Mellin transform $\frac1{\cos(x)-1}$

Asked Sep 24 '16 at 16:55

Active Sep 24 '16 at 23:14

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Is it possible to show the following formulas?

$$\int_0^{\infty } \frac{x^{s-1}}{\cos (x)-1} \, dx=-i^{-s} (2 \pi )^{s-1} \left(-s+(-i)^s i^s\right) \zeta (2-s) \csc \left(\frac{\pi s}{2}\right)$$ and $$\int_0^{\infty } \frac{x^{s-1} \sin (x)}{(\cos (x)-1)^2} \, dx=(-1)^{1-s} \left((-1)^s-1\right) \zeta (s-2) \csc \left(\frac{\pi s}{2}\right) \Gamma (s)$$

edited Sep 24 '16 at 23:14

iamvegan

2,552

asked Sep 24 '16 at 16:55

it should be true only for $Re(s) \in (0,1)$. Expand $\frac{1}{\cos(x)-1}$ as a series in $e^{i nx}$, and show that $\Gamma(s) e^{i \pi s /2} = \int_0^\infty x^{s-1}e^{-i x} dx, Re(s) \in (0,1)$ but you need some complex analysis theorems (analytic continuation, complex integration). – reuns Sep 25 '16 at 13:39
@CarlSchildkraut everything disappeared :) – reuns Sep 25 '16 at 18:01

Mellin transform $\frac1{\cos(x)-1}$

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