Closed form of $\int_{0}^{\infty}\frac{y^{2n-1}\,dy}{\cosh y - \cos(z + \pi)}$

Question

I'm looking for a nice closed form of the below integral

$$\int_{0}^{\infty}\dfrac{y^{2n-1}\,dy}{\cosh y - \cos(z + \pi)}$$

I'm stuck on this for quite some time, I have tried contour integration and integrating by parts and that doesn't seem to get me anywhere. Your insight would be very helpful.

Thanks in advance.

What I mean is is it supposed to be a constant or is it supposed to be $\cos(y+\pi)$? Please offer more details than "um Yeah" — Ninad Munshi, Apr 25 '21 at 06:41
While sparse OP's post does not deserve to be downvoted, it is an nonduplicate with vague explanations of where OP went wrong. — Ninad Munshi, Apr 25 '21 at 07:06

score 3 · Answer 1 · answered Apr 25 '21 at 06:47

3

The integral appears in my comment here. For $|r|<1$ we have $$\sum_{k=1}^\infty r^k\sin kx=\Im\frac1{1-re^{ix}}=\frac{r\sin x}{1-2r\cos x+r^2},$$ so that, if we put $r=e^{-y}$, multiply by $y^{2n-1}$ and integrate, we get $$\frac{\sin x}{2}\int_0^\infty\frac{y^{2n-1}\,dy}{\cosh y-\cos x}=\sum_{k=1}^\infty\sin kx\int_0^\infty y^{2n-1}e^{-ky}\,dy=(2n-1)!\sum_{k=1}^\infty\frac{\sin kx}{k^{2n}},$$ resulting in Clausen functions. (Put $x=z+\pi$.)

answered Apr 25 '21 at 06:47

metamorphy

39,111

so you mean my integral equals $(2n-1)!,\mathrm{Cl}_{2n}(z+\pi)$? – BooleanCoder Apr 25 '21 at 07:19
@BooleanWick: ... multiplied by $2/\sin(z+\pi)=-2/\sin z$. – metamorphy Apr 25 '21 at 07:25

Closed form of $\int_{0}^{\infty}\frac{y^{2n-1}\,dy}{\cosh y - \cos(z + \pi)}$

1 Answers1