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!
No comments:
Post a Comment