Find the value of $1-frac{1}{7}+frac{1}{9}-frac{1}{15}+frac{1}{17}-frac{1}{23}+frac{1}{25}…$
$begingroup$
Find the value of this :$$1-frac{1}{7}+frac{1}{9}-frac{1}{15}+frac{1}{17}-frac{1}{23}+frac{1}{25}....$$
Try: We can write the above series as
$${S} = int^{1}_{0}bigg[1-x^6+x^8-x^{14}+x^{16}-x^{22}+cdotsbigg]dx$$
$$S = int^{1}_{0}(1-x^6)bigg[1+x^{8}+x^{16}+cdots cdots bigg]dx$$
So $$S = int^{1}_{0}frac{1-x^6}{1-x^{8}}dx = int^{1}_{0}frac{x^4+x^2+1}{(x^2+1)(x^4+1)}dx$$
Now i am struck in that integration.
Did not understand how to solve it
could some help me to solve it. Thanks in advance
integration sequences-and-series definite-integrals
edited Dec 19 '18 at 6:04
Asaf Karagila♦
307k33440773
asked Dec 18 '18 at 15:20
DXTDXT
6,0762732
$endgroup$
|
show 2 more comments
$begingroup$
Find the value of this :$$1-frac{1}{7}+frac{1}{9}-frac{1}{15}+frac{1}{17}-frac{1}{23}+frac{1}{25}....$$
Try: We can write the above series as
$${S} = int^{1}_{0}bigg[1-x^6+x^8-x^{14}+x^{16}-x^{22}+cdotsbigg]dx$$
$$S = int^{1}_{0}(1-x^6)bigg[1+x^{8}+x^{16}+cdots cdots bigg]dx$$
So $$S = int^{1}_{0}frac{1-x^6}{1-x^{8}}dx = int^{1}_{0}frac{x^4+x^2+1}{(x^2+1)(x^4+1)}dx$$
Now i am struck in that integration.
Did not understand how to solve it
could some help me to solve it. Thanks in advance
integration sequences-and-series definite-integrals
edited Dec 19 '18 at 6:04
Asaf Karagila♦
307k33440773
asked Dec 18 '18 at 15:20
DXTDXT
6,0762732
$endgroup$
$begingroup$
Hint: Partial fraction decomposition ($x^4+1$ can be factored over the reals).
$endgroup$
– Michael Burr
Dec 18 '18 at 15:24
2
$begingroup$
Replace $x^2=y$ and use mathworld.wolfram.com/PartialFractionDecomposition.html and replace back $x$
$endgroup$
– lab bhattacharjee
Dec 18 '18 at 15:25
1
$begingroup$
See also math.stackexchange.com/questions/2363639/int-fracx2x4x21-dx/…
$endgroup$
– lab bhattacharjee
Dec 18 '18 at 15:26
$begingroup$
Hey, is this a problem from Joseph edwards integral calculus for beginners? I remember solving this. The answer should equal something like $frac{pi}{8}(1+sqrt{2)}$ or something.
$endgroup$
– crskhr
Dec 18 '18 at 15:28
1
$begingroup$
@Stockfish: Absolutely convergent series.
$endgroup$
– Clayton
Dec 18 '18 at 15:32
|
show 2 more comments
$begingroup$
Find the value of this :$$1-frac{1}{7}+frac{1}{9}-frac{1}{15}+frac{1}{17}-frac{1}{23}+frac{1}{25}....$$
Try: We can write the above series as
$${S} = int^{1}_{0}bigg[1-x^6+x^8-x^{14}+x^{16}-x^{22}+cdotsbigg]dx$$
$$S = int^{1}_{0}(1-x^6)bigg[1+x^{8}+x^{16}+cdots cdots bigg]dx$$
So $$S = int^{1}_{0}frac{1-x^6}{1-x^{8}}dx = int^{1}_{0}frac{x^4+x^2+1}{(x^2+1)(x^4+1)}dx$$
Now i am struck in that integration.
Did not understand how to solve it
could some help me to solve it. Thanks in advance
integration sequences-and-series definite-integrals
edited Dec 19 '18 at 6:04
Asaf Karagila♦
307k33440773
asked Dec 18 '18 at 15:20
DXTDXT
6,0762732
$endgroup$
Find the value of this :$$1-frac{1}{7}+frac{1}{9}-frac{1}{15}+frac{1}{17}-frac{1}{23}+frac{1}{25}....$$
Try: We can write the above series as
$${S} = int^{1}_{0}bigg[1-x^6+x^8-x^{14}+x^{16}-x^{22}+cdotsbigg]dx$$
$$S = int^{1}_{0}(1-x^6)bigg[1+x^{8}+x^{16}+cdots cdots bigg]dx$$
So $$S = int^{1}_{0}frac{1-x^6}{1-x^{8}}dx = int^{1}_{0}frac{x^4+x^2+1}{(x^2+1)(x^4+1)}dx$$
Now i am struck in that integration.
Did not understand how to solve it
could some help me to solve it. Thanks in advance
integration sequences-and-series definite-integrals
integration sequences-and-series definite-integrals
edited Dec 19 '18 at 6:04
Asaf Karagila♦
307k33440773
asked Dec 18 '18 at 15:20
DXTDXT
6,0762732
edited Dec 19 '18 at 6:04
Asaf Karagila♦
307k33440773
asked Dec 18 '18 at 15:20
DXTDXT
6,0762732
edited Dec 19 '18 at 6:04
Asaf Karagila♦
307k33440773
edited Dec 19 '18 at 6:04
Asaf Karagila♦
307k33440773
edited Dec 19 '18 at 6:04
Asaf Karagila♦
307k33440773
307k33440773
asked Dec 18 '18 at 15:20
DXTDXT
6,0762732
asked Dec 18 '18 at 15:20
DXTDXT
6,0762732
asked Dec 18 '18 at 15:20
DXTDXT
6,0762732
6,0762732
$begingroup$
Hint: Partial fraction decomposition ($x^4+1$ can be factored over the reals).
$endgroup$
– Michael Burr
Dec 18 '18 at 15:24
2
$begingroup$
Replace $x^2=y$ and use mathworld.wolfram.com/PartialFractionDecomposition.html and replace back $x$
$endgroup$
– lab bhattacharjee
Dec 18 '18 at 15:25
1
$begingroup$
See also math.stackexchange.com/questions/2363639/int-fracx2x4x21-dx/…
$endgroup$
– lab bhattacharjee
Dec 18 '18 at 15:26
$begingroup$
Hey, is this a problem from Joseph edwards integral calculus for beginners? I remember solving this. The answer should equal something like $frac{pi}{8}(1+sqrt{2)}$ or something.
$endgroup$
– crskhr
Dec 18 '18 at 15:28
1
$begingroup$
@Stockfish: Absolutely convergent series.
$endgroup$
– Clayton
Dec 18 '18 at 15:32
|
show 2 more comments
$begingroup$
Hint: Partial fraction decomposition ($x^4+1$ can be factored over the reals).
$endgroup$
– Michael Burr
Dec 18 '18 at 15:24
2
$begingroup$
Replace $x^2=y$ and use mathworld.wolfram.com/PartialFractionDecomposition.html and replace back $x$
$endgroup$
– lab bhattacharjee
Dec 18 '18 at 15:25
1
$begingroup$
See also math.stackexchange.com/questions/2363639/int-fracx2x4x21-dx/…
$endgroup$
– lab bhattacharjee
Dec 18 '18 at 15:26
$begingroup$
Hey, is this a problem from Joseph edwards integral calculus for beginners? I remember solving this. The answer should equal something like $frac{pi}{8}(1+sqrt{2)}$ or something.
$endgroup$
– crskhr
Dec 18 '18 at 15:28
1
$begingroup$
@Stockfish: Absolutely convergent series.
$endgroup$
– Clayton
Dec 18 '18 at 15:32
$begingroup$
Hint: Partial fraction decomposition ($x^4+1$ can be factored over the reals).
$endgroup$
– Michael Burr
Dec 18 '18 at 15:24
$begingroup$
Hint: Partial fraction decomposition ($x^4+1$ can be factored over the reals).
$endgroup$
– Michael Burr
Dec 18 '18 at 15:24
2
2
$begingroup$
Replace $x^2=y$ and use mathworld.wolfram.com/PartialFractionDecomposition.html and replace back $x$
$endgroup$
– lab bhattacharjee
Dec 18 '18 at 15:25
$begingroup$
Replace $x^2=y$ and use mathworld.wolfram.com/PartialFractionDecomposition.html and replace back $x$
$endgroup$
– lab bhattacharjee
Dec 18 '18 at 15:25
1
1
$begingroup$
See also math.stackexchange.com/questions/2363639/int-fracx2x4x21-dx/…
$endgroup$
– lab bhattacharjee
Dec 18 '18 at 15:26
$begingroup$
See also math.stackexchange.com/questions/2363639/int-fracx2x4x21-dx/…
$endgroup$
– lab bhattacharjee
Dec 18 '18 at 15:26
$begingroup$
Hey, is this a problem from Joseph edwards integral calculus for beginners? I remember solving this. The answer should equal something like $frac{pi}{8}(1+sqrt{2)}$ or something.
$endgroup$
– crskhr
Dec 18 '18 at 15:28
$begingroup$
Hey, is this a problem from Joseph edwards integral calculus for beginners? I remember solving this. The answer should equal something like $frac{pi}{8}(1+sqrt{2)}$ or something.
$endgroup$
– crskhr
Dec 18 '18 at 15:28
1
1
$begingroup$
@Stockfish: Absolutely convergent series.
$endgroup$
– Clayton
Dec 18 '18 at 15:32
$begingroup$
@Stockfish: Absolutely convergent series.
$endgroup$
– Clayton
Dec 18 '18 at 15:32
|
show 2 more comments
4 Answers
4
active
oldest
votes
$begingroup$
Another approach, which could be useful to check yours through integrals,
is same as per the similar post, using the Digamma function $psi (z)$.
$$
eqalign{
& 1 - {1 over {8 - 1}} + {1 over {8 + 1}} - {1 over {2 cdot 8 - 1}} + {1 over {2 cdot 8 + 1}} + cdots = cr
& = 1 - sumlimits_{1, le ,n} {{1 over {8n - 1}}} + sumlimits_{1, le ,n} {{1 over {8n + 1}}} = cr
& = 1 - {1 over 8}left( {sumlimits_{1, le ,n} {{1 over {n - 1/8}}} - sumlimits_{1, le ,n} {{1 over {n + 1/8}}} } right) = cr
& = 1 - {1 over 8}left( {sumnolimits_{;n = 7/8}^{;infty } {{1 over n}} - sumnolimits_{;n = 9/8}^{;infty } {{1 over n}} } right) = cr
& = 1 - {1 over 8}left( {sumnolimits_{;n = 7/8}^{;9/8} {{1 over n}} } right) = cr
& = 1 - {1 over 8}left( {psi (9/8) - psi (7/8)} right) = cr
& = 1 - {1 over 8}left( {psi left( {1 + 1/8} right) - psi left( {1 - 1/8} right)} right) = cr
& = 1 - {1 over 8}left( {{1 over {1/8}} + psi left( {1/8} right) - psi left( {1 - 1/8} right)} right) = cr
& = {1 over 8}left( {psi left( {1 - 1/8} right) - psi left( {1/8} right)} right) = cr
& = {1 over 8}left( {pi cot left( {{pi over 8}} right)} right) = cr
& = {{pi left( {1 + sqrt 2 } right)} over 8} cr}
$$
where:
$$
Delta _{,z} psi left( z right) = psi left( {z + 1} right) - psi left( z right) = {1 over z}
$$
is the functional equation for the Digamma;
which implies that Digamma is the Antidelta of $1/z$
$$
eqalign{
& psi left( z right) = Delta _{,z} ^{ - 1} left( {{1 over z}} right) = sumnolimits_z {{1 over z}} quad Rightarrow cr
& Rightarrow quad sumnolimits_{;z = a}^{;b} {{1 over z}} = psi left( b right) - psi left( a right) cr}
$$
and we used the Reflection formula for Digamma
$$
psi left( {1 - z} right) = psi left( z right) + pi cot left( {pi z} right)
$$
Now, the above, suggests a way to solve the integral.
Let's replace $x^8$ with $y$
$$
int_{x = 0}^{,1} {{{1 - x^{,6} } over {1 - x^{,8} }}dx} quad mathop
= limits^{x^{,8} = y} quad {1 over 8}int_{y = 0}^{,1} {{{1 - y^{,3/4} } over {left( {1 - y} right)y^{,7/8} }}dy}
$$
then consider that we have
$$
eqalign{
& {{1 - y^{,3/4} } over {left( {1 - y} right)y^{,7/8} }} = cr
& = left( {left( {1 - y} right)^{ - 1} y^{, - 7/8} - left( {1 - y} right)^{ - 1} y^{, - 1/8} } right) cr
& = mathop {lim }limits_{varepsilon , to ,0} left( {left( {1 - y} right)^{varepsilon - 1} y^{,1/8 - 1}
- left( {1 - y} right)^{varepsilon - 1} y^{,,7/8 - 1} } right) cr}
$$
so we are ready to use the integral and Gamma representation for the Beta function
$$
eqalign{
& 8 ;int_{x = 0}^{,1} {{{1 - x^{,6} } over {1 - x^{,8} }}dx} = int_{y = 0}^{,1} {{{1 - y^{,3/4} } over {left( {1 - y} right)y^{,7/8} }}dy} = cr
& = mathop {lim }limits_{varepsilon , to ,0} left( {int_{y = 0}^{,1} {left( {1 - y} right)^{varepsilon - 1} y^{,1/8 - 1} dy}
- int_{y = 0}^{,1} {left( {1 - y} right)^{varepsilon - 1} y^{,,7/8 - 1} dy} } right) = cr
& = mathop {lim }limits_{varepsilon , to ,0} left( {{rm B}left( {1/8,varepsilon } right) - {rm B}left( {7/8,varepsilon } right)} right) = cr
& = mathop {lim }limits_{varepsilon , to ,0} left( {{{Gamma left( {1/8} right)Gamma left( varepsilon right)}
over {Gamma left( {1/8 + varepsilon } right)}}
- {{Gamma left( {7/8} right)Gamma left( varepsilon right)} over {Gamma left( {7/8 + varepsilon } right)}}} right) = cr
& = mathop {lim }limits_{varepsilon , to ,0} left( {{{Gamma left( {1/8} right)Gamma left( varepsilon right)}
over {psi left( {1/8} right)varepsilon ;Gamma left( {1/8} right)
+ Gamma left( {1/8} right)}} - {{Gamma left( {7/8} right)Gamma left( varepsilon right)}
over {psi left( {7/8} right)varepsilon ;Gamma left( {7/8} right) + Gamma left( {7/8} right)}}} right) = cr
& = mathop {lim }limits_{varepsilon , to ,0} left( {{{Gamma left( varepsilon right)} over {psi left( {1/8} right)varepsilon ; + 1}}
- {{Gamma left( varepsilon right)} over {psi left( {7/8} right)varepsilon ; + 1}}} right) = cr
& = mathop {lim }limits_{varepsilon , to ,0} Gamma left( varepsilon right)left( {left( {1 - psi left( {1/8} right)varepsilon ;} right)
- left( {1 - psi left( {7/8} right)varepsilon ;} right)} right) = cr
& = mathop {lim }limits_{varepsilon , to ,0} varepsilon ,Gamma left( varepsilon right)left( {psi left( {7/8} right) - psi left( {1/8} right)} right) = cr
& = mathop {lim }limits_{varepsilon , to ,0} ,Gamma left( {1 + varepsilon } right)left( {psi left( {7/8} right) - psi left( {1/8} right)} right) = cr
& = psi left( {7/8} right) - psi left( {1/8} right) cr}
$$
which is the same result as above.
Let me add a more straight derivation of the above.
We rewrite the integrand as
$$
eqalign{
& {{1 - y^{,3/4} } over {left( {1 - y} right)y^{,7/8} }} = {{y^{, - 7/8} - y^{, - 1/8} } over {left( {1 - y} right)}} = cr
& = {{1 - y^{, - 1/8} - left( {1 - y^{, - 7/8} } right)} over {left( {1 - y} right)}} cr}
$$
and compare with the integral representation of Digamma
$$
psi (s + 1) = - gamma + int_0^1 {{{1 - x^{,s} } over {1 - x}}dx}
$$
answered Dec 18 '18 at 16:19
G CabG Cab
20.4k31341
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add a comment |
$begingroup$
HINT: Notice that
$$frac{x^4+x^2+1}{(x^2+1)(x^4+1)}=frac{1}{4}frac{1}{x^2+xsqrt{2}+1}+frac{1}{4}frac{1}{x^2-xsqrt{2}+1}+frac{1}{2}frac{1}{x^2+1}$$
Now we just have to evaluate the integrals
$$I_1=frac{1}{4}int_0^1 frac{dx}{x^2+xsqrt{2}+1}$$
$$I_2=frac{1}{4}int_0^1 frac{dx}{x^2-xsqrt{2}+1}$$
$$I_3=frac{1}{2}int_0^1 frac{dx}{x^2+1}$$
and your sum is given by $I_1+I_2+I_3$. Can you evaluate these?
answered Dec 18 '18 at 15:26
FrpzzdFrpzzd
23k841110
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add a comment |
$begingroup$
The integral is tedious. We have, by Sophie Germain, that $x^4+1=(x^2+sqrt 2 x+1)(x^2-sqrt 2 x+1)$. Now, partial fraction and arctan gives the indefinite to be $frac{1}{4}left(2tan^{-1}x+sqrt{2}left(tan^{-1}left(sqrt{2}x+1right)-tan^{-1}left(1-sqrt{2}xright)right)right)$
answered Dec 18 '18 at 15:30
william122william122
54912
$endgroup$
add a comment |
$begingroup$
Hint: your fraction is $(x^2+1)^{-1}+frac{x^2}{x^4+1}$, and $frac{x^2}{x^4+1}=frac{ax+b}{x^2-sqrt{2}x+1}+frac{-ax-b}{x^2+sqrt{2}x+1}$ for some real numbers $a,b$.
answered Dec 18 '18 at 15:28
MindlackMindlack
4,900211
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add a comment |
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4 Answers
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active
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votes
4 Answers
4
active
oldest
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active
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active
oldest
votes
$begingroup$
Another approach, which could be useful to check yours through integrals,
is same as per the similar post, using the Digamma function $psi (z)$.
$$
eqalign{
& 1 - {1 over {8 - 1}} + {1 over {8 + 1}} - {1 over {2 cdot 8 - 1}} + {1 over {2 cdot 8 + 1}} + cdots = cr
& = 1 - sumlimits_{1, le ,n} {{1 over {8n - 1}}} + sumlimits_{1, le ,n} {{1 over {8n + 1}}} = cr
& = 1 - {1 over 8}left( {sumlimits_{1, le ,n} {{1 over {n - 1/8}}} - sumlimits_{1, le ,n} {{1 over {n + 1/8}}} } right) = cr
& = 1 - {1 over 8}left( {sumnolimits_{;n = 7/8}^{;infty } {{1 over n}} - sumnolimits_{;n = 9/8}^{;infty } {{1 over n}} } right) = cr
& = 1 - {1 over 8}left( {sumnolimits_{;n = 7/8}^{;9/8} {{1 over n}} } right) = cr
& = 1 - {1 over 8}left( {psi (9/8) - psi (7/8)} right) = cr
& = 1 - {1 over 8}left( {psi left( {1 + 1/8} right) - psi left( {1 - 1/8} right)} right) = cr
& = 1 - {1 over 8}left( {{1 over {1/8}} + psi left( {1/8} right) - psi left( {1 - 1/8} right)} right) = cr
& = {1 over 8}left( {psi left( {1 - 1/8} right) - psi left( {1/8} right)} right) = cr
& = {1 over 8}left( {pi cot left( {{pi over 8}} right)} right) = cr
& = {{pi left( {1 + sqrt 2 } right)} over 8} cr}
$$
where:
$$
Delta _{,z} psi left( z right) = psi left( {z + 1} right) - psi left( z right) = {1 over z}
$$
is the functional equation for the Digamma;
which implies that Digamma is the Antidelta of $1/z$
$$
eqalign{
& psi left( z right) = Delta _{,z} ^{ - 1} left( {{1 over z}} right) = sumnolimits_z {{1 over z}} quad Rightarrow cr
& Rightarrow quad sumnolimits_{;z = a}^{;b} {{1 over z}} = psi left( b right) - psi left( a right) cr}
$$
and we used the Reflection formula for Digamma
$$
psi left( {1 - z} right) = psi left( z right) + pi cot left( {pi z} right)
$$
Now, the above, suggests a way to solve the integral.
Let's replace $x^8$ with $y$
$$
int_{x = 0}^{,1} {{{1 - x^{,6} } over {1 - x^{,8} }}dx} quad mathop
= limits^{x^{,8} = y} quad {1 over 8}int_{y = 0}^{,1} {{{1 - y^{,3/4} } over {left( {1 - y} right)y^{,7/8} }}dy}
$$
then consider that we have
$$
eqalign{
& {{1 - y^{,3/4} } over {left( {1 - y} right)y^{,7/8} }} = cr
& = left( {left( {1 - y} right)^{ - 1} y^{, - 7/8} - left( {1 - y} right)^{ - 1} y^{, - 1/8} } right) cr
& = mathop {lim }limits_{varepsilon , to ,0} left( {left( {1 - y} right)^{varepsilon - 1} y^{,1/8 - 1}
- left( {1 - y} right)^{varepsilon - 1} y^{,,7/8 - 1} } right) cr}
$$
so we are ready to use the integral and Gamma representation for the Beta function
$$
eqalign{
& 8 ;int_{x = 0}^{,1} {{{1 - x^{,6} } over {1 - x^{,8} }}dx} = int_{y = 0}^{,1} {{{1 - y^{,3/4} } over {left( {1 - y} right)y^{,7/8} }}dy} = cr
& = mathop {lim }limits_{varepsilon , to ,0} left( {int_{y = 0}^{,1} {left( {1 - y} right)^{varepsilon - 1} y^{,1/8 - 1} dy}
- int_{y = 0}^{,1} {left( {1 - y} right)^{varepsilon - 1} y^{,,7/8 - 1} dy} } right) = cr
& = mathop {lim }limits_{varepsilon , to ,0} left( {{rm B}left( {1/8,varepsilon } right) - {rm B}left( {7/8,varepsilon } right)} right) = cr
& = mathop {lim }limits_{varepsilon , to ,0} left( {{{Gamma left( {1/8} right)Gamma left( varepsilon right)}
over {Gamma left( {1/8 + varepsilon } right)}}
- {{Gamma left( {7/8} right)Gamma left( varepsilon right)} over {Gamma left( {7/8 + varepsilon } right)}}} right) = cr
& = mathop {lim }limits_{varepsilon , to ,0} left( {{{Gamma left( {1/8} right)Gamma left( varepsilon right)}
over {psi left( {1/8} right)varepsilon ;Gamma left( {1/8} right)
+ Gamma left( {1/8} right)}} - {{Gamma left( {7/8} right)Gamma left( varepsilon right)}
over {psi left( {7/8} right)varepsilon ;Gamma left( {7/8} right) + Gamma left( {7/8} right)}}} right) = cr
& = mathop {lim }limits_{varepsilon , to ,0} left( {{{Gamma left( varepsilon right)} over {psi left( {1/8} right)varepsilon ; + 1}}
- {{Gamma left( varepsilon right)} over {psi left( {7/8} right)varepsilon ; + 1}}} right) = cr
& = mathop {lim }limits_{varepsilon , to ,0} Gamma left( varepsilon right)left( {left( {1 - psi left( {1/8} right)varepsilon ;} right)
- left( {1 - psi left( {7/8} right)varepsilon ;} right)} right) = cr
& = mathop {lim }limits_{varepsilon , to ,0} varepsilon ,Gamma left( varepsilon right)left( {psi left( {7/8} right) - psi left( {1/8} right)} right) = cr
& = mathop {lim }limits_{varepsilon , to ,0} ,Gamma left( {1 + varepsilon } right)left( {psi left( {7/8} right) - psi left( {1/8} right)} right) = cr
& = psi left( {7/8} right) - psi left( {1/8} right) cr}
$$
which is the same result as above.
Let me add a more straight derivation of the above.
We rewrite the integrand as
$$
eqalign{
& {{1 - y^{,3/4} } over {left( {1 - y} right)y^{,7/8} }} = {{y^{, - 7/8} - y^{, - 1/8} } over {left( {1 - y} right)}} = cr
& = {{1 - y^{, - 1/8} - left( {1 - y^{, - 7/8} } right)} over {left( {1 - y} right)}} cr}
$$
and compare with the integral representation of Digamma
$$
psi (s + 1) = - gamma + int_0^1 {{{1 - x^{,s} } over {1 - x}}dx}
$$
answered Dec 18 '18 at 16:19
G CabG Cab
20.4k31341
$endgroup$
add a comment |
$begingroup$
Another approach, which could be useful to check yours through integrals,
is same as per the similar post, using the Digamma function $psi (z)$.
$$
eqalign{
& 1 - {1 over {8 - 1}} + {1 over {8 + 1}} - {1 over {2 cdot 8 - 1}} + {1 over {2 cdot 8 + 1}} + cdots = cr
& = 1 - sumlimits_{1, le ,n} {{1 over {8n - 1}}} + sumlimits_{1, le ,n} {{1 over {8n + 1}}} = cr
& = 1 - {1 over 8}left( {sumlimits_{1, le ,n} {{1 over {n - 1/8}}} - sumlimits_{1, le ,n} {{1 over {n + 1/8}}} } right) = cr
& = 1 - {1 over 8}left( {sumnolimits_{;n = 7/8}^{;infty } {{1 over n}} - sumnolimits_{;n = 9/8}^{;infty } {{1 over n}} } right) = cr
& = 1 - {1 over 8}left( {sumnolimits_{;n = 7/8}^{;9/8} {{1 over n}} } right) = cr
& = 1 - {1 over 8}left( {psi (9/8) - psi (7/8)} right) = cr
& = 1 - {1 over 8}left( {psi left( {1 + 1/8} right) - psi left( {1 - 1/8} right)} right) = cr
& = 1 - {1 over 8}left( {{1 over {1/8}} + psi left( {1/8} right) - psi left( {1 - 1/8} right)} right) = cr
& = {1 over 8}left( {psi left( {1 - 1/8} right) - psi left( {1/8} right)} right) = cr
& = {1 over 8}left( {pi cot left( {{pi over 8}} right)} right) = cr
& = {{pi left( {1 + sqrt 2 } right)} over 8} cr}
$$
where:
$$
Delta _{,z} psi left( z right) = psi left( {z + 1} right) - psi left( z right) = {1 over z}
$$
is the functional equation for the Digamma;
which implies that Digamma is the Antidelta of $1/z$
$$
eqalign{
& psi left( z right) = Delta _{,z} ^{ - 1} left( {{1 over z}} right) = sumnolimits_z {{1 over z}} quad Rightarrow cr
& Rightarrow quad sumnolimits_{;z = a}^{;b} {{1 over z}} = psi left( b right) - psi left( a right) cr}
$$
and we used the Reflection formula for Digamma
$$
psi left( {1 - z} right) = psi left( z right) + pi cot left( {pi z} right)
$$
Now, the above, suggests a way to solve the integral.
Let's replace $x^8$ with $y$
$$
int_{x = 0}^{,1} {{{1 - x^{,6} } over {1 - x^{,8} }}dx} quad mathop
= limits^{x^{,8} = y} quad {1 over 8}int_{y = 0}^{,1} {{{1 - y^{,3/4} } over {left( {1 - y} right)y^{,7/8} }}dy}
$$
then consider that we have
$$
eqalign{
& {{1 - y^{,3/4} } over {left( {1 - y} right)y^{,7/8} }} = cr
& = left( {left( {1 - y} right)^{ - 1} y^{, - 7/8} - left( {1 - y} right)^{ - 1} y^{, - 1/8} } right) cr
& = mathop {lim }limits_{varepsilon , to ,0} left( {left( {1 - y} right)^{varepsilon - 1} y^{,1/8 - 1}
- left( {1 - y} right)^{varepsilon - 1} y^{,,7/8 - 1} } right) cr}
$$
so we are ready to use the integral and Gamma representation for the Beta function
$$
eqalign{
& 8 ;int_{x = 0}^{,1} {{{1 - x^{,6} } over {1 - x^{,8} }}dx} = int_{y = 0}^{,1} {{{1 - y^{,3/4} } over {left( {1 - y} right)y^{,7/8} }}dy} = cr
& = mathop {lim }limits_{varepsilon , to ,0} left( {int_{y = 0}^{,1} {left( {1 - y} right)^{varepsilon - 1} y^{,1/8 - 1} dy}
- int_{y = 0}^{,1} {left( {1 - y} right)^{varepsilon - 1} y^{,,7/8 - 1} dy} } right) = cr
& = mathop {lim }limits_{varepsilon , to ,0} left( {{rm B}left( {1/8,varepsilon } right) - {rm B}left( {7/8,varepsilon } right)} right) = cr
& = mathop {lim }limits_{varepsilon , to ,0} left( {{{Gamma left( {1/8} right)Gamma left( varepsilon right)}
over {Gamma left( {1/8 + varepsilon } right)}}
- {{Gamma left( {7/8} right)Gamma left( varepsilon right)} over {Gamma left( {7/8 + varepsilon } right)}}} right) = cr
& = mathop {lim }limits_{varepsilon , to ,0} left( {{{Gamma left( {1/8} right)Gamma left( varepsilon right)}
over {psi left( {1/8} right)varepsilon ;Gamma left( {1/8} right)
+ Gamma left( {1/8} right)}} - {{Gamma left( {7/8} right)Gamma left( varepsilon right)}
over {psi left( {7/8} right)varepsilon ;Gamma left( {7/8} right) + Gamma left( {7/8} right)}}} right) = cr
& = mathop {lim }limits_{varepsilon , to ,0} left( {{{Gamma left( varepsilon right)} over {psi left( {1/8} right)varepsilon ; + 1}}
- {{Gamma left( varepsilon right)} over {psi left( {7/8} right)varepsilon ; + 1}}} right) = cr
& = mathop {lim }limits_{varepsilon , to ,0} Gamma left( varepsilon right)left( {left( {1 - psi left( {1/8} right)varepsilon ;} right)
- left( {1 - psi left( {7/8} right)varepsilon ;} right)} right) = cr
& = mathop {lim }limits_{varepsilon , to ,0} varepsilon ,Gamma left( varepsilon right)left( {psi left( {7/8} right) - psi left( {1/8} right)} right) = cr
& = mathop {lim }limits_{varepsilon , to ,0} ,Gamma left( {1 + varepsilon } right)left( {psi left( {7/8} right) - psi left( {1/8} right)} right) = cr
& = psi left( {7/8} right) - psi left( {1/8} right) cr}
$$
which is the same result as above.
Let me add a more straight derivation of the above.
We rewrite the integrand as
$$
eqalign{
& {{1 - y^{,3/4} } over {left( {1 - y} right)y^{,7/8} }} = {{y^{, - 7/8} - y^{, - 1/8} } over {left( {1 - y} right)}} = cr
& = {{1 - y^{, - 1/8} - left( {1 - y^{, - 7/8} } right)} over {left( {1 - y} right)}} cr}
$$
and compare with the integral representation of Digamma
$$
psi (s + 1) = - gamma + int_0^1 {{{1 - x^{,s} } over {1 - x}}dx}
$$
answered Dec 18 '18 at 16:19
G CabG Cab
20.4k31341
$endgroup$
add a comment |
$begingroup$
Another approach, which could be useful to check yours through integrals,
is same as per the similar post, using the Digamma function $psi (z)$.
$$
eqalign{
& 1 - {1 over {8 - 1}} + {1 over {8 + 1}} - {1 over {2 cdot 8 - 1}} + {1 over {2 cdot 8 + 1}} + cdots = cr
& = 1 - sumlimits_{1, le ,n} {{1 over {8n - 1}}} + sumlimits_{1, le ,n} {{1 over {8n + 1}}} = cr
& = 1 - {1 over 8}left( {sumlimits_{1, le ,n} {{1 over {n - 1/8}}} - sumlimits_{1, le ,n} {{1 over {n + 1/8}}} } right) = cr
& = 1 - {1 over 8}left( {sumnolimits_{;n = 7/8}^{;infty } {{1 over n}} - sumnolimits_{;n = 9/8}^{;infty } {{1 over n}} } right) = cr
& = 1 - {1 over 8}left( {sumnolimits_{;n = 7/8}^{;9/8} {{1 over n}} } right) = cr
& = 1 - {1 over 8}left( {psi (9/8) - psi (7/8)} right) = cr
& = 1 - {1 over 8}left( {psi left( {1 + 1/8} right) - psi left( {1 - 1/8} right)} right) = cr
& = 1 - {1 over 8}left( {{1 over {1/8}} + psi left( {1/8} right) - psi left( {1 - 1/8} right)} right) = cr
& = {1 over 8}left( {psi left( {1 - 1/8} right) - psi left( {1/8} right)} right) = cr
& = {1 over 8}left( {pi cot left( {{pi over 8}} right)} right) = cr
& = {{pi left( {1 + sqrt 2 } right)} over 8} cr}
$$
where:
$$
Delta _{,z} psi left( z right) = psi left( {z + 1} right) - psi left( z right) = {1 over z}
$$
is the functional equation for the Digamma;
which implies that Digamma is the Antidelta of $1/z$
$$
eqalign{
& psi left( z right) = Delta _{,z} ^{ - 1} left( {{1 over z}} right) = sumnolimits_z {{1 over z}} quad Rightarrow cr
& Rightarrow quad sumnolimits_{;z = a}^{;b} {{1 over z}} = psi left( b right) - psi left( a right) cr}
$$
and we used the Reflection formula for Digamma
$$
psi left( {1 - z} right) = psi left( z right) + pi cot left( {pi z} right)
$$
Now, the above, suggests a way to solve the integral.
Let's replace $x^8$ with $y$
$$
int_{x = 0}^{,1} {{{1 - x^{,6} } over {1 - x^{,8} }}dx} quad mathop
= limits^{x^{,8} = y} quad {1 over 8}int_{y = 0}^{,1} {{{1 - y^{,3/4} } over {left( {1 - y} right)y^{,7/8} }}dy}
$$
then consider that we have
$$
eqalign{
& {{1 - y^{,3/4} } over {left( {1 - y} right)y^{,7/8} }} = cr
& = left( {left( {1 - y} right)^{ - 1} y^{, - 7/8} - left( {1 - y} right)^{ - 1} y^{, - 1/8} } right) cr
& = mathop {lim }limits_{varepsilon , to ,0} left( {left( {1 - y} right)^{varepsilon - 1} y^{,1/8 - 1}
- left( {1 - y} right)^{varepsilon - 1} y^{,,7/8 - 1} } right) cr}
$$
so we are ready to use the integral and Gamma representation for the Beta function
$$
eqalign{
& 8 ;int_{x = 0}^{,1} {{{1 - x^{,6} } over {1 - x^{,8} }}dx} = int_{y = 0}^{,1} {{{1 - y^{,3/4} } over {left( {1 - y} right)y^{,7/8} }}dy} = cr
& = mathop {lim }limits_{varepsilon , to ,0} left( {int_{y = 0}^{,1} {left( {1 - y} right)^{varepsilon - 1} y^{,1/8 - 1} dy}
- int_{y = 0}^{,1} {left( {1 - y} right)^{varepsilon - 1} y^{,,7/8 - 1} dy} } right) = cr
& = mathop {lim }limits_{varepsilon , to ,0} left( {{rm B}left( {1/8,varepsilon } right) - {rm B}left( {7/8,varepsilon } right)} right) = cr
& = mathop {lim }limits_{varepsilon , to ,0} left( {{{Gamma left( {1/8} right)Gamma left( varepsilon right)}
over {Gamma left( {1/8 + varepsilon } right)}}
- {{Gamma left( {7/8} right)Gamma left( varepsilon right)} over {Gamma left( {7/8 + varepsilon } right)}}} right) = cr
& = mathop {lim }limits_{varepsilon , to ,0} left( {{{Gamma left( {1/8} right)Gamma left( varepsilon right)}
over {psi left( {1/8} right)varepsilon ;Gamma left( {1/8} right)
+ Gamma left( {1/8} right)}} - {{Gamma left( {7/8} right)Gamma left( varepsilon right)}
over {psi left( {7/8} right)varepsilon ;Gamma left( {7/8} right) + Gamma left( {7/8} right)}}} right) = cr
& = mathop {lim }limits_{varepsilon , to ,0} left( {{{Gamma left( varepsilon right)} over {psi left( {1/8} right)varepsilon ; + 1}}
- {{Gamma left( varepsilon right)} over {psi left( {7/8} right)varepsilon ; + 1}}} right) = cr
& = mathop {lim }limits_{varepsilon , to ,0} Gamma left( varepsilon right)left( {left( {1 - psi left( {1/8} right)varepsilon ;} right)
- left( {1 - psi left( {7/8} right)varepsilon ;} right)} right) = cr
& = mathop {lim }limits_{varepsilon , to ,0} varepsilon ,Gamma left( varepsilon right)left( {psi left( {7/8} right) - psi left( {1/8} right)} right) = cr
& = mathop {lim }limits_{varepsilon , to ,0} ,Gamma left( {1 + varepsilon } right)left( {psi left( {7/8} right) - psi left( {1/8} right)} right) = cr
& = psi left( {7/8} right) - psi left( {1/8} right) cr}
$$
which is the same result as above.
Let me add a more straight derivation of the above.
We rewrite the integrand as
$$
eqalign{
& {{1 - y^{,3/4} } over {left( {1 - y} right)y^{,7/8} }} = {{y^{, - 7/8} - y^{, - 1/8} } over {left( {1 - y} right)}} = cr
& = {{1 - y^{, - 1/8} - left( {1 - y^{, - 7/8} } right)} over {left( {1 - y} right)}} cr}
$$
and compare with the integral representation of Digamma
$$
psi (s + 1) = - gamma + int_0^1 {{{1 - x^{,s} } over {1 - x}}dx}
$$
answered Dec 18 '18 at 16:19
G CabG Cab
20.4k31341
$endgroup$
Another approach, which could be useful to check yours through integrals,
is same as per the similar post, using the Digamma function $psi (z)$.
$$
eqalign{
& 1 - {1 over {8 - 1}} + {1 over {8 + 1}} - {1 over {2 cdot 8 - 1}} + {1 over {2 cdot 8 + 1}} + cdots = cr
& = 1 - sumlimits_{1, le ,n} {{1 over {8n - 1}}} + sumlimits_{1, le ,n} {{1 over {8n + 1}}} = cr
& = 1 - {1 over 8}left( {sumlimits_{1, le ,n} {{1 over {n - 1/8}}} - sumlimits_{1, le ,n} {{1 over {n + 1/8}}} } right) = cr
& = 1 - {1 over 8}left( {sumnolimits_{;n = 7/8}^{;infty } {{1 over n}} - sumnolimits_{;n = 9/8}^{;infty } {{1 over n}} } right) = cr
& = 1 - {1 over 8}left( {sumnolimits_{;n = 7/8}^{;9/8} {{1 over n}} } right) = cr
& = 1 - {1 over 8}left( {psi (9/8) - psi (7/8)} right) = cr
& = 1 - {1 over 8}left( {psi left( {1 + 1/8} right) - psi left( {1 - 1/8} right)} right) = cr
& = 1 - {1 over 8}left( {{1 over {1/8}} + psi left( {1/8} right) - psi left( {1 - 1/8} right)} right) = cr
& = {1 over 8}left( {psi left( {1 - 1/8} right) - psi left( {1/8} right)} right) = cr
& = {1 over 8}left( {pi cot left( {{pi over 8}} right)} right) = cr
& = {{pi left( {1 + sqrt 2 } right)} over 8} cr}
$$
where:
$$
Delta _{,z} psi left( z right) = psi left( {z + 1} right) - psi left( z right) = {1 over z}
$$
is the functional equation for the Digamma;
which implies that Digamma is the Antidelta of $1/z$
$$
eqalign{
& psi left( z right) = Delta _{,z} ^{ - 1} left( {{1 over z}} right) = sumnolimits_z {{1 over z}} quad Rightarrow cr
& Rightarrow quad sumnolimits_{;z = a}^{;b} {{1 over z}} = psi left( b right) - psi left( a right) cr}
$$
and we used the Reflection formula for Digamma
$$
psi left( {1 - z} right) = psi left( z right) + pi cot left( {pi z} right)
$$
Now, the above, suggests a way to solve the integral.
Let's replace $x^8$ with $y$
$$
int_{x = 0}^{,1} {{{1 - x^{,6} } over {1 - x^{,8} }}dx} quad mathop
= limits^{x^{,8} = y} quad {1 over 8}int_{y = 0}^{,1} {{{1 - y^{,3/4} } over {left( {1 - y} right)y^{,7/8} }}dy}
$$
then consider that we have
$$
eqalign{
& {{1 - y^{,3/4} } over {left( {1 - y} right)y^{,7/8} }} = cr
& = left( {left( {1 - y} right)^{ - 1} y^{, - 7/8} - left( {1 - y} right)^{ - 1} y^{, - 1/8} } right) cr
& = mathop {lim }limits_{varepsilon , to ,0} left( {left( {1 - y} right)^{varepsilon - 1} y^{,1/8 - 1}
- left( {1 - y} right)^{varepsilon - 1} y^{,,7/8 - 1} } right) cr}
$$
so we are ready to use the integral and Gamma representation for the Beta function
$$
eqalign{
& 8 ;int_{x = 0}^{,1} {{{1 - x^{,6} } over {1 - x^{,8} }}dx} = int_{y = 0}^{,1} {{{1 - y^{,3/4} } over {left( {1 - y} right)y^{,7/8} }}dy} = cr
& = mathop {lim }limits_{varepsilon , to ,0} left( {int_{y = 0}^{,1} {left( {1 - y} right)^{varepsilon - 1} y^{,1/8 - 1} dy}
- int_{y = 0}^{,1} {left( {1 - y} right)^{varepsilon - 1} y^{,,7/8 - 1} dy} } right) = cr
& = mathop {lim }limits_{varepsilon , to ,0} left( {{rm B}left( {1/8,varepsilon } right) - {rm B}left( {7/8,varepsilon } right)} right) = cr
& = mathop {lim }limits_{varepsilon , to ,0} left( {{{Gamma left( {1/8} right)Gamma left( varepsilon right)}
over {Gamma left( {1/8 + varepsilon } right)}}
- {{Gamma left( {7/8} right)Gamma left( varepsilon right)} over {Gamma left( {7/8 + varepsilon } right)}}} right) = cr
& = mathop {lim }limits_{varepsilon , to ,0} left( {{{Gamma left( {1/8} right)Gamma left( varepsilon right)}
over {psi left( {1/8} right)varepsilon ;Gamma left( {1/8} right)
+ Gamma left( {1/8} right)}} - {{Gamma left( {7/8} right)Gamma left( varepsilon right)}
over {psi left( {7/8} right)varepsilon ;Gamma left( {7/8} right) + Gamma left( {7/8} right)}}} right) = cr
& = mathop {lim }limits_{varepsilon , to ,0} left( {{{Gamma left( varepsilon right)} over {psi left( {1/8} right)varepsilon ; + 1}}
- {{Gamma left( varepsilon right)} over {psi left( {7/8} right)varepsilon ; + 1}}} right) = cr
& = mathop {lim }limits_{varepsilon , to ,0} Gamma left( varepsilon right)left( {left( {1 - psi left( {1/8} right)varepsilon ;} right)
- left( {1 - psi left( {7/8} right)varepsilon ;} right)} right) = cr
& = mathop {lim }limits_{varepsilon , to ,0} varepsilon ,Gamma left( varepsilon right)left( {psi left( {7/8} right) - psi left( {1/8} right)} right) = cr
& = mathop {lim }limits_{varepsilon , to ,0} ,Gamma left( {1 + varepsilon } right)left( {psi left( {7/8} right) - psi left( {1/8} right)} right) = cr
& = psi left( {7/8} right) - psi left( {1/8} right) cr}
$$
which is the same result as above.
Let me add a more straight derivation of the above.
We rewrite the integrand as
$$
eqalign{
& {{1 - y^{,3/4} } over {left( {1 - y} right)y^{,7/8} }} = {{y^{, - 7/8} - y^{, - 1/8} } over {left( {1 - y} right)}} = cr
& = {{1 - y^{, - 1/8} - left( {1 - y^{, - 7/8} } right)} over {left( {1 - y} right)}} cr}
$$
and compare with the integral representation of Digamma
$$
psi (s + 1) = - gamma + int_0^1 {{{1 - x^{,s} } over {1 - x}}dx}
$$
answered Dec 18 '18 at 16:19
G CabG Cab
20.4k31341
edited Dec 18 '18 at 22:24
answered Dec 18 '18 at 16:19
G CabG Cab
20.4k31341
answered Dec 18 '18 at 16:19
G CabG Cab
20.4k31341
answered Dec 18 '18 at 16:19
G CabG Cab
20.4k31341
20.4k31341
add a comment |
add a comment |
$begingroup$
HINT: Notice that
$$frac{x^4+x^2+1}{(x^2+1)(x^4+1)}=frac{1}{4}frac{1}{x^2+xsqrt{2}+1}+frac{1}{4}frac{1}{x^2-xsqrt{2}+1}+frac{1}{2}frac{1}{x^2+1}$$
Now we just have to evaluate the integrals
$$I_1=frac{1}{4}int_0^1 frac{dx}{x^2+xsqrt{2}+1}$$
$$I_2=frac{1}{4}int_0^1 frac{dx}{x^2-xsqrt{2}+1}$$
$$I_3=frac{1}{2}int_0^1 frac{dx}{x^2+1}$$
and your sum is given by $I_1+I_2+I_3$. Can you evaluate these?
answered Dec 18 '18 at 15:26
FrpzzdFrpzzd
23k841110
$endgroup$
add a comment |
$begingroup$
HINT: Notice that
$$frac{x^4+x^2+1}{(x^2+1)(x^4+1)}=frac{1}{4}frac{1}{x^2+xsqrt{2}+1}+frac{1}{4}frac{1}{x^2-xsqrt{2}+1}+frac{1}{2}frac{1}{x^2+1}$$
Now we just have to evaluate the integrals
$$I_1=frac{1}{4}int_0^1 frac{dx}{x^2+xsqrt{2}+1}$$
$$I_2=frac{1}{4}int_0^1 frac{dx}{x^2-xsqrt{2}+1}$$
$$I_3=frac{1}{2}int_0^1 frac{dx}{x^2+1}$$
and your sum is given by $I_1+I_2+I_3$. Can you evaluate these?
answered Dec 18 '18 at 15:26
FrpzzdFrpzzd
23k841110
$endgroup$
add a comment |
$begingroup$
HINT: Notice that
$$frac{x^4+x^2+1}{(x^2+1)(x^4+1)}=frac{1}{4}frac{1}{x^2+xsqrt{2}+1}+frac{1}{4}frac{1}{x^2-xsqrt{2}+1}+frac{1}{2}frac{1}{x^2+1}$$
Now we just have to evaluate the integrals
$$I_1=frac{1}{4}int_0^1 frac{dx}{x^2+xsqrt{2}+1}$$
$$I_2=frac{1}{4}int_0^1 frac{dx}{x^2-xsqrt{2}+1}$$
$$I_3=frac{1}{2}int_0^1 frac{dx}{x^2+1}$$
and your sum is given by $I_1+I_2+I_3$. Can you evaluate these?
answered Dec 18 '18 at 15:26
FrpzzdFrpzzd
23k841110
$endgroup$
HINT: Notice that
$$frac{x^4+x^2+1}{(x^2+1)(x^4+1)}=frac{1}{4}frac{1}{x^2+xsqrt{2}+1}+frac{1}{4}frac{1}{x^2-xsqrt{2}+1}+frac{1}{2}frac{1}{x^2+1}$$
Now we just have to evaluate the integrals
$$I_1=frac{1}{4}int_0^1 frac{dx}{x^2+xsqrt{2}+1}$$
$$I_2=frac{1}{4}int_0^1 frac{dx}{x^2-xsqrt{2}+1}$$
$$I_3=frac{1}{2}int_0^1 frac{dx}{x^2+1}$$
and your sum is given by $I_1+I_2+I_3$. Can you evaluate these?
answered Dec 18 '18 at 15:26
FrpzzdFrpzzd
23k841110
answered Dec 18 '18 at 15:26
FrpzzdFrpzzd
23k841110
answered Dec 18 '18 at 15:26
FrpzzdFrpzzd
23k841110
answered Dec 18 '18 at 15:26
FrpzzdFrpzzd
23k841110
23k841110
add a comment |
add a comment |
$begingroup$
The integral is tedious. We have, by Sophie Germain, that $x^4+1=(x^2+sqrt 2 x+1)(x^2-sqrt 2 x+1)$. Now, partial fraction and arctan gives the indefinite to be $frac{1}{4}left(2tan^{-1}x+sqrt{2}left(tan^{-1}left(sqrt{2}x+1right)-tan^{-1}left(1-sqrt{2}xright)right)right)$
answered Dec 18 '18 at 15:30
william122william122
54912
$endgroup$
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$begingroup$
The integral is tedious. We have, by Sophie Germain, that $x^4+1=(x^2+sqrt 2 x+1)(x^2-sqrt 2 x+1)$. Now, partial fraction and arctan gives the indefinite to be $frac{1}{4}left(2tan^{-1}x+sqrt{2}left(tan^{-1}left(sqrt{2}x+1right)-tan^{-1}left(1-sqrt{2}xright)right)right)$
answered Dec 18 '18 at 15:30
william122william122
54912
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The integral is tedious. We have, by Sophie Germain, that $x^4+1=(x^2+sqrt 2 x+1)(x^2-sqrt 2 x+1)$. Now, partial fraction and arctan gives the indefinite to be $frac{1}{4}left(2tan^{-1}x+sqrt{2}left(tan^{-1}left(sqrt{2}x+1right)-tan^{-1}left(1-sqrt{2}xright)right)right)$
answered Dec 18 '18 at 15:30
william122william122
54912
$endgroup$
The integral is tedious. We have, by Sophie Germain, that $x^4+1=(x^2+sqrt 2 x+1)(x^2-sqrt 2 x+1)$. Now, partial fraction and arctan gives the indefinite to be $frac{1}{4}left(2tan^{-1}x+sqrt{2}left(tan^{-1}left(sqrt{2}x+1right)-tan^{-1}left(1-sqrt{2}xright)right)right)$
answered Dec 18 '18 at 15:30
william122william122
54912
answered Dec 18 '18 at 15:30
william122william122
54912
answered Dec 18 '18 at 15:30
william122william122
54912
answered Dec 18 '18 at 15:30
william122william122
54912
54912
add a comment |
add a comment |
$begingroup$
Hint: your fraction is $(x^2+1)^{-1}+frac{x^2}{x^4+1}$, and $frac{x^2}{x^4+1}=frac{ax+b}{x^2-sqrt{2}x+1}+frac{-ax-b}{x^2+sqrt{2}x+1}$ for some real numbers $a,b$.
answered Dec 18 '18 at 15:28
MindlackMindlack
4,900211
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add a comment |
$begingroup$
Hint: your fraction is $(x^2+1)^{-1}+frac{x^2}{x^4+1}$, and $frac{x^2}{x^4+1}=frac{ax+b}{x^2-sqrt{2}x+1}+frac{-ax-b}{x^2+sqrt{2}x+1}$ for some real numbers $a,b$.
answered Dec 18 '18 at 15:28
MindlackMindlack
4,900211
$endgroup$
add a comment |
$begingroup$
Hint: your fraction is $(x^2+1)^{-1}+frac{x^2}{x^4+1}$, and $frac{x^2}{x^4+1}=frac{ax+b}{x^2-sqrt{2}x+1}+frac{-ax-b}{x^2+sqrt{2}x+1}$ for some real numbers $a,b$.
answered Dec 18 '18 at 15:28
MindlackMindlack
4,900211
$endgroup$
Hint: your fraction is $(x^2+1)^{-1}+frac{x^2}{x^4+1}$, and $frac{x^2}{x^4+1}=frac{ax+b}{x^2-sqrt{2}x+1}+frac{-ax-b}{x^2+sqrt{2}x+1}$ for some real numbers $a,b$.
answered Dec 18 '18 at 15:28
MindlackMindlack
4,900211
answered Dec 18 '18 at 15:28
MindlackMindlack
4,900211
answered Dec 18 '18 at 15:28
MindlackMindlack
4,900211
answered Dec 18 '18 at 15:28
MindlackMindlack
4,900211
4,900211
add a comment |
add a comment |
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Hint: Partial fraction decomposition ($x^4+1$ can be factored over the reals).
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– Michael Burr
Dec 18 '18 at 15:24
2
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Replace $x^2=y$ and use mathworld.wolfram.com/PartialFractionDecomposition.html and replace back $x$
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– lab bhattacharjee
Dec 18 '18 at 15:25
1
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See also math.stackexchange.com/questions/2363639/int-fracx2x4x21-dx/…
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– lab bhattacharjee
Dec 18 '18 at 15:26
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Hey, is this a problem from Joseph edwards integral calculus for beginners? I remember solving this. The answer should equal something like $frac{pi}{8}(1+sqrt{2)}$ or something.
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– crskhr
Dec 18 '18 at 15:28
1
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@Stockfish: Absolutely convergent series.
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– Clayton
Dec 18 '18 at 15:32