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I am interested in the integral$$\int_0^1 \frac{\sin(\pi x)}{x^3-1}dx=\frac19\cosh\left(\frac{\sqrt{3}\pi}{2}\right)$$

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.

aleden
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    Have you tried expanding the term $\frac 1{x^3-1}=-\sum_{n=0}^\infty x^{3n}$? – Mark Viola Feb 01 '19 at 00:32
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    Yeah the series for sine won't work – clathratus Feb 01 '19 at 00:35
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    @MarkViola I have but I am left with a very ugly expression involving a sum of incomplete gamma functions. – aleden Feb 01 '19 at 00:41
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    How did you obtain that result? Are you sure that is correct? – Zacky Feb 01 '19 at 00:51
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    @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 – clathratus Feb 01 '19 at 01:01
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    @zacky You're correct. That can't be the correct answer. The LHS is negative and the RHS is positive. – Mark Viola Feb 01 '19 at 01:05
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    @Zacky I used Wolphram but apparently its wrong? – aleden Feb 01 '19 at 01:07
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    Yes the result given seems to be wrong.. I also believe that when we combine a trig function with a polynomial and the bounds aren't $\infty$ we don't get a nice result. Using partial fractions and relying on the sine integral might work, but it won't be nice. – Zacky Feb 01 '19 at 01:07
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    @Zacky Agreed, I was excited because I thought that this integral had surprisingly simple closed form, but WA has let me down. – aleden Feb 01 '19 at 01:09
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    The right side is about $0.8475...$ – FDP Feb 01 '19 at 15:47

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