Evaluating $int_{0}^{2pi}x^2ln^2(1-cos x)dx$
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I learnt that
$$int_{0}^{frac{pi}{2}} x^2 ln^2 cos x dx
= frac{11 pi^5}{1440} + frac{pi^3}{24} ln^2 2 + frac{pi}{2}zeta(3) ln 2$$
from Sangchul Lee's answer on How to evaluate $I=int_0^{pi/2}x^2ln(sin x)ln(cos x) dx$
I did some other calculations, and it appears that
$$I=int_{0}^{2pi}x^2ln^2(1-cos x)~dx = frac{48pizeta(3)ln2+8pi^3ln^22}{3}+frac{52pi^5}{45}.$$
However, I am not sure how to verify the result. What method should I use to calculate $I$?
calculus integration improper-integrals
asked Nov 1 '18 at 22:15
LarryLarry
2,53031131
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add a comment |
$begingroup$
I learnt that
$$int_{0}^{frac{pi}{2}} x^2 ln^2 cos x dx
= frac{11 pi^5}{1440} + frac{pi^3}{24} ln^2 2 + frac{pi}{2}zeta(3) ln 2$$
from Sangchul Lee's answer on How to evaluate $I=int_0^{pi/2}x^2ln(sin x)ln(cos x) dx$
I did some other calculations, and it appears that
$$I=int_{0}^{2pi}x^2ln^2(1-cos x)~dx = frac{48pizeta(3)ln2+8pi^3ln^22}{3}+frac{52pi^5}{45}.$$
However, I am not sure how to verify the result. What method should I use to calculate $I$?
calculus integration improper-integrals
asked Nov 1 '18 at 22:15
LarryLarry
2,53031131
$endgroup$
1
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My bet is on Fourier series ! That ln function has a well known Fourier series !
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– Tolaso
Nov 1 '18 at 22:19
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You can split it into three smaller results that need proving, using $ln (1-cos x)=ln 2+2lnsinfrac{x}{2}$, which at least explains why we get a quadratic in $ln 2$.
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– J.G.
Nov 1 '18 at 22:37
3
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I agree with Tolaso. The Fourier series of $log(1-cos x)$ is straightforward and the Fourier series of $log^2(1-cos x)$ can be computed by convolution and it involves harmonic numbers. Paired with the Fourier series of $x^2$ it converts $I$ into a combination of Euler sums with weight $5$.
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– Jack D'Aurizio
Nov 2 '18 at 2:31
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@Jack: I understand that $ln(1-cos x) = -sum_{n=1}^{infty}frac{cos^n x}{n}$, but how should I compute $ln^2(1-cos x)$ using convolution?
$endgroup$
– Larry
Nov 2 '18 at 13:16
add a comment |
$begingroup$
I learnt that
$$int_{0}^{frac{pi}{2}} x^2 ln^2 cos x dx
= frac{11 pi^5}{1440} + frac{pi^3}{24} ln^2 2 + frac{pi}{2}zeta(3) ln 2$$
from Sangchul Lee's answer on How to evaluate $I=int_0^{pi/2}x^2ln(sin x)ln(cos x) dx$
I did some other calculations, and it appears that
$$I=int_{0}^{2pi}x^2ln^2(1-cos x)~dx = frac{48pizeta(3)ln2+8pi^3ln^22}{3}+frac{52pi^5}{45}.$$
However, I am not sure how to verify the result. What method should I use to calculate $I$?
calculus integration improper-integrals
asked Nov 1 '18 at 22:15
LarryLarry
2,53031131
$endgroup$
I learnt that
$$int_{0}^{frac{pi}{2}} x^2 ln^2 cos x dx
= frac{11 pi^5}{1440} + frac{pi^3}{24} ln^2 2 + frac{pi}{2}zeta(3) ln 2$$
from Sangchul Lee's answer on How to evaluate $I=int_0^{pi/2}x^2ln(sin x)ln(cos x) dx$
I did some other calculations, and it appears that
$$I=int_{0}^{2pi}x^2ln^2(1-cos x)~dx = frac{48pizeta(3)ln2+8pi^3ln^22}{3}+frac{52pi^5}{45}.$$
However, I am not sure how to verify the result. What method should I use to calculate $I$?
calculus integration improper-integrals
calculus integration improper-integrals
asked Nov 1 '18 at 22:15
LarryLarry
2,53031131
asked Nov 1 '18 at 22:15
LarryLarry
2,53031131
edited Dec 19 '18 at 23:21
Larry
asked Nov 1 '18 at 22:15
LarryLarry
2,53031131
asked Nov 1 '18 at 22:15
LarryLarry
2,53031131
asked Nov 1 '18 at 22:15
LarryLarry
2,53031131
2,53031131
1
$begingroup$
My bet is on Fourier series ! That ln function has a well known Fourier series !
$endgroup$
– Tolaso
Nov 1 '18 at 22:19
$begingroup$
You can split it into three smaller results that need proving, using $ln (1-cos x)=ln 2+2lnsinfrac{x}{2}$, which at least explains why we get a quadratic in $ln 2$.
$endgroup$
– J.G.
Nov 1 '18 at 22:37
3
$begingroup$
I agree with Tolaso. The Fourier series of $log(1-cos x)$ is straightforward and the Fourier series of $log^2(1-cos x)$ can be computed by convolution and it involves harmonic numbers. Paired with the Fourier series of $x^2$ it converts $I$ into a combination of Euler sums with weight $5$.
$endgroup$
– Jack D'Aurizio
Nov 2 '18 at 2:31
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@Jack: I understand that $ln(1-cos x) = -sum_{n=1}^{infty}frac{cos^n x}{n}$, but how should I compute $ln^2(1-cos x)$ using convolution?
$endgroup$
– Larry
Nov 2 '18 at 13:16
add a comment |
1
$begingroup$
My bet is on Fourier series ! That ln function has a well known Fourier series !
$endgroup$
– Tolaso
Nov 1 '18 at 22:19
$begingroup$
You can split it into three smaller results that need proving, using $ln (1-cos x)=ln 2+2lnsinfrac{x}{2}$, which at least explains why we get a quadratic in $ln 2$.
$endgroup$
– J.G.
Nov 1 '18 at 22:37
3
$begingroup$
I agree with Tolaso. The Fourier series of $log(1-cos x)$ is straightforward and the Fourier series of $log^2(1-cos x)$ can be computed by convolution and it involves harmonic numbers. Paired with the Fourier series of $x^2$ it converts $I$ into a combination of Euler sums with weight $5$.
$endgroup$
– Jack D'Aurizio
Nov 2 '18 at 2:31
$begingroup$
@Jack: I understand that $ln(1-cos x) = -sum_{n=1}^{infty}frac{cos^n x}{n}$, but how should I compute $ln^2(1-cos x)$ using convolution?
$endgroup$
– Larry
Nov 2 '18 at 13:16
1
1
$begingroup$
My bet is on Fourier series ! That ln function has a well known Fourier series !
$endgroup$
– Tolaso
Nov 1 '18 at 22:19
$begingroup$
My bet is on Fourier series ! That ln function has a well known Fourier series !
$endgroup$
– Tolaso
Nov 1 '18 at 22:19
$begingroup$
You can split it into three smaller results that need proving, using $ln (1-cos x)=ln 2+2lnsinfrac{x}{2}$, which at least explains why we get a quadratic in $ln 2$.
$endgroup$
– J.G.
Nov 1 '18 at 22:37
$begingroup$
You can split it into three smaller results that need proving, using $ln (1-cos x)=ln 2+2lnsinfrac{x}{2}$, which at least explains why we get a quadratic in $ln 2$.
$endgroup$
– J.G.
Nov 1 '18 at 22:37
3
3
$begingroup$
I agree with Tolaso. The Fourier series of $log(1-cos x)$ is straightforward and the Fourier series of $log^2(1-cos x)$ can be computed by convolution and it involves harmonic numbers. Paired with the Fourier series of $x^2$ it converts $I$ into a combination of Euler sums with weight $5$.
$endgroup$
– Jack D'Aurizio
Nov 2 '18 at 2:31
$begingroup$
I agree with Tolaso. The Fourier series of $log(1-cos x)$ is straightforward and the Fourier series of $log^2(1-cos x)$ can be computed by convolution and it involves harmonic numbers. Paired with the Fourier series of $x^2$ it converts $I$ into a combination of Euler sums with weight $5$.
$endgroup$
– Jack D'Aurizio
Nov 2 '18 at 2:31
$begingroup$
@Jack: I understand that $ln(1-cos x) = -sum_{n=1}^{infty}frac{cos^n x}{n}$, but how should I compute $ln^2(1-cos x)$ using convolution?
$endgroup$
– Larry
Nov 2 '18 at 13:16
$begingroup$
@Jack: I understand that $ln(1-cos x) = -sum_{n=1}^{infty}frac{cos^n x}{n}$, but how should I compute $ln^2(1-cos x)$ using convolution?
$endgroup$
– Larry
Nov 2 '18 at 13:16
add a comment |
2 Answers
2
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After spending a lot of time I've reached the answer (not without help of "MathStackExchangians"). I'll continue the derivation by Larry starting from $I_2$.
I will use the following integral:
$$int_0^pi x^2cos(2kx)~dx=fracpi{2k^2}$$
We have
$$smallbegin{align}
I_2 &= 32int_{0}^{pi}x^2ln^2(sin x)~dx\
&= 32int_{0}^{pi}x^2left(ln(2)+sum_{n=1}^{infty}frac{cos (2nx)}{n}right)^2~dx\
&= 32int_{0}^{pi}x^2ln^2(2)~dx+64ln(2)int_{0}^{pi}x^2sum_{n=1}^{infty}frac{cos (2nx)}{n}~dx+32int_{0}^{pi}x^2left(sum_{n=1}^{infty}frac{cos (2nx)}{n}right)^2~dx\
&=frac{32}{3}pi^3ln^2(2)+sum_{n=1}^{infty}frac{64ln(2)}{n}int_{0}^{pi}x^2cos(2nx)~dx+32underbrace{int_{0}^{pi}x^2left(sum_{n=1}^{infty}frac{cos (2nx)}{n}right)^2~dx}_{J}
end{align}$$
$$smallbegin{align}
J &= int_{0}^{pi}x^2left(sum_{n=1}^{infty}frac{cos (2nx)}{n}right)^2~dx\
&=int_{0}^{pi}x^2left(sum_{n=1}^{infty}frac{cos^2 (2nx)}{n^2}+sum_{m,n=1;mneq n}^{infty}frac{cos (2mx)cos (2nx)}{mn}right)~dx\
&=sum_{n=1}^{infty}frac1{n^2}int_{0}^{pi}x^2cos^2 (2nx)~dx+sum_{m,n=1;mneq n}^{infty}frac1{mn}int_{0}^{pi}x^2cos (2mx)cos (2nx)~dx\
&=sum_{n=1}^{infty}frac1{2n^2}int_{0}^{pi}x^2(1+cos (4nx))~dx+sum_{m,n=1;mneq n}^{infty}frac1{2mn}int_{0}^{pi}x^2(cos (2(m+n)x)+cos (2(m-n)x))~dx\
&=sum_{n=1}^{infty}frac1{2n^2}left(int_{0}^{pi}x^2~dx+int_{0}^{pi}x^2cos (4nx)~dxright)+sum_{m,n=1;mneq n}^{infty}frac1{2mn}left(int_{0}^{pi}x^2cos (2(m+n)x)~dx+int_{0}^{pi}x^2cos (2(m-n)x)~dxright)\
&=sum_{n=1}^{infty}frac1{2n^2}left(frac{pi^3}3+fracpi{2(2n)^2}right)+sum_{m,n=1;mneq n}^{infty}frac1{2mn}left(fracpi{2(m+n)^2}+fracpi{2(m-n)^2}right)\
&=frac{pi^3}6sum_{n=1}^{infty}frac1{n^2}+fracpi{16}sum_{n=1}^{infty}frac1{n^4}+fracpi2sum_{m,n=1;mneq n}^{infty}frac{m^2+n^2}{mn(m^2-n^2)^2}\
&=frac{pi^3}6frac{pi^2}6+frac{pi}{16}frac{pi^4}{90}+fracpi2frac{11pi^4}{720}=frac{13pi^5}{360}
end{align}
$$
The last sum is evaluated (my thanks to Robert Z and Zvi) in this question
Finally we have
$$smallbegin{align}
I_2 &= frac{32}{3}pi^3ln^2(2)+sum_{n=1}^{infty}frac{64ln(2)}{n}int_{0}^{pi}x^2cos(2nx)~dx+32frac{13pi^5}{360}\
&= frac{32}{3}pi^3ln^2(2)+sum_{n=1}^{infty}frac{64ln(2)}{n}fracpi{2n^2}+frac{52pi^5}{45}\
&=frac{32}{3}pi^3ln^2(2)+32ln(2)pisum_{n=1}^{infty}frac1{n^3}+frac{52pi^5}{45}\
&= pi^3ln^2(2)+32piln(2)zeta(3)+frac{52pi^5}{45}
end{align}$$
And thus
$$begin{align}
I&=I_1+I_2\
&= -8pi^3ln^2(2)-16piln(2)zeta(3)+frac{32}{3}pi^3ln^2(2)+32piln(2)zeta(3)+frac{52pi^5}{45}\
&= 16pizeta(3)ln(2)+frac{8pi^3ln^2(2)}{3}+frac{52pi^5}{45}
end{align}$$
answered Nov 6 '18 at 13:06
Mikalai ParshutsichMikalai Parshutsich
473315
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Well done. I am not sure whether you were aware of this question I posted on AoPS or not. Howsoever I just wanted to add the link to a slightly different evaluation of the integral $J$ :)
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– mrtaurho
Nov 7 '18 at 17:46
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@Mikalai: Nice solution, thank you for your answer.
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– Larry
Nov 7 '18 at 20:48
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@mrtaurho: I used the same method that you had showed me in another question. I also checked out your question in AOPS. It is a clear solution.
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– Larry
Nov 7 '18 at 20:54
1
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@Larry Yes, I saw it within your own answer. I am happy that I was able to help :)
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– mrtaurho
Nov 7 '18 at 20:57
add a comment |
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Attempt:
By using the trigonometric identity
$$1-cos(x)=2sin^2left(frac x2right)$$
The given integral becomes
$$smallbegin{align}
I = int_{0}^{2pi}x^2ln^2 (1-cos x)~dx &= int_{0}^{2pi}x^2 ln^2left(2sin^2left(frac x2right)right)~dx\
&=int_{0}^{2pi}x^2left(ln(2)+2lnleft(sin frac{x}{2}right)right)^2dx\
&=int_{0}^{2pi}x^2 left(ln^2(2)+4ln(2)lnleft(sin frac{x}{2}right)+4ln^2left(sin frac{x}{2}right)right)dx\
&=frac{8pi^3}{3}ln^2(2)+32ln(2)int_0^{pi}x^2ln(sin x)~dx+4int_{0}^{2pi}x^2ln^2left(sin frac{x}{2}right)~dx
end{align}$$
where within the second integral the substitution $x=frac x2$ was used. Now use the Fourier series expansion
$$ln(sin x)=-ln(2)-sum_{n=1}^{infty}frac{cos(2nx)}{n}$$
to further get
$$smallbegin{align}
I_1 = frac{8pi^3}{3}ln^2(2)+32ln(2)int_0^{pi}x^2ln(sin x)~dx&=frac{8pi^3}{3}ln^2(2)+32ln(2)int_0^{pi}x^2left[-ln(2)-sum_{n=1}^{infty}frac{cos(2nx)}{n}right]~dx\
&=frac{8pi^3}{3}ln^2(2)-32ln^2(2)int_0^{pi}x^2~dx-sum_{n=1}^{infty}frac{32ln(2)}nint_0^{pi}x^2cos(2nx)~dx\
&=-8pi^3ln^2(2)-sum_{n=1}^{infty}frac{32ln(2)}nint_0^{pi}x^2cos(2nx)~dx
end{align}$$
Using integration by part, we get
$$smallbegin{align}
I_1 &= -8pi^3ln^2(2)-sum_{n=1}^{infty}frac{32ln(2)}nint_0^{pi}x^2cos(2nx)~dx\
&=-8pi^3ln^2(2)-sum_{n=1}^{infty}frac{32ln(2)}nleft[-x^2frac{sin(2nx)}{2n}+frac{2xcos(2nx)}{4n^2}-frac{2sin(2nx)}{8n^3}right]_{0}^{pi}\
&=-8pi^3ln^2(2)-sum_{n=1}^{infty}frac{32ln(2)}{n}frac{2pi}{4n^2}\
&= -8pi^3ln^2(2)-16piln(2)sum_{n=1}^{infty}frac{1}{n^3}\
&= -8pi^3ln^2(2)-16piln(2)zeta(3)
end{align}$$
Let
$$I_2 = 4int_{0}^{2pi}x^2ln^2left(sin frac{x}{2}right)~dx$$
Again, use $x = frac{x}{2}$
$$smallbegin{align}
I_2 &= 32int_{0}^{pi}x^2ln^2(sin x)~dx\
&= 32int_{0}^{pi}x^2left(ln(2)+sum_{n=1}^{infty}frac{cos (2nx)}{n}right)^2~dx\
&= 32int_{0}^{pi}x^2ln^2(2)~dx+64ln(2)int_{0}^{pi}x^2sum_{n=1}^{infty}frac{cos (2nx)}{n}~dx+32int_{0}^{pi}x^2left(sum_{n=1}^{infty}frac{cos (2nx)}{n}right)^2~dx\
&=frac{32}{3}pi^3ln^2(2)+sum_{n=1}^{infty}frac{64ln(2)}{n}int_{0}^{pi}x^2cos(2nx)~dx+32int_{0}^{pi}x^2left(sum_{n=1}^{infty}frac{cos (2nx)}{n}right)^2~dx\
&=frac{32}{3}pi^3ln^2(2)+32piln(2)zeta(3)+32int_{0}^{pi}x^2sum_{n=1}^{infty}frac{cos(2nx)}{n}sum_{n=1}^{infty}frac{cos(2nx)}{n}~dx\
&=frac{32}{3}pi^3ln^2(2)+32piln(2)zeta(3)+sum_{n=1}^{infty}frac{32}{n}int_{0}^{pi}x^2cos(2nx)sum_{n=1}^{infty}frac{cos(2nx)}{n}~dx\
&=frac{32}{3}pi^3ln^2(2)+32piln(2)zeta(3)+sum_{n=1}^{infty}frac{32}{n}sum_{n=1}^{infty}frac{1}{n}int_{0}^{pi}x^2cos^2(2nx)~dx\
&=frac{32}{3}pi^3ln^2(2)+32piln(2)zeta(3)+sum_{n=1}^{infty}frac{32}{n^2}left[frac{x^2sin(4nx)}{8n}-frac{sin(4nx)}{64n^3}+frac{xcos(4nx)}{16n^2}+frac{x^3}{6}right]_{0}^{pi}tag{a}\
&=frac{32}{3}pi^3ln^2(2)+32piln(2)zeta(3)+sum_{n=1}^{infty}frac{32}{n^2}left(frac{pi}{16n^2}+frac{pi^3}{6}right)\
&=frac{32}{3}pi^3ln^2(2)+32piln(2)zeta(3)+sum_{n=1}^{infty}frac{2pi}{n^4}+frac{16pi^3}{3n^2}\
&=frac{32}{3}pi^3ln^2(2)+32piln(2)zeta(3)+frac{pi^5}{45}+frac{8pi^5}{9}\
&=frac{32}{3}pi^3ln^2(2)+32piln(2)zeta(3)+frac{41pi^5}{45}
end{align}$$
Note that
$$begin{align}
I&=I_1+I_2\
&= -8pi^3ln^2(2)-16piln(2)zeta(3)+frac{32}{3}pi^3ln^2(2)+32piln(2)zeta(3)+frac{41pi^5}{45}\
&= frac{48pizeta(3)ln(2)+8pi^3ln^2(2)}{3}+frac{41pi^5}{45}
end{align}$$
However, the last term should be $frac{52pi^5}{45}$. I think I did something wrong on step (a).
answered Nov 2 '18 at 22:56
LarryLarry
2,53031131
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1
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You've expanded the square of the sum in wrong way. You've resulted $left(sum_{n=1}^{infty}frac{cos (2nx)}{n}right)^2=sum_{n=1}^{infty}left(frac{cos (2nx)}{n}right)^2$
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– Mikalai Parshutsich
Nov 5 '18 at 14:07
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That's what I thought, but I am not sure how to correct my mistake.
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– Larry
Nov 5 '18 at 18:17
add a comment |
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2 Answers
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2 Answers
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$begingroup$
After spending a lot of time I've reached the answer (not without help of "MathStackExchangians"). I'll continue the derivation by Larry starting from $I_2$.
I will use the following integral:
$$int_0^pi x^2cos(2kx)~dx=fracpi{2k^2}$$
We have
$$smallbegin{align}
I_2 &= 32int_{0}^{pi}x^2ln^2(sin x)~dx\
&= 32int_{0}^{pi}x^2left(ln(2)+sum_{n=1}^{infty}frac{cos (2nx)}{n}right)^2~dx\
&= 32int_{0}^{pi}x^2ln^2(2)~dx+64ln(2)int_{0}^{pi}x^2sum_{n=1}^{infty}frac{cos (2nx)}{n}~dx+32int_{0}^{pi}x^2left(sum_{n=1}^{infty}frac{cos (2nx)}{n}right)^2~dx\
&=frac{32}{3}pi^3ln^2(2)+sum_{n=1}^{infty}frac{64ln(2)}{n}int_{0}^{pi}x^2cos(2nx)~dx+32underbrace{int_{0}^{pi}x^2left(sum_{n=1}^{infty}frac{cos (2nx)}{n}right)^2~dx}_{J}
end{align}$$
$$smallbegin{align}
J &= int_{0}^{pi}x^2left(sum_{n=1}^{infty}frac{cos (2nx)}{n}right)^2~dx\
&=int_{0}^{pi}x^2left(sum_{n=1}^{infty}frac{cos^2 (2nx)}{n^2}+sum_{m,n=1;mneq n}^{infty}frac{cos (2mx)cos (2nx)}{mn}right)~dx\
&=sum_{n=1}^{infty}frac1{n^2}int_{0}^{pi}x^2cos^2 (2nx)~dx+sum_{m,n=1;mneq n}^{infty}frac1{mn}int_{0}^{pi}x^2cos (2mx)cos (2nx)~dx\
&=sum_{n=1}^{infty}frac1{2n^2}int_{0}^{pi}x^2(1+cos (4nx))~dx+sum_{m,n=1;mneq n}^{infty}frac1{2mn}int_{0}^{pi}x^2(cos (2(m+n)x)+cos (2(m-n)x))~dx\
&=sum_{n=1}^{infty}frac1{2n^2}left(int_{0}^{pi}x^2~dx+int_{0}^{pi}x^2cos (4nx)~dxright)+sum_{m,n=1;mneq n}^{infty}frac1{2mn}left(int_{0}^{pi}x^2cos (2(m+n)x)~dx+int_{0}^{pi}x^2cos (2(m-n)x)~dxright)\
&=sum_{n=1}^{infty}frac1{2n^2}left(frac{pi^3}3+fracpi{2(2n)^2}right)+sum_{m,n=1;mneq n}^{infty}frac1{2mn}left(fracpi{2(m+n)^2}+fracpi{2(m-n)^2}right)\
&=frac{pi^3}6sum_{n=1}^{infty}frac1{n^2}+fracpi{16}sum_{n=1}^{infty}frac1{n^4}+fracpi2sum_{m,n=1;mneq n}^{infty}frac{m^2+n^2}{mn(m^2-n^2)^2}\
&=frac{pi^3}6frac{pi^2}6+frac{pi}{16}frac{pi^4}{90}+fracpi2frac{11pi^4}{720}=frac{13pi^5}{360}
end{align}
$$
The last sum is evaluated (my thanks to Robert Z and Zvi) in this question
Finally we have
$$smallbegin{align}
I_2 &= frac{32}{3}pi^3ln^2(2)+sum_{n=1}^{infty}frac{64ln(2)}{n}int_{0}^{pi}x^2cos(2nx)~dx+32frac{13pi^5}{360}\
&= frac{32}{3}pi^3ln^2(2)+sum_{n=1}^{infty}frac{64ln(2)}{n}fracpi{2n^2}+frac{52pi^5}{45}\
&=frac{32}{3}pi^3ln^2(2)+32ln(2)pisum_{n=1}^{infty}frac1{n^3}+frac{52pi^5}{45}\
&= pi^3ln^2(2)+32piln(2)zeta(3)+frac{52pi^5}{45}
end{align}$$
And thus
$$begin{align}
I&=I_1+I_2\
&= -8pi^3ln^2(2)-16piln(2)zeta(3)+frac{32}{3}pi^3ln^2(2)+32piln(2)zeta(3)+frac{52pi^5}{45}\
&= 16pizeta(3)ln(2)+frac{8pi^3ln^2(2)}{3}+frac{52pi^5}{45}
end{align}$$
answered Nov 6 '18 at 13:06
Mikalai ParshutsichMikalai Parshutsich
473315
$endgroup$
$begingroup$
Well done. I am not sure whether you were aware of this question I posted on AoPS or not. Howsoever I just wanted to add the link to a slightly different evaluation of the integral $J$ :)
$endgroup$
– mrtaurho
Nov 7 '18 at 17:46
$begingroup$
@Mikalai: Nice solution, thank you for your answer.
$endgroup$
– Larry
Nov 7 '18 at 20:48
$begingroup$
@mrtaurho: I used the same method that you had showed me in another question. I also checked out your question in AOPS. It is a clear solution.
$endgroup$
– Larry
Nov 7 '18 at 20:54
1
$begingroup$
@Larry Yes, I saw it within your own answer. I am happy that I was able to help :)
$endgroup$
– mrtaurho
Nov 7 '18 at 20:57
add a comment |
$begingroup$
After spending a lot of time I've reached the answer (not without help of "MathStackExchangians"). I'll continue the derivation by Larry starting from $I_2$.
I will use the following integral:
$$int_0^pi x^2cos(2kx)~dx=fracpi{2k^2}$$
We have
$$smallbegin{align}
I_2 &= 32int_{0}^{pi}x^2ln^2(sin x)~dx\
&= 32int_{0}^{pi}x^2left(ln(2)+sum_{n=1}^{infty}frac{cos (2nx)}{n}right)^2~dx\
&= 32int_{0}^{pi}x^2ln^2(2)~dx+64ln(2)int_{0}^{pi}x^2sum_{n=1}^{infty}frac{cos (2nx)}{n}~dx+32int_{0}^{pi}x^2left(sum_{n=1}^{infty}frac{cos (2nx)}{n}right)^2~dx\
&=frac{32}{3}pi^3ln^2(2)+sum_{n=1}^{infty}frac{64ln(2)}{n}int_{0}^{pi}x^2cos(2nx)~dx+32underbrace{int_{0}^{pi}x^2left(sum_{n=1}^{infty}frac{cos (2nx)}{n}right)^2~dx}_{J}
end{align}$$
$$smallbegin{align}
J &= int_{0}^{pi}x^2left(sum_{n=1}^{infty}frac{cos (2nx)}{n}right)^2~dx\
&=int_{0}^{pi}x^2left(sum_{n=1}^{infty}frac{cos^2 (2nx)}{n^2}+sum_{m,n=1;mneq n}^{infty}frac{cos (2mx)cos (2nx)}{mn}right)~dx\
&=sum_{n=1}^{infty}frac1{n^2}int_{0}^{pi}x^2cos^2 (2nx)~dx+sum_{m,n=1;mneq n}^{infty}frac1{mn}int_{0}^{pi}x^2cos (2mx)cos (2nx)~dx\
&=sum_{n=1}^{infty}frac1{2n^2}int_{0}^{pi}x^2(1+cos (4nx))~dx+sum_{m,n=1;mneq n}^{infty}frac1{2mn}int_{0}^{pi}x^2(cos (2(m+n)x)+cos (2(m-n)x))~dx\
&=sum_{n=1}^{infty}frac1{2n^2}left(int_{0}^{pi}x^2~dx+int_{0}^{pi}x^2cos (4nx)~dxright)+sum_{m,n=1;mneq n}^{infty}frac1{2mn}left(int_{0}^{pi}x^2cos (2(m+n)x)~dx+int_{0}^{pi}x^2cos (2(m-n)x)~dxright)\
&=sum_{n=1}^{infty}frac1{2n^2}left(frac{pi^3}3+fracpi{2(2n)^2}right)+sum_{m,n=1;mneq n}^{infty}frac1{2mn}left(fracpi{2(m+n)^2}+fracpi{2(m-n)^2}right)\
&=frac{pi^3}6sum_{n=1}^{infty}frac1{n^2}+fracpi{16}sum_{n=1}^{infty}frac1{n^4}+fracpi2sum_{m,n=1;mneq n}^{infty}frac{m^2+n^2}{mn(m^2-n^2)^2}\
&=frac{pi^3}6frac{pi^2}6+frac{pi}{16}frac{pi^4}{90}+fracpi2frac{11pi^4}{720}=frac{13pi^5}{360}
end{align}
$$
The last sum is evaluated (my thanks to Robert Z and Zvi) in this question
Finally we have
$$smallbegin{align}
I_2 &= frac{32}{3}pi^3ln^2(2)+sum_{n=1}^{infty}frac{64ln(2)}{n}int_{0}^{pi}x^2cos(2nx)~dx+32frac{13pi^5}{360}\
&= frac{32}{3}pi^3ln^2(2)+sum_{n=1}^{infty}frac{64ln(2)}{n}fracpi{2n^2}+frac{52pi^5}{45}\
&=frac{32}{3}pi^3ln^2(2)+32ln(2)pisum_{n=1}^{infty}frac1{n^3}+frac{52pi^5}{45}\
&= pi^3ln^2(2)+32piln(2)zeta(3)+frac{52pi^5}{45}
end{align}$$
And thus
$$begin{align}
I&=I_1+I_2\
&= -8pi^3ln^2(2)-16piln(2)zeta(3)+frac{32}{3}pi^3ln^2(2)+32piln(2)zeta(3)+frac{52pi^5}{45}\
&= 16pizeta(3)ln(2)+frac{8pi^3ln^2(2)}{3}+frac{52pi^5}{45}
end{align}$$
answered Nov 6 '18 at 13:06
Mikalai ParshutsichMikalai Parshutsich
473315
$endgroup$
$begingroup$
Well done. I am not sure whether you were aware of this question I posted on AoPS or not. Howsoever I just wanted to add the link to a slightly different evaluation of the integral $J$ :)
$endgroup$
– mrtaurho
Nov 7 '18 at 17:46
$begingroup$
@Mikalai: Nice solution, thank you for your answer.
$endgroup$
– Larry
Nov 7 '18 at 20:48
$begingroup$
@mrtaurho: I used the same method that you had showed me in another question. I also checked out your question in AOPS. It is a clear solution.
$endgroup$
– Larry
Nov 7 '18 at 20:54
1
$begingroup$
@Larry Yes, I saw it within your own answer. I am happy that I was able to help :)
$endgroup$
– mrtaurho
Nov 7 '18 at 20:57
add a comment |
$begingroup$
After spending a lot of time I've reached the answer (not without help of "MathStackExchangians"). I'll continue the derivation by Larry starting from $I_2$.
I will use the following integral:
$$int_0^pi x^2cos(2kx)~dx=fracpi{2k^2}$$
We have
$$smallbegin{align}
I_2 &= 32int_{0}^{pi}x^2ln^2(sin x)~dx\
&= 32int_{0}^{pi}x^2left(ln(2)+sum_{n=1}^{infty}frac{cos (2nx)}{n}right)^2~dx\
&= 32int_{0}^{pi}x^2ln^2(2)~dx+64ln(2)int_{0}^{pi}x^2sum_{n=1}^{infty}frac{cos (2nx)}{n}~dx+32int_{0}^{pi}x^2left(sum_{n=1}^{infty}frac{cos (2nx)}{n}right)^2~dx\
&=frac{32}{3}pi^3ln^2(2)+sum_{n=1}^{infty}frac{64ln(2)}{n}int_{0}^{pi}x^2cos(2nx)~dx+32underbrace{int_{0}^{pi}x^2left(sum_{n=1}^{infty}frac{cos (2nx)}{n}right)^2~dx}_{J}
end{align}$$
$$smallbegin{align}
J &= int_{0}^{pi}x^2left(sum_{n=1}^{infty}frac{cos (2nx)}{n}right)^2~dx\
&=int_{0}^{pi}x^2left(sum_{n=1}^{infty}frac{cos^2 (2nx)}{n^2}+sum_{m,n=1;mneq n}^{infty}frac{cos (2mx)cos (2nx)}{mn}right)~dx\
&=sum_{n=1}^{infty}frac1{n^2}int_{0}^{pi}x^2cos^2 (2nx)~dx+sum_{m,n=1;mneq n}^{infty}frac1{mn}int_{0}^{pi}x^2cos (2mx)cos (2nx)~dx\
&=sum_{n=1}^{infty}frac1{2n^2}int_{0}^{pi}x^2(1+cos (4nx))~dx+sum_{m,n=1;mneq n}^{infty}frac1{2mn}int_{0}^{pi}x^2(cos (2(m+n)x)+cos (2(m-n)x))~dx\
&=sum_{n=1}^{infty}frac1{2n^2}left(int_{0}^{pi}x^2~dx+int_{0}^{pi}x^2cos (4nx)~dxright)+sum_{m,n=1;mneq n}^{infty}frac1{2mn}left(int_{0}^{pi}x^2cos (2(m+n)x)~dx+int_{0}^{pi}x^2cos (2(m-n)x)~dxright)\
&=sum_{n=1}^{infty}frac1{2n^2}left(frac{pi^3}3+fracpi{2(2n)^2}right)+sum_{m,n=1;mneq n}^{infty}frac1{2mn}left(fracpi{2(m+n)^2}+fracpi{2(m-n)^2}right)\
&=frac{pi^3}6sum_{n=1}^{infty}frac1{n^2}+fracpi{16}sum_{n=1}^{infty}frac1{n^4}+fracpi2sum_{m,n=1;mneq n}^{infty}frac{m^2+n^2}{mn(m^2-n^2)^2}\
&=frac{pi^3}6frac{pi^2}6+frac{pi}{16}frac{pi^4}{90}+fracpi2frac{11pi^4}{720}=frac{13pi^5}{360}
end{align}
$$
The last sum is evaluated (my thanks to Robert Z and Zvi) in this question
Finally we have
$$smallbegin{align}
I_2 &= frac{32}{3}pi^3ln^2(2)+sum_{n=1}^{infty}frac{64ln(2)}{n}int_{0}^{pi}x^2cos(2nx)~dx+32frac{13pi^5}{360}\
&= frac{32}{3}pi^3ln^2(2)+sum_{n=1}^{infty}frac{64ln(2)}{n}fracpi{2n^2}+frac{52pi^5}{45}\
&=frac{32}{3}pi^3ln^2(2)+32ln(2)pisum_{n=1}^{infty}frac1{n^3}+frac{52pi^5}{45}\
&= pi^3ln^2(2)+32piln(2)zeta(3)+frac{52pi^5}{45}
end{align}$$
And thus
$$begin{align}
I&=I_1+I_2\
&= -8pi^3ln^2(2)-16piln(2)zeta(3)+frac{32}{3}pi^3ln^2(2)+32piln(2)zeta(3)+frac{52pi^5}{45}\
&= 16pizeta(3)ln(2)+frac{8pi^3ln^2(2)}{3}+frac{52pi^5}{45}
end{align}$$
answered Nov 6 '18 at 13:06
Mikalai ParshutsichMikalai Parshutsich
473315
$endgroup$
After spending a lot of time I've reached the answer (not without help of "MathStackExchangians"). I'll continue the derivation by Larry starting from $I_2$.
I will use the following integral:
$$int_0^pi x^2cos(2kx)~dx=fracpi{2k^2}$$
We have
$$smallbegin{align}
I_2 &= 32int_{0}^{pi}x^2ln^2(sin x)~dx\
&= 32int_{0}^{pi}x^2left(ln(2)+sum_{n=1}^{infty}frac{cos (2nx)}{n}right)^2~dx\
&= 32int_{0}^{pi}x^2ln^2(2)~dx+64ln(2)int_{0}^{pi}x^2sum_{n=1}^{infty}frac{cos (2nx)}{n}~dx+32int_{0}^{pi}x^2left(sum_{n=1}^{infty}frac{cos (2nx)}{n}right)^2~dx\
&=frac{32}{3}pi^3ln^2(2)+sum_{n=1}^{infty}frac{64ln(2)}{n}int_{0}^{pi}x^2cos(2nx)~dx+32underbrace{int_{0}^{pi}x^2left(sum_{n=1}^{infty}frac{cos (2nx)}{n}right)^2~dx}_{J}
end{align}$$
$$smallbegin{align}
J &= int_{0}^{pi}x^2left(sum_{n=1}^{infty}frac{cos (2nx)}{n}right)^2~dx\
&=int_{0}^{pi}x^2left(sum_{n=1}^{infty}frac{cos^2 (2nx)}{n^2}+sum_{m,n=1;mneq n}^{infty}frac{cos (2mx)cos (2nx)}{mn}right)~dx\
&=sum_{n=1}^{infty}frac1{n^2}int_{0}^{pi}x^2cos^2 (2nx)~dx+sum_{m,n=1;mneq n}^{infty}frac1{mn}int_{0}^{pi}x^2cos (2mx)cos (2nx)~dx\
&=sum_{n=1}^{infty}frac1{2n^2}int_{0}^{pi}x^2(1+cos (4nx))~dx+sum_{m,n=1;mneq n}^{infty}frac1{2mn}int_{0}^{pi}x^2(cos (2(m+n)x)+cos (2(m-n)x))~dx\
&=sum_{n=1}^{infty}frac1{2n^2}left(int_{0}^{pi}x^2~dx+int_{0}^{pi}x^2cos (4nx)~dxright)+sum_{m,n=1;mneq n}^{infty}frac1{2mn}left(int_{0}^{pi}x^2cos (2(m+n)x)~dx+int_{0}^{pi}x^2cos (2(m-n)x)~dxright)\
&=sum_{n=1}^{infty}frac1{2n^2}left(frac{pi^3}3+fracpi{2(2n)^2}right)+sum_{m,n=1;mneq n}^{infty}frac1{2mn}left(fracpi{2(m+n)^2}+fracpi{2(m-n)^2}right)\
&=frac{pi^3}6sum_{n=1}^{infty}frac1{n^2}+fracpi{16}sum_{n=1}^{infty}frac1{n^4}+fracpi2sum_{m,n=1;mneq n}^{infty}frac{m^2+n^2}{mn(m^2-n^2)^2}\
&=frac{pi^3}6frac{pi^2}6+frac{pi}{16}frac{pi^4}{90}+fracpi2frac{11pi^4}{720}=frac{13pi^5}{360}
end{align}
$$
The last sum is evaluated (my thanks to Robert Z and Zvi) in this question
Finally we have
$$smallbegin{align}
I_2 &= frac{32}{3}pi^3ln^2(2)+sum_{n=1}^{infty}frac{64ln(2)}{n}int_{0}^{pi}x^2cos(2nx)~dx+32frac{13pi^5}{360}\
&= frac{32}{3}pi^3ln^2(2)+sum_{n=1}^{infty}frac{64ln(2)}{n}fracpi{2n^2}+frac{52pi^5}{45}\
&=frac{32}{3}pi^3ln^2(2)+32ln(2)pisum_{n=1}^{infty}frac1{n^3}+frac{52pi^5}{45}\
&= pi^3ln^2(2)+32piln(2)zeta(3)+frac{52pi^5}{45}
end{align}$$
And thus
$$begin{align}
I&=I_1+I_2\
&= -8pi^3ln^2(2)-16piln(2)zeta(3)+frac{32}{3}pi^3ln^2(2)+32piln(2)zeta(3)+frac{52pi^5}{45}\
&= 16pizeta(3)ln(2)+frac{8pi^3ln^2(2)}{3}+frac{52pi^5}{45}
end{align}$$
answered Nov 6 '18 at 13:06
Mikalai ParshutsichMikalai Parshutsich
473315
answered Nov 6 '18 at 13:06
Mikalai ParshutsichMikalai Parshutsich
473315
answered Nov 6 '18 at 13:06
Mikalai ParshutsichMikalai Parshutsich
473315
answered Nov 6 '18 at 13:06
Mikalai ParshutsichMikalai Parshutsich
473315
473315
$begingroup$
Well done. I am not sure whether you were aware of this question I posted on AoPS or not. Howsoever I just wanted to add the link to a slightly different evaluation of the integral $J$ :)
$endgroup$
– mrtaurho
Nov 7 '18 at 17:46
$begingroup$
@Mikalai: Nice solution, thank you for your answer.
$endgroup$
– Larry
Nov 7 '18 at 20:48
$begingroup$
@mrtaurho: I used the same method that you had showed me in another question. I also checked out your question in AOPS. It is a clear solution.
$endgroup$
– Larry
Nov 7 '18 at 20:54
1
$begingroup$
@Larry Yes, I saw it within your own answer. I am happy that I was able to help :)
$endgroup$
– mrtaurho
Nov 7 '18 at 20:57
add a comment |
$begingroup$
Well done. I am not sure whether you were aware of this question I posted on AoPS or not. Howsoever I just wanted to add the link to a slightly different evaluation of the integral $J$ :)
$endgroup$
– mrtaurho
Nov 7 '18 at 17:46
$begingroup$
@Mikalai: Nice solution, thank you for your answer.
$endgroup$
– Larry
Nov 7 '18 at 20:48
$begingroup$
@mrtaurho: I used the same method that you had showed me in another question. I also checked out your question in AOPS. It is a clear solution.
$endgroup$
– Larry
Nov 7 '18 at 20:54
1
$begingroup$
@Larry Yes, I saw it within your own answer. I am happy that I was able to help :)
$endgroup$
– mrtaurho
Nov 7 '18 at 20:57
$begingroup$
Well done. I am not sure whether you were aware of this question I posted on AoPS or not. Howsoever I just wanted to add the link to a slightly different evaluation of the integral $J$ :)
$endgroup$
– mrtaurho
Nov 7 '18 at 17:46
$begingroup$
Well done. I am not sure whether you were aware of this question I posted on AoPS or not. Howsoever I just wanted to add the link to a slightly different evaluation of the integral $J$ :)
$endgroup$
– mrtaurho
Nov 7 '18 at 17:46
$begingroup$
@Mikalai: Nice solution, thank you for your answer.
$endgroup$
– Larry
Nov 7 '18 at 20:48
$begingroup$
@Mikalai: Nice solution, thank you for your answer.
$endgroup$
– Larry
Nov 7 '18 at 20:48
$begingroup$
@mrtaurho: I used the same method that you had showed me in another question. I also checked out your question in AOPS. It is a clear solution.
$endgroup$
– Larry
Nov 7 '18 at 20:54
$begingroup$
@mrtaurho: I used the same method that you had showed me in another question. I also checked out your question in AOPS. It is a clear solution.
$endgroup$
– Larry
Nov 7 '18 at 20:54
1
1
$begingroup$
@Larry Yes, I saw it within your own answer. I am happy that I was able to help :)
$endgroup$
– mrtaurho
Nov 7 '18 at 20:57
$begingroup$
@Larry Yes, I saw it within your own answer. I am happy that I was able to help :)
$endgroup$
– mrtaurho
Nov 7 '18 at 20:57
add a comment |
$begingroup$
Attempt:
By using the trigonometric identity
$$1-cos(x)=2sin^2left(frac x2right)$$
The given integral becomes
$$smallbegin{align}
I = int_{0}^{2pi}x^2ln^2 (1-cos x)~dx &= int_{0}^{2pi}x^2 ln^2left(2sin^2left(frac x2right)right)~dx\
&=int_{0}^{2pi}x^2left(ln(2)+2lnleft(sin frac{x}{2}right)right)^2dx\
&=int_{0}^{2pi}x^2 left(ln^2(2)+4ln(2)lnleft(sin frac{x}{2}right)+4ln^2left(sin frac{x}{2}right)right)dx\
&=frac{8pi^3}{3}ln^2(2)+32ln(2)int_0^{pi}x^2ln(sin x)~dx+4int_{0}^{2pi}x^2ln^2left(sin frac{x}{2}right)~dx
end{align}$$
where within the second integral the substitution $x=frac x2$ was used. Now use the Fourier series expansion
$$ln(sin x)=-ln(2)-sum_{n=1}^{infty}frac{cos(2nx)}{n}$$
to further get
$$smallbegin{align}
I_1 = frac{8pi^3}{3}ln^2(2)+32ln(2)int_0^{pi}x^2ln(sin x)~dx&=frac{8pi^3}{3}ln^2(2)+32ln(2)int_0^{pi}x^2left[-ln(2)-sum_{n=1}^{infty}frac{cos(2nx)}{n}right]~dx\
&=frac{8pi^3}{3}ln^2(2)-32ln^2(2)int_0^{pi}x^2~dx-sum_{n=1}^{infty}frac{32ln(2)}nint_0^{pi}x^2cos(2nx)~dx\
&=-8pi^3ln^2(2)-sum_{n=1}^{infty}frac{32ln(2)}nint_0^{pi}x^2cos(2nx)~dx
end{align}$$
Using integration by part, we get
$$smallbegin{align}
I_1 &= -8pi^3ln^2(2)-sum_{n=1}^{infty}frac{32ln(2)}nint_0^{pi}x^2cos(2nx)~dx\
&=-8pi^3ln^2(2)-sum_{n=1}^{infty}frac{32ln(2)}nleft[-x^2frac{sin(2nx)}{2n}+frac{2xcos(2nx)}{4n^2}-frac{2sin(2nx)}{8n^3}right]_{0}^{pi}\
&=-8pi^3ln^2(2)-sum_{n=1}^{infty}frac{32ln(2)}{n}frac{2pi}{4n^2}\
&= -8pi^3ln^2(2)-16piln(2)sum_{n=1}^{infty}frac{1}{n^3}\
&= -8pi^3ln^2(2)-16piln(2)zeta(3)
end{align}$$
Let
$$I_2 = 4int_{0}^{2pi}x^2ln^2left(sin frac{x}{2}right)~dx$$
Again, use $x = frac{x}{2}$
$$smallbegin{align}
I_2 &= 32int_{0}^{pi}x^2ln^2(sin x)~dx\
&= 32int_{0}^{pi}x^2left(ln(2)+sum_{n=1}^{infty}frac{cos (2nx)}{n}right)^2~dx\
&= 32int_{0}^{pi}x^2ln^2(2)~dx+64ln(2)int_{0}^{pi}x^2sum_{n=1}^{infty}frac{cos (2nx)}{n}~dx+32int_{0}^{pi}x^2left(sum_{n=1}^{infty}frac{cos (2nx)}{n}right)^2~dx\
&=frac{32}{3}pi^3ln^2(2)+sum_{n=1}^{infty}frac{64ln(2)}{n}int_{0}^{pi}x^2cos(2nx)~dx+32int_{0}^{pi}x^2left(sum_{n=1}^{infty}frac{cos (2nx)}{n}right)^2~dx\
&=frac{32}{3}pi^3ln^2(2)+32piln(2)zeta(3)+32int_{0}^{pi}x^2sum_{n=1}^{infty}frac{cos(2nx)}{n}sum_{n=1}^{infty}frac{cos(2nx)}{n}~dx\
&=frac{32}{3}pi^3ln^2(2)+32piln(2)zeta(3)+sum_{n=1}^{infty}frac{32}{n}int_{0}^{pi}x^2cos(2nx)sum_{n=1}^{infty}frac{cos(2nx)}{n}~dx\
&=frac{32}{3}pi^3ln^2(2)+32piln(2)zeta(3)+sum_{n=1}^{infty}frac{32}{n}sum_{n=1}^{infty}frac{1}{n}int_{0}^{pi}x^2cos^2(2nx)~dx\
&=frac{32}{3}pi^3ln^2(2)+32piln(2)zeta(3)+sum_{n=1}^{infty}frac{32}{n^2}left[frac{x^2sin(4nx)}{8n}-frac{sin(4nx)}{64n^3}+frac{xcos(4nx)}{16n^2}+frac{x^3}{6}right]_{0}^{pi}tag{a}\
&=frac{32}{3}pi^3ln^2(2)+32piln(2)zeta(3)+sum_{n=1}^{infty}frac{32}{n^2}left(frac{pi}{16n^2}+frac{pi^3}{6}right)\
&=frac{32}{3}pi^3ln^2(2)+32piln(2)zeta(3)+sum_{n=1}^{infty}frac{2pi}{n^4}+frac{16pi^3}{3n^2}\
&=frac{32}{3}pi^3ln^2(2)+32piln(2)zeta(3)+frac{pi^5}{45}+frac{8pi^5}{9}\
&=frac{32}{3}pi^3ln^2(2)+32piln(2)zeta(3)+frac{41pi^5}{45}
end{align}$$
Note that
$$begin{align}
I&=I_1+I_2\
&= -8pi^3ln^2(2)-16piln(2)zeta(3)+frac{32}{3}pi^3ln^2(2)+32piln(2)zeta(3)+frac{41pi^5}{45}\
&= frac{48pizeta(3)ln(2)+8pi^3ln^2(2)}{3}+frac{41pi^5}{45}
end{align}$$
However, the last term should be $frac{52pi^5}{45}$. I think I did something wrong on step (a).
answered Nov 2 '18 at 22:56
LarryLarry
2,53031131
$endgroup$
1
$begingroup$
You've expanded the square of the sum in wrong way. You've resulted $left(sum_{n=1}^{infty}frac{cos (2nx)}{n}right)^2=sum_{n=1}^{infty}left(frac{cos (2nx)}{n}right)^2$
$endgroup$
– Mikalai Parshutsich
Nov 5 '18 at 14:07
$begingroup$
That's what I thought, but I am not sure how to correct my mistake.
$endgroup$
– Larry
Nov 5 '18 at 18:17
add a comment |
$begingroup$
Attempt:
By using the trigonometric identity
$$1-cos(x)=2sin^2left(frac x2right)$$
The given integral becomes
$$smallbegin{align}
I = int_{0}^{2pi}x^2ln^2 (1-cos x)~dx &= int_{0}^{2pi}x^2 ln^2left(2sin^2left(frac x2right)right)~dx\
&=int_{0}^{2pi}x^2left(ln(2)+2lnleft(sin frac{x}{2}right)right)^2dx\
&=int_{0}^{2pi}x^2 left(ln^2(2)+4ln(2)lnleft(sin frac{x}{2}right)+4ln^2left(sin frac{x}{2}right)right)dx\
&=frac{8pi^3}{3}ln^2(2)+32ln(2)int_0^{pi}x^2ln(sin x)~dx+4int_{0}^{2pi}x^2ln^2left(sin frac{x}{2}right)~dx
end{align}$$
where within the second integral the substitution $x=frac x2$ was used. Now use the Fourier series expansion
$$ln(sin x)=-ln(2)-sum_{n=1}^{infty}frac{cos(2nx)}{n}$$
to further get
$$smallbegin{align}
I_1 = frac{8pi^3}{3}ln^2(2)+32ln(2)int_0^{pi}x^2ln(sin x)~dx&=frac{8pi^3}{3}ln^2(2)+32ln(2)int_0^{pi}x^2left[-ln(2)-sum_{n=1}^{infty}frac{cos(2nx)}{n}right]~dx\
&=frac{8pi^3}{3}ln^2(2)-32ln^2(2)int_0^{pi}x^2~dx-sum_{n=1}^{infty}frac{32ln(2)}nint_0^{pi}x^2cos(2nx)~dx\
&=-8pi^3ln^2(2)-sum_{n=1}^{infty}frac{32ln(2)}nint_0^{pi}x^2cos(2nx)~dx
end{align}$$
Using integration by part, we get
$$smallbegin{align}
I_1 &= -8pi^3ln^2(2)-sum_{n=1}^{infty}frac{32ln(2)}nint_0^{pi}x^2cos(2nx)~dx\
&=-8pi^3ln^2(2)-sum_{n=1}^{infty}frac{32ln(2)}nleft[-x^2frac{sin(2nx)}{2n}+frac{2xcos(2nx)}{4n^2}-frac{2sin(2nx)}{8n^3}right]_{0}^{pi}\
&=-8pi^3ln^2(2)-sum_{n=1}^{infty}frac{32ln(2)}{n}frac{2pi}{4n^2}\
&= -8pi^3ln^2(2)-16piln(2)sum_{n=1}^{infty}frac{1}{n^3}\
&= -8pi^3ln^2(2)-16piln(2)zeta(3)
end{align}$$
Let
$$I_2 = 4int_{0}^{2pi}x^2ln^2left(sin frac{x}{2}right)~dx$$
Again, use $x = frac{x}{2}$
$$smallbegin{align}
I_2 &= 32int_{0}^{pi}x^2ln^2(sin x)~dx\
&= 32int_{0}^{pi}x^2left(ln(2)+sum_{n=1}^{infty}frac{cos (2nx)}{n}right)^2~dx\
&= 32int_{0}^{pi}x^2ln^2(2)~dx+64ln(2)int_{0}^{pi}x^2sum_{n=1}^{infty}frac{cos (2nx)}{n}~dx+32int_{0}^{pi}x^2left(sum_{n=1}^{infty}frac{cos (2nx)}{n}right)^2~dx\
&=frac{32}{3}pi^3ln^2(2)+sum_{n=1}^{infty}frac{64ln(2)}{n}int_{0}^{pi}x^2cos(2nx)~dx+32int_{0}^{pi}x^2left(sum_{n=1}^{infty}frac{cos (2nx)}{n}right)^2~dx\
&=frac{32}{3}pi^3ln^2(2)+32piln(2)zeta(3)+32int_{0}^{pi}x^2sum_{n=1}^{infty}frac{cos(2nx)}{n}sum_{n=1}^{infty}frac{cos(2nx)}{n}~dx\
&=frac{32}{3}pi^3ln^2(2)+32piln(2)zeta(3)+sum_{n=1}^{infty}frac{32}{n}int_{0}^{pi}x^2cos(2nx)sum_{n=1}^{infty}frac{cos(2nx)}{n}~dx\
&=frac{32}{3}pi^3ln^2(2)+32piln(2)zeta(3)+sum_{n=1}^{infty}frac{32}{n}sum_{n=1}^{infty}frac{1}{n}int_{0}^{pi}x^2cos^2(2nx)~dx\
&=frac{32}{3}pi^3ln^2(2)+32piln(2)zeta(3)+sum_{n=1}^{infty}frac{32}{n^2}left[frac{x^2sin(4nx)}{8n}-frac{sin(4nx)}{64n^3}+frac{xcos(4nx)}{16n^2}+frac{x^3}{6}right]_{0}^{pi}tag{a}\
&=frac{32}{3}pi^3ln^2(2)+32piln(2)zeta(3)+sum_{n=1}^{infty}frac{32}{n^2}left(frac{pi}{16n^2}+frac{pi^3}{6}right)\
&=frac{32}{3}pi^3ln^2(2)+32piln(2)zeta(3)+sum_{n=1}^{infty}frac{2pi}{n^4}+frac{16pi^3}{3n^2}\
&=frac{32}{3}pi^3ln^2(2)+32piln(2)zeta(3)+frac{pi^5}{45}+frac{8pi^5}{9}\
&=frac{32}{3}pi^3ln^2(2)+32piln(2)zeta(3)+frac{41pi^5}{45}
end{align}$$
Note that
$$begin{align}
I&=I_1+I_2\
&= -8pi^3ln^2(2)-16piln(2)zeta(3)+frac{32}{3}pi^3ln^2(2)+32piln(2)zeta(3)+frac{41pi^5}{45}\
&= frac{48pizeta(3)ln(2)+8pi^3ln^2(2)}{3}+frac{41pi^5}{45}
end{align}$$
However, the last term should be $frac{52pi^5}{45}$. I think I did something wrong on step (a).
answered Nov 2 '18 at 22:56
LarryLarry
2,53031131
$endgroup$
1
$begingroup$
You've expanded the square of the sum in wrong way. You've resulted $left(sum_{n=1}^{infty}frac{cos (2nx)}{n}right)^2=sum_{n=1}^{infty}left(frac{cos (2nx)}{n}right)^2$
$endgroup$
– Mikalai Parshutsich
Nov 5 '18 at 14:07
$begingroup$
That's what I thought, but I am not sure how to correct my mistake.
$endgroup$
– Larry
Nov 5 '18 at 18:17
add a comment |
$begingroup$
Attempt:
By using the trigonometric identity
$$1-cos(x)=2sin^2left(frac x2right)$$
The given integral becomes
$$smallbegin{align}
I = int_{0}^{2pi}x^2ln^2 (1-cos x)~dx &= int_{0}^{2pi}x^2 ln^2left(2sin^2left(frac x2right)right)~dx\
&=int_{0}^{2pi}x^2left(ln(2)+2lnleft(sin frac{x}{2}right)right)^2dx\
&=int_{0}^{2pi}x^2 left(ln^2(2)+4ln(2)lnleft(sin frac{x}{2}right)+4ln^2left(sin frac{x}{2}right)right)dx\
&=frac{8pi^3}{3}ln^2(2)+32ln(2)int_0^{pi}x^2ln(sin x)~dx+4int_{0}^{2pi}x^2ln^2left(sin frac{x}{2}right)~dx
end{align}$$
where within the second integral the substitution $x=frac x2$ was used. Now use the Fourier series expansion
$$ln(sin x)=-ln(2)-sum_{n=1}^{infty}frac{cos(2nx)}{n}$$
to further get
$$smallbegin{align}
I_1 = frac{8pi^3}{3}ln^2(2)+32ln(2)int_0^{pi}x^2ln(sin x)~dx&=frac{8pi^3}{3}ln^2(2)+32ln(2)int_0^{pi}x^2left[-ln(2)-sum_{n=1}^{infty}frac{cos(2nx)}{n}right]~dx\
&=frac{8pi^3}{3}ln^2(2)-32ln^2(2)int_0^{pi}x^2~dx-sum_{n=1}^{infty}frac{32ln(2)}nint_0^{pi}x^2cos(2nx)~dx\
&=-8pi^3ln^2(2)-sum_{n=1}^{infty}frac{32ln(2)}nint_0^{pi}x^2cos(2nx)~dx
end{align}$$
Using integration by part, we get
$$smallbegin{align}
I_1 &= -8pi^3ln^2(2)-sum_{n=1}^{infty}frac{32ln(2)}nint_0^{pi}x^2cos(2nx)~dx\
&=-8pi^3ln^2(2)-sum_{n=1}^{infty}frac{32ln(2)}nleft[-x^2frac{sin(2nx)}{2n}+frac{2xcos(2nx)}{4n^2}-frac{2sin(2nx)}{8n^3}right]_{0}^{pi}\
&=-8pi^3ln^2(2)-sum_{n=1}^{infty}frac{32ln(2)}{n}frac{2pi}{4n^2}\
&= -8pi^3ln^2(2)-16piln(2)sum_{n=1}^{infty}frac{1}{n^3}\
&= -8pi^3ln^2(2)-16piln(2)zeta(3)
end{align}$$
Let
$$I_2 = 4int_{0}^{2pi}x^2ln^2left(sin frac{x}{2}right)~dx$$
Again, use $x = frac{x}{2}$
$$smallbegin{align}
I_2 &= 32int_{0}^{pi}x^2ln^2(sin x)~dx\
&= 32int_{0}^{pi}x^2left(ln(2)+sum_{n=1}^{infty}frac{cos (2nx)}{n}right)^2~dx\
&= 32int_{0}^{pi}x^2ln^2(2)~dx+64ln(2)int_{0}^{pi}x^2sum_{n=1}^{infty}frac{cos (2nx)}{n}~dx+32int_{0}^{pi}x^2left(sum_{n=1}^{infty}frac{cos (2nx)}{n}right)^2~dx\
&=frac{32}{3}pi^3ln^2(2)+sum_{n=1}^{infty}frac{64ln(2)}{n}int_{0}^{pi}x^2cos(2nx)~dx+32int_{0}^{pi}x^2left(sum_{n=1}^{infty}frac{cos (2nx)}{n}right)^2~dx\
&=frac{32}{3}pi^3ln^2(2)+32piln(2)zeta(3)+32int_{0}^{pi}x^2sum_{n=1}^{infty}frac{cos(2nx)}{n}sum_{n=1}^{infty}frac{cos(2nx)}{n}~dx\
&=frac{32}{3}pi^3ln^2(2)+32piln(2)zeta(3)+sum_{n=1}^{infty}frac{32}{n}int_{0}^{pi}x^2cos(2nx)sum_{n=1}^{infty}frac{cos(2nx)}{n}~dx\
&=frac{32}{3}pi^3ln^2(2)+32piln(2)zeta(3)+sum_{n=1}^{infty}frac{32}{n}sum_{n=1}^{infty}frac{1}{n}int_{0}^{pi}x^2cos^2(2nx)~dx\
&=frac{32}{3}pi^3ln^2(2)+32piln(2)zeta(3)+sum_{n=1}^{infty}frac{32}{n^2}left[frac{x^2sin(4nx)}{8n}-frac{sin(4nx)}{64n^3}+frac{xcos(4nx)}{16n^2}+frac{x^3}{6}right]_{0}^{pi}tag{a}\
&=frac{32}{3}pi^3ln^2(2)+32piln(2)zeta(3)+sum_{n=1}^{infty}frac{32}{n^2}left(frac{pi}{16n^2}+frac{pi^3}{6}right)\
&=frac{32}{3}pi^3ln^2(2)+32piln(2)zeta(3)+sum_{n=1}^{infty}frac{2pi}{n^4}+frac{16pi^3}{3n^2}\
&=frac{32}{3}pi^3ln^2(2)+32piln(2)zeta(3)+frac{pi^5}{45}+frac{8pi^5}{9}\
&=frac{32}{3}pi^3ln^2(2)+32piln(2)zeta(3)+frac{41pi^5}{45}
end{align}$$
Note that
$$begin{align}
I&=I_1+I_2\
&= -8pi^3ln^2(2)-16piln(2)zeta(3)+frac{32}{3}pi^3ln^2(2)+32piln(2)zeta(3)+frac{41pi^5}{45}\
&= frac{48pizeta(3)ln(2)+8pi^3ln^2(2)}{3}+frac{41pi^5}{45}
end{align}$$
However, the last term should be $frac{52pi^5}{45}$. I think I did something wrong on step (a).
answered Nov 2 '18 at 22:56
LarryLarry
2,53031131
$endgroup$
Attempt:
By using the trigonometric identity
$$1-cos(x)=2sin^2left(frac x2right)$$
The given integral becomes
$$smallbegin{align}
I = int_{0}^{2pi}x^2ln^2 (1-cos x)~dx &= int_{0}^{2pi}x^2 ln^2left(2sin^2left(frac x2right)right)~dx\
&=int_{0}^{2pi}x^2left(ln(2)+2lnleft(sin frac{x}{2}right)right)^2dx\
&=int_{0}^{2pi}x^2 left(ln^2(2)+4ln(2)lnleft(sin frac{x}{2}right)+4ln^2left(sin frac{x}{2}right)right)dx\
&=frac{8pi^3}{3}ln^2(2)+32ln(2)int_0^{pi}x^2ln(sin x)~dx+4int_{0}^{2pi}x^2ln^2left(sin frac{x}{2}right)~dx
end{align}$$
where within the second integral the substitution $x=frac x2$ was used. Now use the Fourier series expansion
$$ln(sin x)=-ln(2)-sum_{n=1}^{infty}frac{cos(2nx)}{n}$$
to further get
$$smallbegin{align}
I_1 = frac{8pi^3}{3}ln^2(2)+32ln(2)int_0^{pi}x^2ln(sin x)~dx&=frac{8pi^3}{3}ln^2(2)+32ln(2)int_0^{pi}x^2left[-ln(2)-sum_{n=1}^{infty}frac{cos(2nx)}{n}right]~dx\
&=frac{8pi^3}{3}ln^2(2)-32ln^2(2)int_0^{pi}x^2~dx-sum_{n=1}^{infty}frac{32ln(2)}nint_0^{pi}x^2cos(2nx)~dx\
&=-8pi^3ln^2(2)-sum_{n=1}^{infty}frac{32ln(2)}nint_0^{pi}x^2cos(2nx)~dx
end{align}$$
Using integration by part, we get
$$smallbegin{align}
I_1 &= -8pi^3ln^2(2)-sum_{n=1}^{infty}frac{32ln(2)}nint_0^{pi}x^2cos(2nx)~dx\
&=-8pi^3ln^2(2)-sum_{n=1}^{infty}frac{32ln(2)}nleft[-x^2frac{sin(2nx)}{2n}+frac{2xcos(2nx)}{4n^2}-frac{2sin(2nx)}{8n^3}right]_{0}^{pi}\
&=-8pi^3ln^2(2)-sum_{n=1}^{infty}frac{32ln(2)}{n}frac{2pi}{4n^2}\
&= -8pi^3ln^2(2)-16piln(2)sum_{n=1}^{infty}frac{1}{n^3}\
&= -8pi^3ln^2(2)-16piln(2)zeta(3)
end{align}$$
Let
$$I_2 = 4int_{0}^{2pi}x^2ln^2left(sin frac{x}{2}right)~dx$$
Again, use $x = frac{x}{2}$
$$smallbegin{align}
I_2 &= 32int_{0}^{pi}x^2ln^2(sin x)~dx\
&= 32int_{0}^{pi}x^2left(ln(2)+sum_{n=1}^{infty}frac{cos (2nx)}{n}right)^2~dx\
&= 32int_{0}^{pi}x^2ln^2(2)~dx+64ln(2)int_{0}^{pi}x^2sum_{n=1}^{infty}frac{cos (2nx)}{n}~dx+32int_{0}^{pi}x^2left(sum_{n=1}^{infty}frac{cos (2nx)}{n}right)^2~dx\
&=frac{32}{3}pi^3ln^2(2)+sum_{n=1}^{infty}frac{64ln(2)}{n}int_{0}^{pi}x^2cos(2nx)~dx+32int_{0}^{pi}x^2left(sum_{n=1}^{infty}frac{cos (2nx)}{n}right)^2~dx\
&=frac{32}{3}pi^3ln^2(2)+32piln(2)zeta(3)+32int_{0}^{pi}x^2sum_{n=1}^{infty}frac{cos(2nx)}{n}sum_{n=1}^{infty}frac{cos(2nx)}{n}~dx\
&=frac{32}{3}pi^3ln^2(2)+32piln(2)zeta(3)+sum_{n=1}^{infty}frac{32}{n}int_{0}^{pi}x^2cos(2nx)sum_{n=1}^{infty}frac{cos(2nx)}{n}~dx\
&=frac{32}{3}pi^3ln^2(2)+32piln(2)zeta(3)+sum_{n=1}^{infty}frac{32}{n}sum_{n=1}^{infty}frac{1}{n}int_{0}^{pi}x^2cos^2(2nx)~dx\
&=frac{32}{3}pi^3ln^2(2)+32piln(2)zeta(3)+sum_{n=1}^{infty}frac{32}{n^2}left[frac{x^2sin(4nx)}{8n}-frac{sin(4nx)}{64n^3}+frac{xcos(4nx)}{16n^2}+frac{x^3}{6}right]_{0}^{pi}tag{a}\
&=frac{32}{3}pi^3ln^2(2)+32piln(2)zeta(3)+sum_{n=1}^{infty}frac{32}{n^2}left(frac{pi}{16n^2}+frac{pi^3}{6}right)\
&=frac{32}{3}pi^3ln^2(2)+32piln(2)zeta(3)+sum_{n=1}^{infty}frac{2pi}{n^4}+frac{16pi^3}{3n^2}\
&=frac{32}{3}pi^3ln^2(2)+32piln(2)zeta(3)+frac{pi^5}{45}+frac{8pi^5}{9}\
&=frac{32}{3}pi^3ln^2(2)+32piln(2)zeta(3)+frac{41pi^5}{45}
end{align}$$
Note that
$$begin{align}
I&=I_1+I_2\
&= -8pi^3ln^2(2)-16piln(2)zeta(3)+frac{32}{3}pi^3ln^2(2)+32piln(2)zeta(3)+frac{41pi^5}{45}\
&= frac{48pizeta(3)ln(2)+8pi^3ln^2(2)}{3}+frac{41pi^5}{45}
end{align}$$
However, the last term should be $frac{52pi^5}{45}$. I think I did something wrong on step (a).
answered Nov 2 '18 at 22:56
LarryLarry
2,53031131
edited Nov 3 '18 at 23:49
answered Nov 2 '18 at 22:56
LarryLarry
2,53031131
answered Nov 2 '18 at 22:56
LarryLarry
2,53031131
answered Nov 2 '18 at 22:56
LarryLarry
2,53031131
2,53031131
1
$begingroup$
You've expanded the square of the sum in wrong way. You've resulted $left(sum_{n=1}^{infty}frac{cos (2nx)}{n}right)^2=sum_{n=1}^{infty}left(frac{cos (2nx)}{n}right)^2$
$endgroup$
– Mikalai Parshutsich
Nov 5 '18 at 14:07
$begingroup$
That's what I thought, but I am not sure how to correct my mistake.
$endgroup$
– Larry
Nov 5 '18 at 18:17
add a comment |
1
$begingroup$
You've expanded the square of the sum in wrong way. You've resulted $left(sum_{n=1}^{infty}frac{cos (2nx)}{n}right)^2=sum_{n=1}^{infty}left(frac{cos (2nx)}{n}right)^2$
$endgroup$
– Mikalai Parshutsich
Nov 5 '18 at 14:07
$begingroup$
That's what I thought, but I am not sure how to correct my mistake.
$endgroup$
– Larry
Nov 5 '18 at 18:17
1
1
$begingroup$
You've expanded the square of the sum in wrong way. You've resulted $left(sum_{n=1}^{infty}frac{cos (2nx)}{n}right)^2=sum_{n=1}^{infty}left(frac{cos (2nx)}{n}right)^2$
$endgroup$
– Mikalai Parshutsich
Nov 5 '18 at 14:07
$begingroup$
You've expanded the square of the sum in wrong way. You've resulted $left(sum_{n=1}^{infty}frac{cos (2nx)}{n}right)^2=sum_{n=1}^{infty}left(frac{cos (2nx)}{n}right)^2$
$endgroup$
– Mikalai Parshutsich
Nov 5 '18 at 14:07
$begingroup$
That's what I thought, but I am not sure how to correct my mistake.
$endgroup$
– Larry
Nov 5 '18 at 18:17
$begingroup$
That's what I thought, but I am not sure how to correct my mistake.
$endgroup$
– Larry
Nov 5 '18 at 18:17
add a comment |
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My bet is on Fourier series ! That ln function has a well known Fourier series !
$endgroup$
– Tolaso
Nov 1 '18 at 22:19
$begingroup$
You can split it into three smaller results that need proving, using $ln (1-cos x)=ln 2+2lnsinfrac{x}{2}$, which at least explains why we get a quadratic in $ln 2$.
$endgroup$
– J.G.
Nov 1 '18 at 22:37
3
$begingroup$
I agree with Tolaso. The Fourier series of $log(1-cos x)$ is straightforward and the Fourier series of $log^2(1-cos x)$ can be computed by convolution and it involves harmonic numbers. Paired with the Fourier series of $x^2$ it converts $I$ into a combination of Euler sums with weight $5$.
$endgroup$
– Jack D'Aurizio
Nov 2 '18 at 2:31
$begingroup$
@Jack: I understand that $ln(1-cos x) = -sum_{n=1}^{infty}frac{cos^n x}{n}$, but how should I compute $ln^2(1-cos x)$ using convolution?
$endgroup$
– Larry
Nov 2 '18 at 13:16