}}{n in text{even}}} frac{1}{n^2} sin left (frac{pi n}{2} right )\
&= sum_{n = 0}^infty frac{(-1)^n}{(2n + 1)^2}\
&= mathbf{G},
end{align}
after a shift of the index $n mapsto 2n + 1$ in the odd sum has been made while the even sum is identically equal to zero.
For the first of the sums, as it converges absolutely we can split it up as folows:
$$S_1 = sum_{stackrel{{Large{n = 1}}}{n in 1,5,9,ldots}}^infty frac{1}{(2n - 1)^2} sin left [frac{pi}{4} (2k - 1) right ] + sum_{stackrel{{Large{n = 1}}}{n in 2,6,10,ldots}}^infty frac{1}{(2n - 1)^2} sin left [frac{pi}{4} (2k - 1) right ] + sum_{stackrel{{Large{n = 1}}}{n in 3,7,11,ldots}}^infty frac{1}{(2n - 1)^2} sin left [frac{pi}{4} (2k - 1) right ] + sum_{stackrel{{Large{n = 1}}}{n in 4,8,12,ldots}}^infty frac{1}{(2n - 1)^2} sin left [frac{pi}{4} (2k - 1) right ].$$
Shifting the indices as follows: $n mapsto 4n - 3, n mapsto 4n - 2, n mapsto 4n - 1, n mapsto 4n$ leads to
begin{align}
S_1 &= frac{1}{sqrt{2}} left [sum_{n = 1}^infty frac{1}{(8n - 7)^2} + sum_{n = 1}^infty frac{1}{(8n - 5)^2} - sum_{n = 1}^infty frac{1}{(8n - 3)^2} - sum_{n = 1}^infty frac{1}{(8n - 1)^2} right ]\
&= frac{1}{sqrt{2}} left [sum_{n = 0}^infty frac{1}{(8n + 1)^2} + sum_{n = 0}^infty frac{1}{(8n + 3)^2} - sum_{n = 0}^infty frac{1}{(8n + 5)^2} - sum_{n = 0}^infty frac{1}{(8n + 7)^2} right ]\
&= frac{1}{64 sqrt{2}} left [sum_{n = 0}^infty frac{1}{(n + 1/8)^2} + sum_{n = 0}^infty frac{1}{(n + 3/8)^2} - sum_{n = 0}^infty frac{1}{(n + 5/8)^2} - sum_{n = 0}^infty frac{1}{(n + 7/8)^2} right ]\
&= frac{1}{64 sqrt{2}} left [psi^{(1)} left (frac{1}{8} right ) + psi^{(1)} left (frac{3}{8} right ) - psi^{(1)} left (frac{5}{8} right ) - psi^{(1)} left (frac{7}{8} right ) right ],
end{align}
where we have made use of the series representation for the polygamma function of order one (as known as the trigamma function). So the value for $I_1$ is:
$$I_1 = -frac{pi}{4} ln 2 - frac{1}{2} mathbf{G} + frac{1}{32 sqrt{2}} left [psi^{(1)} left (frac{1}{8} right ) + psi^{(1)} left (frac{3}{8} right ) - psi^{(1)} left (frac{5}{8} right ) - psi^{(1)} left (frac{7}{8} right ) right ],$$
leading to a final result of
$$int_{-pi/4}^{pi/4} x , text{cosec} , x , dx = -frac{pi}{2} ln (1 + sqrt{2}) + frac{1}{16 sqrt{2}} left [psi^{(1)} left (frac{1}{8} right ) + psi^{(1)} left (frac{3}{8} right ) - psi^{(1)} left (frac{5}{8} right ) - psi^{(1)} left (frac{7}{8} right ) right ].$$
Converting the trigamma functions to Clausen functions of order two
Inspired by the answer given by Zacky in terms of the Clausen function of order two, here I will show how to convert my answer in terms of the 4 trigamma functions into 2 Clausen functions of order 2.
The relation between the Clausen function of order two and the trigamma function is given by (a proof of this can be found here)
$$text{Cl}_2 left (frac{q pi}{p} right ) = frac{1}{(2p)^{2m} (2m - 1)!} sum_{n = 1}^p sin left (frac{qnpi}{p} right ) left [psi^{(1)} left (frac{n}{2p} right ) + (-1)^q psi^{(1)} left (frac{n + p}{2p} right ) right ].$$
Setting $m = 1, q = 1, p = 4$ gives
begin{align}
text{Cl}_2 left (frac{pi}{4} right ) &= frac{1}{64} left [frac{1}{sqrt{2}} left {psi^{(1)} left (frac{1}{8} right ) - psi^{(1)} left (frac{5}{8} right ) right } + psi^{(1)} left (frac{1}{4} right ) - psi^{(1)} left (frac{3}{4} right ) right.\
& qquad left. + frac{1}{sqrt{2}} left {psi^{(1)} left (frac{3}{8} right ) - psi^{(1)} left (frac{7}{8} right ) right } right ], qquad (*)
end{align}
and setting $m = 1, q = 3, p = 4$ gives
begin{align}
text{Cl}_2 left (frac{3pi}{4} right ) &= frac{1}{64} left [frac{1}{sqrt{2}} left {psi^{(1)} left (frac{1}{8} right ) - psi^{(1)} left (frac{5}{8} right ) right } - psi^{(1)} left (frac{1}{4} right ) + psi^{(1)} left (frac{3}{4} right ) right.\
& qquad left. + frac{1}{sqrt{2}} left {psi^{(1)} left (frac{3}{8} right ) - psi^{(1)} left (frac{7}{8} right ) right } right ]. qquad (**)
end{align}
On adding ($*$) to ($**$) we see that
$$psi^{(1)} left (frac{1}{8} right ) + psi^{(1)} left (frac{3}{8} right ) - psi^{(1)} left (frac{5}{8} right ) - psi^{(1)} left (frac{7}{8} right ) = 32 sqrt{2} left [text{Cl}_2 left (frac{pi}{4} right ) + text{Cl}_2 left (frac{3pi}{4} right ) right ],$$
giving
$$int_{-pi/4}^{pi/4} frac{x}{sin x} , dx = -frac{pi}{2} ln (1 + sqrt{2}) + 2 , text{Cl}_2 left (frac{pi}{4} right ) + 2 , text{Cl}_2 left (frac{3pi}{4} right ).$$
answered Dec 21 '18 at 10:45
omegadotomegadot
6,2692829
$endgroup$
Let
$$I = int_{-pi/4}^{pi/4} x , text{cosec} , x , dx = 2 int_0^{pi/4} x , text{cosec} , x , dx.$$
After integrating by parts we have
begin{align}
I &= -frac{pi}{2} ln (1 + sqrt{2}) + 2 int_0^{pi/4} ln (text{cosec} , x + cot x) , dx\
&= -frac{pi}{2} ln (1 + sqrt{2}) + 2 int_0^{pi/4} ln (1 + cos x) , dx - 2 int_0^{pi/4} ln (sin x) , dx.
end{align}
The second of these integrals is perhaps (?) reasonably well known (for an evaluation, see here). The result is:
$$int_0^{pi/4} ln (sin x) , dx = -frac{1}{2} mathbf{G} - frac{pi}{4} ln 2,$$
where $mathbf{G}$ is Catalan's constant. Thus
$$I = -frac{pi}{2} ln (1 + sqrt{2}) + mathbf{G} + frac{pi}{2} ln 2 + 2I_1.$$
For the first of the integrals we will make use of the following formula, a proof of which can be found here
$$ln (1 + cos x) = 2 sum_{n = 1}^infty (-1)^{n + 1} frac{cos (nx)}{n} - ln 2.$$
So
begin{align}
I_1 &= int_0^{pi/4} ln (1 + cos x) , dx\
&= 2 sum_{n = 1}^infty frac{(-1)^{n + 1}}{n} int_0^{pi/4} cos (nx) , dx - ln 2 int_0^{pi/4} dx\
&= -frac{pi}{4} ln 2 + 2 sum_{n = 1}^infty frac{(-1)^{n + 1}}{n^2} sin left (frac{n pi}{4} right )\
&= -frac{pi}{4} ln 2 + 2 sum_{stackrel{{Large{n = 1}}}{n in text{odd}}}^infty frac{(-1)^{n + 1}}{n^2} sin left (frac{n pi}{4} right ) + + 2 sum_{stackrel{{Large{n = 1}}}{n in text{even}}}^infty frac{(-1)^{n + 1}}{n^2} sin left (frac{n pi}{4} right )\
&= -frac{pi}{4} ln 2 + 2 sum_{n = 1}^infty frac{1}{(2n - 1)^2} sin left [frac{pi}{4} (2n - 1) right ] - frac{1}{2} sum_{n = 1}^infty frac{1}{n^2} sin left (frac{n pi}{2} right )\
&= -frac{pi}{4} ln 2 + 2 S_1 - frac{1}{2} S_2.
end{align}
For the second of these sums,
begin{align}
S_2 &= sum_{stackrel{{Large{n = 1}}}{n in text{odd}}}^infty frac{1}{n^2} sin left (frac{pi n}{2} right ) + sum_{stackrel{{Large{n = 1}}}{n in text{even}}} frac{1}{n^2} sin left (frac{pi n}{2} right )\
&= sum_{n = 0}^infty frac{(-1)^n}{(2n + 1)^2}\
&= mathbf{G},
end{align}
after a shift of the index $n mapsto 2n + 1$ in the odd sum has been made while the even sum is identically equal to zero.
For the first of the sums, as it converges absolutely we can split it up as folows:
$$S_1 = sum_{stackrel{{Large{n = 1}}}{n in 1,5,9,ldots}}^infty frac{1}{(2n - 1)^2} sin left [frac{pi}{4} (2k - 1) right ] + sum_{stackrel{{Large{n = 1}}}{n in 2,6,10,ldots}}^infty frac{1}{(2n - 1)^2} sin left [frac{pi}{4} (2k - 1) right ] + sum_{stackrel{{Large{n = 1}}}{n in 3,7,11,ldots}}^infty frac{1}{(2n - 1)^2} sin left [frac{pi}{4} (2k - 1) right ] + sum_{stackrel{{Large{n = 1}}}{n in 4,8,12,ldots}}^infty frac{1}{(2n - 1)^2} sin left [frac{pi}{4} (2k - 1) right ].$$
Shifting the indices as follows: $n mapsto 4n - 3, n mapsto 4n - 2, n mapsto 4n - 1, n mapsto 4n$ leads to
begin{align}
S_1 &= frac{1}{sqrt{2}} left [sum_{n = 1}^infty frac{1}{(8n - 7)^2} + sum_{n = 1}^infty frac{1}{(8n - 5)^2} - sum_{n = 1}^infty frac{1}{(8n - 3)^2} - sum_{n = 1}^infty frac{1}{(8n - 1)^2} right ]\
&= frac{1}{sqrt{2}} left [sum_{n = 0}^infty frac{1}{(8n + 1)^2} + sum_{n = 0}^infty frac{1}{(8n + 3)^2} - sum_{n = 0}^infty frac{1}{(8n + 5)^2} - sum_{n = 0}^infty frac{1}{(8n + 7)^2} right ]\
&= frac{1}{64 sqrt{2}} left [sum_{n = 0}^infty frac{1}{(n + 1/8)^2} + sum_{n = 0}^infty frac{1}{(n + 3/8)^2} - sum_{n = 0}^infty frac{1}{(n + 5/8)^2} - sum_{n = 0}^infty frac{1}{(n + 7/8)^2} right ]\
&= frac{1}{64 sqrt{2}} left [psi^{(1)} left (frac{1}{8} right ) + psi^{(1)} left (frac{3}{8} right ) - psi^{(1)} left (frac{5}{8} right ) - psi^{(1)} left (frac{7}{8} right ) right ],
end{align}
where we have made use of the series representation for the polygamma function of order one (as known as the trigamma function). So the value for $I_1$ is:
$$I_1 = -frac{pi}{4} ln 2 - frac{1}{2} mathbf{G} + frac{1}{32 sqrt{2}} left [psi^{(1)} left (frac{1}{8} right ) + psi^{(1)} left (frac{3}{8} right ) - psi^{(1)} left (frac{5}{8} right ) - psi^{(1)} left (frac{7}{8} right ) right ],$$
leading to a final result of
$$int_{-pi/4}^{pi/4} x , text{cosec} , x , dx = -frac{pi}{2} ln (1 + sqrt{2}) + frac{1}{16 sqrt{2}} left [psi^{(1)} left (frac{1}{8} right ) + psi^{(1)} left (frac{3}{8} right ) - psi^{(1)} left (frac{5}{8} right ) - psi^{(1)} left (frac{7}{8} right ) right ].$$
Converting the trigamma functions to Clausen functions of order two
Inspired by the answer given by Zacky in terms of the Clausen function of order two, here I will show how to convert my answer in terms of the 4 trigamma functions into 2 Clausen functions of order 2.
The relation between the Clausen function of order two and the trigamma function is given by (a proof of this can be found here)
$$text{Cl}_2 left (frac{q pi}{p} right ) = frac{1}{(2p)^{2m} (2m - 1)!} sum_{n = 1}^p sin left (frac{qnpi}{p} right ) left [psi^{(1)} left (frac{n}{2p} right ) + (-1)^q psi^{(1)} left (frac{n + p}{2p} right ) right ].$$
Setting $m = 1, q = 1, p = 4$ gives
begin{align}
text{Cl}_2 left (frac{pi}{4} right ) &= frac{1}{64} left [frac{1}{sqrt{2}} left {psi^{(1)} left (frac{1}{8} right ) - psi^{(1)} left (frac{5}{8} right ) right } + psi^{(1)} left (frac{1}{4} right ) - psi^{(1)} left (frac{3}{4} right ) right.\
& qquad left. + frac{1}{sqrt{2}} left {psi^{(1)} left (frac{3}{8} right ) - psi^{(1)} left (frac{7}{8} right ) right } right ], qquad (*)
end{align}
and setting $m = 1, q = 3, p = 4$ gives
begin{align}
text{Cl}_2 left (frac{3pi}{4} right ) &= frac{1}{64} left [frac{1}{sqrt{2}} left {psi^{(1)} left (frac{1}{8} right ) - psi^{(1)} left (frac{5}{8} right ) right } - psi^{(1)} left (frac{1}{4} right ) + psi^{(1)} left (frac{3}{4} right ) right.\
& qquad left. + frac{1}{sqrt{2}} left {psi^{(1)} left (frac{3}{8} right ) - psi^{(1)} left (frac{7}{8} right ) right } right ]. qquad (**)
end{align}
On adding ($*$) to ($**$) we see that
$$psi^{(1)} left (frac{1}{8} right ) + psi^{(1)} left (frac{3}{8} right ) - psi^{(1)} left (frac{5}{8} right ) - psi^{(1)} left (frac{7}{8} right ) = 32 sqrt{2} left [text{Cl}_2 left (frac{pi}{4} right ) + text{Cl}_2 left (frac{3pi}{4} right ) right ],$$
giving
$$int_{-pi/4}^{pi/4} frac{x}{sin x} , dx = -frac{pi}{2} ln (1 + sqrt{2}) + 2 , text{Cl}_2 left (frac{pi}{4} right ) + 2 , text{Cl}_2 left (frac{3pi}{4} right ).$$
answered Dec 21 '18 at 10:45
omegadotomegadot
6,2692829
edited Dec 22 '18 at 10:13
answered Dec 21 '18 at 10:45
omegadotomegadot
6,2692829
answered Dec 21 '18 at 10:45
omegadotomegadot
6,2692829
answered Dec 21 '18 at 10:45
omegadotomegadot
6,2692829
6,2692829
$begingroup$
Nice answer! We might be able to reduce those $4$ values of the trigamma function using: $$psi_1left(xright)+psi_1left(x+frac12right)=4psi_1left(2xright)$$ I mean: $$psi_1left(frac18right)+left(frac58right) = 4psi_1left(frac14right)$$ And same with: $$psi_1left(frac38right)+left(frac78right) = 4psi_1left(frac34right)$$ And the values of $psi_1left(frac14right)$ and $psi_1left(frac34right)$ are known from here: mathworld.wolfram.com/TrigammaFunction.html
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– Zacky
Dec 21 '18 at 12:06
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Perfect beautiful wonderful amazing thank you for a really great answer :)
$endgroup$
– clathratus
Dec 21 '18 at 19:04
$begingroup$
Also: does the closed form I found simplify to the answer you give?
$endgroup$
– clathratus
Dec 21 '18 at 19:56
add a comment |
$begingroup$
Nice answer! We might be able to reduce those $4$ values of the trigamma function using: $$psi_1left(xright)+psi_1left(x+frac12right)=4psi_1left(2xright)$$ I mean: $$psi_1left(frac18right)+left(frac58right) = 4psi_1left(frac14right)$$ And same with: $$psi_1left(frac38right)+left(frac78right) = 4psi_1left(frac34right)$$ And the values of $psi_1left(frac14right)$ and $psi_1left(frac34right)$ are known from here: mathworld.wolfram.com/TrigammaFunction.html
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– Zacky
Dec 21 '18 at 12:06
$begingroup$
Perfect beautiful wonderful amazing thank you for a really great answer :)
$endgroup$
– clathratus
Dec 21 '18 at 19:04
$begingroup$
Also: does the closed form I found simplify to the answer you give?
$endgroup$
– clathratus
Dec 21 '18 at 19:56
$begingroup$
Nice answer! We might be able to reduce those $4$ values of the trigamma function using: $$psi_1left(xright)+psi_1left(x+frac12right)=4psi_1left(2xright)$$ I mean: $$psi_1left(frac18right)+left(frac58right) = 4psi_1left(frac14right)$$ And same with: $$psi_1left(frac38right)+left(frac78right) = 4psi_1left(frac34right)$$ And the values of $psi_1left(frac14right)$ and $psi_1left(frac34right)$ are known from here: mathworld.wolfram.com/TrigammaFunction.html
$endgroup$
– Zacky
Dec 21 '18 at 12:06
$begingroup$
Nice answer! We might be able to reduce those $4$ values of the trigamma function using: $$psi_1left(xright)+psi_1left(x+frac12right)=4psi_1left(2xright)$$ I mean: $$psi_1left(frac18right)+left(frac58right) = 4psi_1left(frac14right)$$ And same with: $$psi_1left(frac38right)+left(frac78right) = 4psi_1left(frac34right)$$ And the values of $psi_1left(frac14right)$ and $psi_1left(frac34right)$ are known from here: mathworld.wolfram.com/TrigammaFunction.html
$endgroup$
– Zacky
Dec 21 '18 at 12:06
$begingroup$
Perfect beautiful wonderful amazing thank you for a really great answer :)
$endgroup$
– clathratus
Dec 21 '18 at 19:04
$begingroup$
Perfect beautiful wonderful amazing thank you for a really great answer :)
$endgroup$
– clathratus
Dec 21 '18 at 19:04
$begingroup$
Also: does the closed form I found simplify to the answer you give?
$endgroup$
– clathratus
Dec 21 '18 at 19:56
$begingroup$
Also: does the closed form I found simplify to the answer you give?
$endgroup$
– clathratus
Dec 21 '18 at 19:56
add a comment |
$begingroup$
For the purpose of an alternative method, mostly relying on Clausen function. We have:
$$I=int_{-pi/4}^{pi/4}frac{x}{sin x} dx=2int_0^{pi/4}xleft(lnleft(tanfrac{x}{2}right)right)' dx=2xlnleft(tanfrac{x}{2}right)bigg|_0^frac{pi}{4}-2int_0^frac{pi}{4}lnleft(tanfrac{x}{2}right)dx=$$
$$=frac{pi}{2}ln(sqrt 2-1)-2int_0^frac{pi}{4}left(lnleft(2sinfrac{x}{2}right)-lnleft(2cosfrac{x}{2}right)right)dx=$$$$=frac{pi}{2}ln(sqrt 2-1)+2text{Cl}_2left(frac{pi}{4}right)+2text{Cl}_2left(frac{3pi}{4}right)$$
The last two integrals can be found on the first link. Now is up to the reader if using Clausen function gives any satisfaction, because in desguise is still a series, but same goes with the trigamma function.
An interesting question might be: For what values of $phi$ does the following integral have an elementary answer (Catalan's constant included)?
$ displaystyle{I(phi)=int_0^phi frac{x}{sin x}dx}$. So far I only know about $Ileft(frac{pi}{6}right)$.
answered Dec 21 '18 at 11:46
ZackyZacky
7,84511062
$endgroup$
1
$begingroup$
I would be happy with the 2 order two Clausen functions as an answer. Far more compact compared to the answer I gave in terms of 4 trigamma functions. And besides, are the elementary functions such as $e^x$ and $sin x$ not themselves just series in disguise?
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– omegadot
Dec 21 '18 at 12:01
$begingroup$
You are definetly right. In my mind, by elementary I reffer of something that is learnt in school (which maybe is not quite the best description of something elementary). Maybe Clausen and Trigamma function are studied somewhere, but I only heard of them from the internet.
$endgroup$
– Zacky
Dec 21 '18 at 12:12
1
$begingroup$
I love special functions. Thanks Zacky
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– clathratus
Dec 21 '18 at 18:58
1
$begingroup$
$Ileft(frac{pi}{2}right)=2G$ , $G$ is the Catalan constant.
$endgroup$
– FDP
Dec 21 '18 at 20:48
add a comment |
$begingroup$
For the purpose of an alternative method, mostly relying on Clausen function. We have:
$$I=int_{-pi/4}^{pi/4}frac{x}{sin x} dx=2int_0^{pi/4}xleft(lnleft(tanfrac{x}{2}right)right)' dx=2xlnleft(tanfrac{x}{2}right)bigg|_0^frac{pi}{4}-2int_0^frac{pi}{4}lnleft(tanfrac{x}{2}right)dx=$$
$$=frac{pi}{2}ln(sqrt 2-1)-2int_0^frac{pi}{4}left(lnleft(2sinfrac{x}{2}right)-lnleft(2cosfrac{x}{2}right)right)dx=$$$$=frac{pi}{2}ln(sqrt 2-1)+2text{Cl}_2left(frac{pi}{4}right)+2text{Cl}_2left(frac{3pi}{4}right)$$
The last two integrals can be found on the first link. Now is up to the reader if using Clausen function gives any satisfaction, because in desguise is still a series, but same goes with the trigamma function.
An interesting question might be: For what values of $phi$ does the following integral have an elementary answer (Catalan's constant included)?
$ displaystyle{I(phi)=int_0^phi frac{x}{sin x}dx}$. So far I only know about $Ileft(frac{pi}{6}right)$.
answered Dec 21 '18 at 11:46
ZackyZacky
7,84511062
$endgroup$
1
$begingroup$
I would be happy with the 2 order two Clausen functions as an answer. Far more compact compared to the answer I gave in terms of 4 trigamma functions. And besides, are the elementary functions such as $e^x$ and $sin x$ not themselves just series in disguise?
$endgroup$
– omegadot
Dec 21 '18 at 12:01
$begingroup$
You are definetly right. In my mind, by elementary I reffer of something that is learnt in school (which maybe is not quite the best description of something elementary). Maybe Clausen and Trigamma function are studied somewhere, but I only heard of them from the internet.
$endgroup$
– Zacky
Dec 21 '18 at 12:12
1
$begingroup$
I love special functions. Thanks Zacky
$endgroup$
– clathratus
Dec 21 '18 at 18:58
1
$begingroup$
$Ileft(frac{pi}{2}right)=2G$ , $G$ is the Catalan constant.
$endgroup$
– FDP
Dec 21 '18 at 20:48
add a comment |
$begingroup$
For the purpose of an alternative method, mostly relying on Clausen function. We have:
$$I=int_{-pi/4}^{pi/4}frac{x}{sin x} dx=2int_0^{pi/4}xleft(lnleft(tanfrac{x}{2}right)right)' dx=2xlnleft(tanfrac{x}{2}right)bigg|_0^frac{pi}{4}-2int_0^frac{pi}{4}lnleft(tanfrac{x}{2}right)dx=$$
$$=frac{pi}{2}ln(sqrt 2-1)-2int_0^frac{pi}{4}left(lnleft(2sinfrac{x}{2}right)-lnleft(2cosfrac{x}{2}right)right)dx=$$$$=frac{pi}{2}ln(sqrt 2-1)+2text{Cl}_2left(frac{pi}{4}right)+2text{Cl}_2left(frac{3pi}{4}right)$$
The last two integrals can be found on the first link. Now is up to the reader if using Clausen function gives any satisfaction, because in desguise is still a series, but same goes with the trigamma function.
An interesting question might be: For what values of $phi$ does the following integral have an elementary answer (Catalan's constant included)?
$ displaystyle{I(phi)=int_0^phi frac{x}{sin x}dx}$. So far I only know about $Ileft(frac{pi}{6}right)$.
answered Dec 21 '18 at 11:46
ZackyZacky
7,84511062
$endgroup$
For the purpose of an alternative method, mostly relying on Clausen function. We have:
$$I=int_{-pi/4}^{pi/4}frac{x}{sin x} dx=2int_0^{pi/4}xleft(lnleft(tanfrac{x}{2}right)right)' dx=2xlnleft(tanfrac{x}{2}right)bigg|_0^frac{pi}{4}-2int_0^frac{pi}{4}lnleft(tanfrac{x}{2}right)dx=$$
$$=frac{pi}{2}ln(sqrt 2-1)-2int_0^frac{pi}{4}left(lnleft(2sinfrac{x}{2}right)-lnleft(2cosfrac{x}{2}right)right)dx=$$$$=frac{pi}{2}ln(sqrt 2-1)+2text{Cl}_2left(frac{pi}{4}right)+2text{Cl}_2left(frac{3pi}{4}right)$$
The last two integrals can be found on the first link. Now is up to the reader if using Clausen function gives any satisfaction, because in desguise is still a series, but same goes with the trigamma function.
An interesting question might be: For what values of $phi$ does the following integral have an elementary answer (Catalan's constant included)?
$ displaystyle{I(phi)=int_0^phi frac{x}{sin x}dx}$. So far I only know about $Ileft(frac{pi}{6}right)$.
answered Dec 21 '18 at 11:46
ZackyZacky
7,84511062
edited Dec 21 '18 at 11:57
answered Dec 21 '18 at 11:46
ZackyZacky
7,84511062
answered Dec 21 '18 at 11:46
ZackyZacky
7,84511062
answered Dec 21 '18 at 11:46
ZackyZacky
7,84511062
7,84511062
1
$begingroup$
I would be happy with the 2 order two Clausen functions as an answer. Far more compact compared to the answer I gave in terms of 4 trigamma functions. And besides, are the elementary functions such as $e^x$ and $sin x$ not themselves just series in disguise?
$endgroup$
– omegadot
Dec 21 '18 at 12:01
$begingroup$
You are definetly right. In my mind, by elementary I reffer of something that is learnt in school (which maybe is not quite the best description of something elementary). Maybe Clausen and Trigamma function are studied somewhere, but I only heard of them from the internet.
$endgroup$
– Zacky
Dec 21 '18 at 12:12
1
$begingroup$
I love special functions. Thanks Zacky
$endgroup$
– clathratus
Dec 21 '18 at 18:58
1
$begingroup$
$Ileft(frac{pi}{2}right)=2G$ , $G$ is the Catalan constant.
$endgroup$
– FDP
Dec 21 '18 at 20:48
add a comment |
1
$begingroup$
I would be happy with the 2 order two Clausen functions as an answer. Far more compact compared to the answer I gave in terms of 4 trigamma functions. And besides, are the elementary functions such as $e^x$ and $sin x$ not themselves just series in disguise?
$endgroup$
– omegadot
Dec 21 '18 at 12:01
$begingroup$
You are definetly right. In my mind, by elementary I reffer of something that is learnt in school (which maybe is not quite the best description of something elementary). Maybe Clausen and Trigamma function are studied somewhere, but I only heard of them from the internet.
$endgroup$
– Zacky
Dec 21 '18 at 12:12
1
$begingroup$
I love special functions. Thanks Zacky
$endgroup$
– clathratus
Dec 21 '18 at 18:58
1
$begingroup$
$Ileft(frac{pi}{2}right)=2G$ , $G$ is the Catalan constant.
$endgroup$
– FDP
Dec 21 '18 at 20:48
1
1
$begingroup$
I would be happy with the 2 order two Clausen functions as an answer. Far more compact compared to the answer I gave in terms of 4 trigamma functions. And besides, are the elementary functions such as $e^x$ and $sin x$ not themselves just series in disguise?
$endgroup$
– omegadot
Dec 21 '18 at 12:01
$begingroup$
I would be happy with the 2 order two Clausen functions as an answer. Far more compact compared to the answer I gave in terms of 4 trigamma functions. And besides, are the elementary functions such as $e^x$ and $sin x$ not themselves just series in disguise?
$endgroup$
– omegadot
Dec 21 '18 at 12:01
$begingroup$
You are definetly right. In my mind, by elementary I reffer of something that is learnt in school (which maybe is not quite the best description of something elementary). Maybe Clausen and Trigamma function are studied somewhere, but I only heard of them from the internet.
$endgroup$
– Zacky
Dec 21 '18 at 12:12
$begingroup$
You are definetly right. In my mind, by elementary I reffer of something that is learnt in school (which maybe is not quite the best description of something elementary). Maybe Clausen and Trigamma function are studied somewhere, but I only heard of them from the internet.
$endgroup$
– Zacky
Dec 21 '18 at 12:12
1
1
$begingroup$
I love special functions. Thanks Zacky
$endgroup$
– clathratus
Dec 21 '18 at 18:58
$begingroup$
I love special functions. Thanks Zacky
$endgroup$
– clathratus
Dec 21 '18 at 18:58
1
1
$begingroup$
$Ileft(frac{pi}{2}right)=2G$ , $G$ is the Catalan constant.
$endgroup$
– FDP
Dec 21 '18 at 20:48
$begingroup$
$Ileft(frac{pi}{2}right)=2G$ , $G$ is the Catalan constant.
$endgroup$
– FDP
Dec 21 '18 at 20:48
add a comment |
$begingroup$
Too long for comments.
$$I=4sum_{ngeq0}frac{(-1)^n}{(2n+1)^2}(sqrt{2}-1)^{2n+1}=4 left(sqrt{2}-1right) ,
_3F_2left(frac{1}{2},frac{1}{2},1;frac{3}{2},frac{3}{2};2
sqrt{2}-3right)$$
On the other hand, working from Wolfram Alpha expression for the antiderivative and trying to simplify as much as I could the integral, I got
$$I=-frac{pi}{4} , log left(3+2 sqrt{2}right)+frac 1 {16 sqrt 2} left(psi ^{(1)}left(frac{1}{8}right)+psi ^{(1)}left(frac{3}{8}right)-psi
^{(1)}left(frac{5}{8}right)-psi ^{(1)}left(frac{7}{8}right) right)$$
What is interesting to mention is that, doing calculations with another CAS, intermediate steps show $G$ appearing a few times.
answered Dec 21 '18 at 6:49
Claude LeiboviciClaude Leibovici
125k1158135
$endgroup$
$begingroup$
Yes $mathbf{G}$ does indeed appear twice before canceling out as can be seen in the answer I provide.
$endgroup$
– omegadot
Dec 21 '18 at 10:58
$begingroup$
@omegadot. Using another way, it did appear four times in complex expressions and ran away ! By the way, your answer is real nice ! $to +1$. Cheers.
$endgroup$
– Claude Leibovici
Dec 21 '18 at 11:22
$begingroup$
So does the closed form I found simplify to this?
$endgroup$
– clathratus
Dec 21 '18 at 19:02
add a comment |
$begingroup$
Too long for comments.
$$I=4sum_{ngeq0}frac{(-1)^n}{(2n+1)^2}(sqrt{2}-1)^{2n+1}=4 left(sqrt{2}-1right) ,
_3F_2left(frac{1}{2},frac{1}{2},1;frac{3}{2},frac{3}{2};2
sqrt{2}-3right)$$
On the other hand, working from Wolfram Alpha expression for the antiderivative and trying to simplify as much as I could the integral, I got
$$I=-frac{pi}{4} , log left(3+2 sqrt{2}right)+frac 1 {16 sqrt 2} left(psi ^{(1)}left(frac{1}{8}right)+psi ^{(1)}left(frac{3}{8}right)-psi
^{(1)}left(frac{5}{8}right)-psi ^{(1)}left(frac{7}{8}right) right)$$
What is interesting to mention is that, doing calculations with another CAS, intermediate steps show $G$ appearing a few times.
answered Dec 21 '18 at 6:49
Claude LeiboviciClaude Leibovici
125k1158135
$endgroup$
$begingroup$
Yes $mathbf{G}$ does indeed appear twice before canceling out as can be seen in the answer I provide.
$endgroup$
– omegadot
Dec 21 '18 at 10:58
$begingroup$
@omegadot. Using another way, it did appear four times in complex expressions and ran away ! By the way, your answer is real nice ! $to +1$. Cheers.
$endgroup$
– Claude Leibovici
Dec 21 '18 at 11:22
$begingroup$
So does the closed form I found simplify to this?
$endgroup$
– clathratus
Dec 21 '18 at 19:02
add a comment |
$begingroup$
Too long for comments.
$$I=4sum_{ngeq0}frac{(-1)^n}{(2n+1)^2}(sqrt{2}-1)^{2n+1}=4 left(sqrt{2}-1right) ,
_3F_2left(frac{1}{2},frac{1}{2},1;frac{3}{2},frac{3}{2};2
sqrt{2}-3right)$$
On the other hand, working from Wolfram Alpha expression for the antiderivative and trying to simplify as much as I could the integral, I got
$$I=-frac{pi}{4} , log left(3+2 sqrt{2}right)+frac 1 {16 sqrt 2} left(psi ^{(1)}left(frac{1}{8}right)+psi ^{(1)}left(frac{3}{8}right)-psi
^{(1)}left(frac{5}{8}right)-psi ^{(1)}left(frac{7}{8}right) right)$$
What is interesting to mention is that, doing calculations with another CAS, intermediate steps show $G$ appearing a few times.
answered Dec 21 '18 at 6:49
Claude LeiboviciClaude Leibovici
125k1158135
$endgroup$
Too long for comments.
$$I=4sum_{ngeq0}frac{(-1)^n}{(2n+1)^2}(sqrt{2}-1)^{2n+1}=4 left(sqrt{2}-1right) ,
_3F_2left(frac{1}{2},frac{1}{2},1;frac{3}{2},frac{3}{2};2
sqrt{2}-3right)$$
On the other hand, working from Wolfram Alpha expression for the antiderivative and trying to simplify as much as I could the integral, I got
$$I=-frac{pi}{4} , log left(3+2 sqrt{2}right)+frac 1 {16 sqrt 2} left(psi ^{(1)}left(frac{1}{8}right)+psi ^{(1)}left(frac{3}{8}right)-psi
^{(1)}left(frac{5}{8}right)-psi ^{(1)}left(frac{7}{8}right) right)$$
What is interesting to mention is that, doing calculations with another CAS, intermediate steps show $G$ appearing a few times.
answered Dec 21 '18 at 6:49
Claude LeiboviciClaude Leibovici
125k1158135
edited Dec 21 '18 at 7:02
answered Dec 21 '18 at 6:49
Claude LeiboviciClaude Leibovici
125k1158135
answered Dec 21 '18 at 6:49
Claude LeiboviciClaude Leibovici
125k1158135
answered Dec 21 '18 at 6:49
Claude LeiboviciClaude Leibovici
125k1158135
125k1158135
$begingroup$
Yes $mathbf{G}$ does indeed appear twice before canceling out as can be seen in the answer I provide.
$endgroup$
– omegadot
Dec 21 '18 at 10:58
$begingroup$
@omegadot. Using another way, it did appear four times in complex expressions and ran away ! By the way, your answer is real nice ! $to +1$. Cheers.
$endgroup$
– Claude Leibovici
Dec 21 '18 at 11:22
$begingroup$
So does the closed form I found simplify to this?
$endgroup$
– clathratus
Dec 21 '18 at 19:02
add a comment |
$begingroup$
Yes $mathbf{G}$ does indeed appear twice before canceling out as can be seen in the answer I provide.
$endgroup$
– omegadot
Dec 21 '18 at 10:58
$begingroup$
@omegadot. Using another way, it did appear four times in complex expressions and ran away ! By the way, your answer is real nice ! $to +1$. Cheers.
$endgroup$
– Claude Leibovici
Dec 21 '18 at 11:22
$begingroup$
So does the closed form I found simplify to this?
$endgroup$
– clathratus
Dec 21 '18 at 19:02
$begingroup$
Yes $mathbf{G}$ does indeed appear twice before canceling out as can be seen in the answer I provide.
$endgroup$
– omegadot
Dec 21 '18 at 10:58
$begingroup$
Yes $mathbf{G}$ does indeed appear twice before canceling out as can be seen in the answer I provide.
$endgroup$
– omegadot
Dec 21 '18 at 10:58
$begingroup$
@omegadot. Using another way, it did appear four times in complex expressions and ran away ! By the way, your answer is real nice ! $to +1$. Cheers.
$endgroup$
– Claude Leibovici
Dec 21 '18 at 11:22
$begingroup$
@omegadot. Using another way, it did appear four times in complex expressions and ran away ! By the way, your answer is real nice ! $to +1$. Cheers.
$endgroup$
– Claude Leibovici
Dec 21 '18 at 11:22
$begingroup$
So does the closed form I found simplify to this?
$endgroup$
– clathratus
Dec 21 '18 at 19:02
$begingroup$
So does the closed form I found simplify to this?
$endgroup$
– clathratus
Dec 21 '18 at 19:02
add a comment |
$begingroup$
$newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
newcommand{dd}{mathrm{d}}
newcommand{ds}[1]{displaystyle{#1}}
newcommand{expo}[1]{,mathrm{e}^{#1},}
newcommand{ic}{mathrm{i}}
newcommand{mc}[1]{mathcal{#1}}
newcommand{mrm}[1]{mathrm{#1}}
newcommand{pars}[1]{left(,{#1},right)}
newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
newcommand{root}[2]{,sqrt[#1]{,{#2},},}
newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
newcommand{verts}[1]{leftvert,{#1},rightvert}$
begin{align}
I & equiv int_{-pi/4}^{pi/4}{x over sinpars{x}},dd x,
pars{~mbox{OP already shows that}
I = bbox[10px,#ffd]{!!!!! 4!int_{0}^{root{2} - 1}!!{arctanpars{x} over x},dd x}~}
end{align}
Then,
begin{align}
I & = bbox[10px,#ffd]{4int_{0}^{root{2} - 1}{arctanpars{x} over x},dd x} =
4,Imint_{0}^{root{2} - 1}{lnpars{1 + ic x} over x},dd x
\[5mm] &
stackrel{{large x = ic t} atop {large t = -ic x}}{=},,,
4,Imint_{0}^{-pars{root{2} - 1}ic}
{lnpars{1 - t} over t},dd t
\[5mm] & =
-4,Imint_{0}^{-pars{root{2} - 1}ic}
mrm{Li}_{2}'pars{t},dd t =
-4,Immrm{Li}_{2}pars{-bracks{root{2} - 1}ic}
\[5mm] & =
bbx{4,Immrm{Li}_{2}pars{bracks{root{2} - 1}ic}}
approx 1.6271
end{align}
answered Dec 21 '18 at 15:14
Felix MarinFelix Marin
68.9k7110147
$endgroup$
$begingroup$
Your final expression $Immrm{Li}_2([sqrt 2 - 1]i)$ is missing a factor of $-4$ infront otherwise it does not approx to $1.6271$.
$endgroup$
– mrtaurho
Dec 21 '18 at 16:41
1
$begingroup$
@mrtaurho Thanks. Indeed it's $displaystylecolor{red}{+}4$ because I already switched the $displaystylemathrm{Li}_{2} $ argument sign. Note that $displaystyleoverline{mathrm{Li}_{2}left(,{z},right)} = mathrm{Li}_{2}left(,{overline{z}},right)$.
$endgroup$
– Felix Marin
Dec 21 '18 at 17:17
$begingroup$
I appreciate the quick and easy approach. Thanks
$endgroup$
– clathratus
Dec 21 '18 at 19:03
$begingroup$
@clathratus You're welcome !!!.
$endgroup$
– Felix Marin
Dec 22 '18 at 16:22
add a comment |
$begingroup$
$newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
newcommand{dd}{mathrm{d}}
newcommand{ds}[1]{displaystyle{#1}}
newcommand{expo}[1]{,mathrm{e}^{#1},}
newcommand{ic}{mathrm{i}}
newcommand{mc}[1]{mathcal{#1}}
newcommand{mrm}[1]{mathrm{#1}}
newcommand{pars}[1]{left(,{#1},right)}
newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
newcommand{root}[2]{,sqrt[#1]{,{#2},},}
newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
newcommand{verts}[1]{leftvert,{#1},rightvert}$
begin{align}
I & equiv int_{-pi/4}^{pi/4}{x over sinpars{x}},dd x,
pars{~mbox{OP already shows that}
I = bbox[10px,#ffd]{!!!!! 4!int_{0}^{root{2} - 1}!!{arctanpars{x} over x},dd x}~}
end{align}
Then,
begin{align}
I & = bbox[10px,#ffd]{4int_{0}^{root{2} - 1}{arctanpars{x} over x},dd x} =
4,Imint_{0}^{root{2} - 1}{lnpars{1 + ic x} over x},dd x
\[5mm] &
stackrel{{large x = ic t} atop {large t = -ic x}}{=},,,
4,Imint_{0}^{-pars{root{2} - 1}ic}
{lnpars{1 - t} over t},dd t
\[5mm] & =
-4,Imint_{0}^{-pars{root{2} - 1}ic}
mrm{Li}_{2}'pars{t},dd t =
-4,Immrm{Li}_{2}pars{-bracks{root{2} - 1}ic}
\[5mm] & =
bbx{4,Immrm{Li}_{2}pars{bracks{root{2} - 1}ic}}
approx 1.6271
end{align}
answered Dec 21 '18 at 15:14
Felix MarinFelix Marin
68.9k7110147
$endgroup$
$begingroup$
Your final expression $Immrm{Li}_2([sqrt 2 - 1]i)$ is missing a factor of $-4$ infront otherwise it does not approx to $1.6271$.
$endgroup$
– mrtaurho
Dec 21 '18 at 16:41
1
$begingroup$
@mrtaurho Thanks. Indeed it's $displaystylecolor{red}{+}4$ because I already switched the $displaystylemathrm{Li}_{2} $ argument sign. Note that $displaystyleoverline{mathrm{Li}_{2}left(,{z},right)} = mathrm{Li}_{2}left(,{overline{z}},right)$.
$endgroup$
– Felix Marin
Dec 21 '18 at 17:17
$begingroup$
I appreciate the quick and easy approach. Thanks
$endgroup$
– clathratus
Dec 21 '18 at 19:03
$begingroup$
@clathratus You're welcome !!!.
$endgroup$
– Felix Marin
Dec 22 '18 at 16:22
add a comment |
$begingroup$
$newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
newcommand{dd}{mathrm{d}}
newcommand{ds}[1]{displaystyle{#1}}
newcommand{expo}[1]{,mathrm{e}^{#1},}
newcommand{ic}{mathrm{i}}
newcommand{mc}[1]{mathcal{#1}}
newcommand{mrm}[1]{mathrm{#1}}
newcommand{pars}[1]{left(,{#1},right)}
newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
newcommand{root}[2]{,sqrt[#1]{,{#2},},}
newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
newcommand{verts}[1]{leftvert,{#1},rightvert}$
begin{align}
I & equiv int_{-pi/4}^{pi/4}{x over sinpars{x}},dd x,
pars{~mbox{OP already shows that}
I = bbox[10px,#ffd]{!!!!! 4!int_{0}^{root{2} - 1}!!{arctanpars{x} over x},dd x}~}
end{align}
Then,
begin{align}
I & = bbox[10px,#ffd]{4int_{0}^{root{2} - 1}{arctanpars{x} over x},dd x} =
4,Imint_{0}^{root{2} - 1}{lnpars{1 + ic x} over x},dd x
\[5mm] &
stackrel{{large x = ic t} atop {large t = -ic x}}{=},,,
4,Imint_{0}^{-pars{root{2} - 1}ic}
{lnpars{1 - t} over t},dd t
\[5mm] & =
-4,Imint_{0}^{-pars{root{2} - 1}ic}
mrm{Li}_{2}'pars{t},dd t =
-4,Immrm{Li}_{2}pars{-bracks{root{2} - 1}ic}
\[5mm] & =
bbx{4,Immrm{Li}_{2}pars{bracks{root{2} - 1}ic}}
approx 1.6271
end{align}
answered Dec 21 '18 at 15:14
Felix MarinFelix Marin
68.9k7110147
$endgroup$
$newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
newcommand{dd}{mathrm{d}}
newcommand{ds}[1]{displaystyle{#1}}
newcommand{expo}[1]{,mathrm{e}^{#1},}
newcommand{ic}{mathrm{i}}
newcommand{mc}[1]{mathcal{#1}}
newcommand{mrm}[1]{mathrm{#1}}
newcommand{pars}[1]{left(,{#1},right)}
newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
newcommand{root}[2]{,sqrt[#1]{,{#2},},}
newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
newcommand{verts}[1]{leftvert,{#1},rightvert}$
begin{align}
I & equiv int_{-pi/4}^{pi/4}{x over sinpars{x}},dd x,
pars{~mbox{OP already shows that}
I = bbox[10px,#ffd]{!!!!! 4!int_{0}^{root{2} - 1}!!{arctanpars{x} over x},dd x}~}
end{align}
Then,
begin{align}
I & = bbox[10px,#ffd]{4int_{0}^{root{2} - 1}{arctanpars{x} over x},dd x} =
4,Imint_{0}^{root{2} - 1}{lnpars{1 + ic x} over x},dd x
\[5mm] &
stackrel{{large x = ic t} atop {large t = -ic x}}{=},,,
4,Imint_{0}^{-pars{root{2} - 1}ic}
{lnpars{1 - t} over t},dd t
\[5mm] & =
-4,Imint_{0}^{-pars{root{2} - 1}ic}
mrm{Li}_{2}'pars{t},dd t =
-4,Immrm{Li}_{2}pars{-bracks{root{2} - 1}ic}
\[5mm] & =
bbx{4,Immrm{Li}_{2}pars{bracks{root{2} - 1}ic}}
approx 1.6271
end{align}
answered Dec 21 '18 at 15:14
Felix MarinFelix Marin
68.9k7110147
edited Dec 21 '18 at 17:12
answered Dec 21 '18 at 15:14
Felix MarinFelix Marin
68.9k7110147
answered Dec 21 '18 at 15:14
Felix MarinFelix Marin
68.9k7110147
answered Dec 21 '18 at 15:14
Felix MarinFelix Marin
68.9k7110147
68.9k7110147
$begingroup$
Your final expression $Immrm{Li}_2([sqrt 2 - 1]i)$ is missing a factor of $-4$ infront otherwise it does not approx to $1.6271$.
$endgroup$
– mrtaurho
Dec 21 '18 at 16:41
1
$begingroup$
@mrtaurho Thanks. Indeed it's $displaystylecolor{red}{+}4$ because I already switched the $displaystylemathrm{Li}_{2} $ argument sign. Note that $displaystyleoverline{mathrm{Li}_{2}left(,{z},right)} = mathrm{Li}_{2}left(,{overline{z}},right)$.
$endgroup$
– Felix Marin
Dec 21 '18 at 17:17
$begingroup$
I appreciate the quick and easy approach. Thanks
$endgroup$
– clathratus
Dec 21 '18 at 19:03
$begingroup$
@clathratus You're welcome !!!.
$endgroup$
– Felix Marin
Dec 22 '18 at 16:22
add a comment |
$begingroup$
Your final expression $Immrm{Li}_2([sqrt 2 - 1]i)$ is missing a factor of $-4$ infront otherwise it does not approx to $1.6271$.
$endgroup$
– mrtaurho
Dec 21 '18 at 16:41
1
$begingroup$
@mrtaurho Thanks. Indeed it's $displaystylecolor{red}{+}4$ because I already switched the $displaystylemathrm{Li}_{2} $ argument sign. Note that $displaystyleoverline{mathrm{Li}_{2}left(,{z},right)} = mathrm{Li}_{2}left(,{overline{z}},right)$.
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– Felix Marin
Dec 21 '18 at 17:17
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I appreciate the quick and easy approach. Thanks
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– clathratus
Dec 21 '18 at 19:03
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@clathratus You're welcome !!!.
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– Felix Marin
Dec 22 '18 at 16:22
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Your final expression $Immrm{Li}_2([sqrt 2 - 1]i)$ is missing a factor of $-4$ infront otherwise it does not approx to $1.6271$.
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– mrtaurho
Dec 21 '18 at 16:41
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Your final expression $Immrm{Li}_2([sqrt 2 - 1]i)$ is missing a factor of $-4$ infront otherwise it does not approx to $1.6271$.
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– mrtaurho
Dec 21 '18 at 16:41
1
1
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@mrtaurho Thanks. Indeed it's $displaystylecolor{red}{+}4$ because I already switched the $displaystylemathrm{Li}_{2} $ argument sign. Note that $displaystyleoverline{mathrm{Li}_{2}left(,{z},right)} = mathrm{Li}_{2}left(,{overline{z}},right)$.
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– Felix Marin
Dec 21 '18 at 17:17
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@mrtaurho Thanks. Indeed it's $displaystylecolor{red}{+}4$ because I already switched the $displaystylemathrm{Li}_{2} $ argument sign. Note that $displaystyleoverline{mathrm{Li}_{2}left(,{z},right)} = mathrm{Li}_{2}left(,{overline{z}},right)$.
$endgroup$
– Felix Marin
Dec 21 '18 at 17:17
$begingroup$
I appreciate the quick and easy approach. Thanks
$endgroup$
– clathratus
Dec 21 '18 at 19:03
$begingroup$
I appreciate the quick and easy approach. Thanks
$endgroup$
– clathratus
Dec 21 '18 at 19:03
$begingroup$
@clathratus You're welcome !!!.
$endgroup$
– Felix Marin
Dec 22 '18 at 16:22
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@clathratus You're welcome !!!.
$endgroup$
– Felix Marin
Dec 22 '18 at 16:22
add a comment |
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begin{align}I&=2int_0^{pi/4}frac{x}{sin x}mathrm{d}x\
&=2Big[xlnleft(tanleft(frac{x}{2}right)right)Big]_0^{pi/4}-2int_0^{pi/4}lnleft(tanleft(frac{x}{2}right)right),dx\
&=frac{pi}{2}ln(sqrt{2}-1)-2int_0^{pi/4}lnleft(tanleft(frac{x}{2}right)right),dx\
end{align}
In the latter integral perform the change of variable $y=dfrac{x}{2}$,
begin{align}I&=frac{pi}{2}ln(sqrt{2}-1)-4int_0^{pi/8}lnleft(tanleft(xright)right),dx\
&=frac{pi}{2}ln(sqrt{2}-1)-4Big[frac{1}{2}text{i}big(text{Li}_2(text{i}tan x)-text{Li}_2(-text{i}tan x)big)+xlnleft(tan xright)Big]_0^{pi/8}\
&=boxed{2text{i}left(text{Li}_2left(text{i}left(1-sqrt{2}right)right)-text{Li}_2left(text{i}left(sqrt{2}-1right)right)right)}
end{align}
answered Dec 21 '18 at 21:17
FDPFDP
6,10211929
$endgroup$
add a comment |
$begingroup$
begin{align}I&=2int_0^{pi/4}frac{x}{sin x}mathrm{d}x\
&=2Big[xlnleft(tanleft(frac{x}{2}right)right)Big]_0^{pi/4}-2int_0^{pi/4}lnleft(tanleft(frac{x}{2}right)right),dx\
&=frac{pi}{2}ln(sqrt{2}-1)-2int_0^{pi/4}lnleft(tanleft(frac{x}{2}right)right),dx\
end{align}
In the latter integral perform the change of variable $y=dfrac{x}{2}$,
begin{align}I&=frac{pi}{2}ln(sqrt{2}-1)-4int_0^{pi/8}lnleft(tanleft(xright)right),dx\
&=frac{pi}{2}ln(sqrt{2}-1)-4Big[frac{1}{2}text{i}big(text{Li}_2(text{i}tan x)-text{Li}_2(-text{i}tan x)big)+xlnleft(tan xright)Big]_0^{pi/8}\
&=boxed{2text{i}left(text{Li}_2left(text{i}left(1-sqrt{2}right)right)-text{Li}_2left(text{i}left(sqrt{2}-1right)right)right)}
end{align}
answered Dec 21 '18 at 21:17
FDPFDP
6,10211929
$endgroup$
add a comment |
$begingroup$
begin{align}I&=2int_0^{pi/4}frac{x}{sin x}mathrm{d}x\
&=2Big[xlnleft(tanleft(frac{x}{2}right)right)Big]_0^{pi/4}-2int_0^{pi/4}lnleft(tanleft(frac{x}{2}right)right),dx\
&=frac{pi}{2}ln(sqrt{2}-1)-2int_0^{pi/4}lnleft(tanleft(frac{x}{2}right)right),dx\
end{align}
In the latter integral perform the change of variable $y=dfrac{x}{2}$,
begin{align}I&=frac{pi}{2}ln(sqrt{2}-1)-4int_0^{pi/8}lnleft(tanleft(xright)right),dx\
&=frac{pi}{2}ln(sqrt{2}-1)-4Big[frac{1}{2}text{i}big(text{Li}_2(text{i}tan x)-text{Li}_2(-text{i}tan x)big)+xlnleft(tan xright)Big]_0^{pi/8}\
&=boxed{2text{i}left(text{Li}_2left(text{i}left(1-sqrt{2}right)right)-text{Li}_2left(text{i}left(sqrt{2}-1right)right)right)}
end{align}
answered Dec 21 '18 at 21:17
FDPFDP
6,10211929
$endgroup$
begin{align}I&=2int_0^{pi/4}frac{x}{sin x}mathrm{d}x\
&=2Big[xlnleft(tanleft(frac{x}{2}right)right)Big]_0^{pi/4}-2int_0^{pi/4}lnleft(tanleft(frac{x}{2}right)right),dx\
&=frac{pi}{2}ln(sqrt{2}-1)-2int_0^{pi/4}lnleft(tanleft(frac{x}{2}right)right),dx\
end{align}
In the latter integral perform the change of variable $y=dfrac{x}{2}$,
begin{align}I&=frac{pi}{2}ln(sqrt{2}-1)-4int_0^{pi/8}lnleft(tanleft(xright)right),dx\
&=frac{pi}{2}ln(sqrt{2}-1)-4Big[frac{1}{2}text{i}big(text{Li}_2(text{i}tan x)-text{Li}_2(-text{i}tan x)big)+xlnleft(tan xright)Big]_0^{pi/8}\
&=boxed{2text{i}left(text{Li}_2left(text{i}left(1-sqrt{2}right)right)-text{Li}_2left(text{i}left(sqrt{2}-1right)right)right)}
end{align}
answered Dec 21 '18 at 21:17
FDPFDP
6,10211929
answered Dec 21 '18 at 21:17
FDPFDP
6,10211929
answered Dec 21 '18 at 21:17
FDPFDP
6,10211929
answered Dec 21 '18 at 21:17
FDPFDP
6,10211929
6,10211929
add a comment |
add a comment |
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I'm not sure why you are confident that your integral has a closed form. Not that Wolfram Alpha is the last word on integration, but it is usually pretty good at recognizing constants, and in this case it does not.
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– DeficientMathDude
Dec 21 '18 at 2:49
1
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@DeficientMathDude - You just contradicted yourself.
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– user150203
Dec 21 '18 at 3:31
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Wolframalpha numerical value for integral = 1.62706 and 2*G = 1.83193118
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– user150203
Dec 21 '18 at 3:39
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You can write $frac{x}{sin(x)}$ as $xcsc(x)$. See where I'm going with this?
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– Aniruddh Venkatesan
Dec 21 '18 at 3:42
1
$begingroup$
@DeficientMathDude see the edit I just made
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– clathratus
Dec 21 '18 at 3:50