Question: If $a$ is in an arbitrary commensurable ratio to $\pi$, that is $a=\tfrac {m\pi}n$, then if $m+n$ is odd$$\int\limits_0^1\mathrm dy\,\frac {y^x\sin a}{1+2y\cos a+y^2}=\frac 12\sum\limits_{i=1}^{n-1}(-1)^{i-1}\sin ia\left[\frac {\mathrm d\Gamma\left(\frac {x+n+i}{2n}\right)}{\mathrm dx}-\frac {\mathrm d\Gamma\left(\frac {x+i}{2n}\right)}{\mathrm dx}\right]$$and when $m+n$ is even$$\int\limits_0^1\mathrm dy\,\frac {y^x\sin a}{1+2y\cos a+y^2}=\frac 12\sum\limits_{i=1}^{(n-1)/2}(-1)^{i-1}\sin ia\left[\frac {\mathrm d\Gamma\left(\frac {x+n+i}n\right)}{\mathrm dx}-\frac {\mathrm d\Gamma\left(\frac {x+i}n\right)}{\mathrm dx}\right]$$

I’m just having difficulty finding out where to start. Since the integral equals an infinite sum, it might be wise to start off with a taylor expansion of some sort. However, which function to expand I’m not very sure.

If you guys have any idea, I would be happy to hear them. Thanks!


Hint: Expand the $\dfrac{\sin a}{1+2y\cos a+y^2}$ and show $$\frac {y}{1+2y\cos a+y^2}=\sum_{n\geq1}(-1)^{n-1}\dfrac{\sin na}{\sin a}y^n$$

  • $\begingroup$ Is the last line supposed to say$$\frac 1{1+2y\cos a+y^2}=\sum\limits_{n\geq1}(-1)^{n-1}\frac {\sin na}{\sin a}y^n$$ $\endgroup$ – Frank W. Dec 17 '18 at 16:13
  • $\begingroup$ @Nosrati Okay I've replaced $2\cos a$ with $e^{ia}+e^{-ia}$. Then, I factored the denominator and expanded to get the $a_n$ term as$$a_n=\sum\limits_{k=0}^{n}(-1)^k e^{(n-2k)ia}=\frac {\sin a(n+1)}{\sin a}$$Therefore, the sum equals$$\int\limits_0^1\mathrm dy\,\frac {y^x\sin a}{1+2y\cos a+y^2}=\sum\limits_{n\geq1}\frac {(-1)^{n-1}\sin an}{x+n}$$I can see that the two are very close, but how do I arrive at the gamma functions? Additionally, how does the infinite sum which we have become finite? $\endgroup$ – Crescendo Dec 18 '18 at 5:21
  • $\begingroup$ Additionally, $\sum_{i=1}^\infty=\sum_{i=1}^{n}+\sum_{i=n+1}^{2n}+\cdots=\sum_{n=1}^{\infty}\sum_{i=1}^{n}$ $\endgroup$ – Nosrati Dec 18 '18 at 5:41

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