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How to derive $int_0^1 frac{sin(pi x)}{x^3-1}dx=frac19coshleft(frac{sqrt{3}pi}{2}right)$?
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I am interested in the integral$$int_0^1 frac{sin(pi x)}{x^3-1}dx=frac19coshleft(frac{sqrt{3}pi}{2}right)$$
I thought about approaching this by expanding $sin(pi x)$ into its taylor series:
$$I=int_0^1 frac{sin(pi x)}{x^3-1}dx=sum_{n=0}^infty frac{(-1)^n pi^{2n+1}}{(2n+1)!}int_0^1 frac{x^{2n+1}}{x^3 -1}dx$$
However I'm not sure where to continue from here. I also considered the general integral $$I(a)=int_0^1 frac{sin(ax)}{x^3-1}dx$$ however Wolfram Alpha says that the general case is divergent so I am not sure that is the right approach. If anyone can help me with this integral I would be very grateful.
EDIT: The result given by Wolfram Alpha is wrong and as such I can wager to say that this integral probably does not have a closed form. Feel free to prove me wrong though.
definite-integrals
asked Feb 1 at 0:07
aledenaleden
2,5351511
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|
show 6 more comments
$begingroup$
I am interested in the integral$$int_0^1 frac{sin(pi x)}{x^3-1}dx=frac19coshleft(frac{sqrt{3}pi}{2}right)$$
I thought about approaching this by expanding $sin(pi x)$ into its taylor series:
$$I=int_0^1 frac{sin(pi x)}{x^3-1}dx=sum_{n=0}^infty frac{(-1)^n pi^{2n+1}}{(2n+1)!}int_0^1 frac{x^{2n+1}}{x^3 -1}dx$$
However I'm not sure where to continue from here. I also considered the general integral $$I(a)=int_0^1 frac{sin(ax)}{x^3-1}dx$$ however Wolfram Alpha says that the general case is divergent so I am not sure that is the right approach. If anyone can help me with this integral I would be very grateful.
EDIT: The result given by Wolfram Alpha is wrong and as such I can wager to say that this integral probably does not have a closed form. Feel free to prove me wrong though.
definite-integrals
asked Feb 1 at 0:07
aledenaleden
2,5351511
$endgroup$
2
$begingroup$
Have you tried expanding the term $frac 1{x^3-1}=-sum_{n=0}^infty x^{3n}$?
$endgroup$
– Mark Viola
Feb 1 at 0:32
1
$begingroup$
Yeah the series for sine won't work
$endgroup$
– clathratus
Feb 1 at 0:35
1
$begingroup$
@MarkViola I have but I am left with a very ugly expression involving a sum of incomplete gamma functions.
$endgroup$
– aleden
Feb 1 at 0:41
3
$begingroup$
How did you obtain that result? Are you sure that is correct?
$endgroup$
– Zacky
Feb 1 at 0:51
1
$begingroup$
@Zacky WA gives the result but every other CAS puts the integral at $−0.924004640359...$ which is very different from the WA result. The decimal expansion seems to be accurate however, because the integrand is purely negative on $[0,1)$... Very strange indeed
$endgroup$
– clathratus
Feb 1 at 1:01
|
show 6 more comments
$begingroup$
I am interested in the integral$$int_0^1 frac{sin(pi x)}{x^3-1}dx=frac19coshleft(frac{sqrt{3}pi}{2}right)$$
I thought about approaching this by expanding $sin(pi x)$ into its taylor series:
$$I=int_0^1 frac{sin(pi x)}{x^3-1}dx=sum_{n=0}^infty frac{(-1)^n pi^{2n+1}}{(2n+1)!}int_0^1 frac{x^{2n+1}}{x^3 -1}dx$$
However I'm not sure where to continue from here. I also considered the general integral $$I(a)=int_0^1 frac{sin(ax)}{x^3-1}dx$$ however Wolfram Alpha says that the general case is divergent so I am not sure that is the right approach. If anyone can help me with this integral I would be very grateful.
EDIT: The result given by Wolfram Alpha is wrong and as such I can wager to say that this integral probably does not have a closed form. Feel free to prove me wrong though.
definite-integrals
asked Feb 1 at 0:07
aledenaleden
2,5351511
$endgroup$
I am interested in the integral$$int_0^1 frac{sin(pi x)}{x^3-1}dx=frac19coshleft(frac{sqrt{3}pi}{2}right)$$
I thought about approaching this by expanding $sin(pi x)$ into its taylor series:
$$I=int_0^1 frac{sin(pi x)}{x^3-1}dx=sum_{n=0}^infty frac{(-1)^n pi^{2n+1}}{(2n+1)!}int_0^1 frac{x^{2n+1}}{x^3 -1}dx$$
However I'm not sure where to continue from here. I also considered the general integral $$I(a)=int_0^1 frac{sin(ax)}{x^3-1}dx$$ however Wolfram Alpha says that the general case is divergent so I am not sure that is the right approach. If anyone can help me with this integral I would be very grateful.
EDIT: The result given by Wolfram Alpha is wrong and as such I can wager to say that this integral probably does not have a closed form. Feel free to prove me wrong though.
definite-integrals
definite-integrals
asked Feb 1 at 0:07
aledenaleden
2,5351511
asked Feb 1 at 0:07
aledenaleden
2,5351511
edited Feb 1 at 1:12
aleden
asked Feb 1 at 0:07
aledenaleden
2,5351511
asked Feb 1 at 0:07
aledenaleden
2,5351511
asked Feb 1 at 0:07
aledenaleden
2,5351511
2,5351511
2
$begingroup$
Have you tried expanding the term $frac 1{x^3-1}=-sum_{n=0}^infty x^{3n}$?
$endgroup$
– Mark Viola
Feb 1 at 0:32
1
$begingroup$
Yeah the series for sine won't work
$endgroup$
– clathratus
Feb 1 at 0:35
1
$begingroup$
@MarkViola I have but I am left with a very ugly expression involving a sum of incomplete gamma functions.
$endgroup$
– aleden
Feb 1 at 0:41
3
$begingroup$
How did you obtain that result? Are you sure that is correct?
$endgroup$
– Zacky
Feb 1 at 0:51
1
$begingroup$
@Zacky WA gives the result but every other CAS puts the integral at $−0.924004640359...$ which is very different from the WA result. The decimal expansion seems to be accurate however, because the integrand is purely negative on $[0,1)$... Very strange indeed
$endgroup$
– clathratus
Feb 1 at 1:01
|
show 6 more comments
2
$begingroup$
Have you tried expanding the term $frac 1{x^3-1}=-sum_{n=0}^infty x^{3n}$?
$endgroup$
– Mark Viola
Feb 1 at 0:32
1
$begingroup$
Yeah the series for sine won't work
$endgroup$
– clathratus
Feb 1 at 0:35
1
$begingroup$
@MarkViola I have but I am left with a very ugly expression involving a sum of incomplete gamma functions.
$endgroup$
– aleden
Feb 1 at 0:41
3
$begingroup$
How did you obtain that result? Are you sure that is correct?
$endgroup$
– Zacky
Feb 1 at 0:51
1
$begingroup$
@Zacky WA gives the result but every other CAS puts the integral at $−0.924004640359...$ which is very different from the WA result. The decimal expansion seems to be accurate however, because the integrand is purely negative on $[0,1)$... Very strange indeed
$endgroup$
– clathratus
Feb 1 at 1:01
2
2
$begingroup$
Have you tried expanding the term $frac 1{x^3-1}=-sum_{n=0}^infty x^{3n}$?
$endgroup$
– Mark Viola
Feb 1 at 0:32
$begingroup$
Have you tried expanding the term $frac 1{x^3-1}=-sum_{n=0}^infty x^{3n}$?
$endgroup$
– Mark Viola
Feb 1 at 0:32
1
1
$begingroup$
Yeah the series for sine won't work
$endgroup$
– clathratus
Feb 1 at 0:35
$begingroup$
Yeah the series for sine won't work
$endgroup$
– clathratus
Feb 1 at 0:35
1
1
$begingroup$
@MarkViola I have but I am left with a very ugly expression involving a sum of incomplete gamma functions.
$endgroup$
– aleden
Feb 1 at 0:41
$begingroup$
@MarkViola I have but I am left with a very ugly expression involving a sum of incomplete gamma functions.
$endgroup$
– aleden
Feb 1 at 0:41
3
3
$begingroup$
How did you obtain that result? Are you sure that is correct?
$endgroup$
– Zacky
Feb 1 at 0:51
$begingroup$
How did you obtain that result? Are you sure that is correct?
$endgroup$
– Zacky
Feb 1 at 0:51
1
1
$begingroup$
@Zacky WA gives the result but every other CAS puts the integral at $−0.924004640359...$ which is very different from the WA result. The decimal expansion seems to be accurate however, because the integrand is purely negative on $[0,1)$... Very strange indeed
$endgroup$
– clathratus
Feb 1 at 1:01
$begingroup$
@Zacky WA gives the result but every other CAS puts the integral at $−0.924004640359...$ which is very different from the WA result. The decimal expansion seems to be accurate however, because the integrand is purely negative on $[0,1)$... Very strange indeed
$endgroup$
– clathratus
Feb 1 at 1:01
|
show 6 more comments
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2
$begingroup$
Have you tried expanding the term $frac 1{x^3-1}=-sum_{n=0}^infty x^{3n}$?
$endgroup$
– Mark Viola
Feb 1 at 0:32
1
$begingroup$
Yeah the series for sine won't work
$endgroup$
– clathratus
Feb 1 at 0:35
1
$begingroup$
@MarkViola I have but I am left with a very ugly expression involving a sum of incomplete gamma functions.
$endgroup$
– aleden
Feb 1 at 0:41
3
$begingroup$
How did you obtain that result? Are you sure that is correct?
$endgroup$
– Zacky
Feb 1 at 0:51
1
$begingroup$
@Zacky WA gives the result but every other CAS puts the integral at $−0.924004640359...$ which is very different from the WA result. The decimal expansion seems to be accurate however, because the integrand is purely negative on $[0,1)$... Very strange indeed
$endgroup$
– clathratus
Feb 1 at 1:01