Mellin transform of $\frac{\sinh (\pi x)}{(\cosh (\pi x)-1)^3}$

Question

I would like to know if the following holds true $$\int_0^{\infty } \frac{x^{s-1} \sinh (\pi x)}{(\cosh (\pi x)-1)^3} \, dx= \frac{1}{3} \pi ^{-s} (\zeta (4-s)-\zeta (2-s)) \Gamma (s)$$ and a route to prove it.

Answer 1

Hint. One may expand the integrand then one is allowed to integrate termwise ($0<s<1$): $$ \begin{align} \int_0^{\infty } \frac{x^{s-1} \sinh (\pi x)}{(\cosh (\pi x)-1)^3} \, dx&=4\int_0^{\infty } \frac{x^{s-1}\left( e^{\pi x}-e^{-\pi x}\right)}{\left( e^{\pi x}+e^{-\pi x}-2\right)^3} \, dx \\&= 4 \int_{0}^{\infty} x^{s-1} \sum_{n=0}^{\infty} \frac{n^4-n^2}{12}e^{-n \pi x}\:dx \\ & = 4\sum_{n=0}^{\infty}\frac{n^4-n^2}{12} \int_{0}^{\infty} x^{s-1} e^{-n \pi x}\:dx \\ &= \frac13 \sum_{n=1}^{\infty}(n^4-n^2)\frac{ \pi^{-s} \Gamma(s)}{n^s} \\ &= \frac{1}{3} \pi ^{-s} (\zeta (4-s)-\zeta (2-s)) \Gamma (s) \end{align}$$ where we have used the standard integral representation of the Euler gamma function.

Answer 2

\begin{align*} \int_0^{\infty}\frac{x^{s-1}\sinh(\pi x)}{\left(\cosh(\pi x)-1\right)^3}\;dx &= \int_0^{\infty}\frac{x^{s-1}\left(\frac{e^{\pi x}-e^{-\pi x}}2\right)}{\left(\frac{e^{\pi x}+e^{-\pi x}}2-1\right)^3}\;dx \\ &= 4\int_0^{\infty}\frac{x^{s-1}\left(e^{\pi x}-e^{-\pi x}\right)}{\left(e^{\pi x}+e^{-\pi x}-2\right)^3}\;dx \\ &= 4\int_0^{\infty}\frac{x^{s-1}\left(e^{-2\pi x}-e^{-4\pi x}\right)}{\left(1-e^{-\pi x}\right)^6}\;dx \\ &= 4\int_0^{\infty}x^{s-1}\left(e^{-2\pi x}-e^{-4\pi x}\right)\left[\sum_{n=0}^{\infty}\binom{n+5}5 e^{-n\pi x}\right]\;dx \\ &= 4\sum_{n=0}^{\infty}\binom{n+5}5 \int_0^{\infty}x^{s-1}\left(e^{-(n+2)\pi x}-e^{-(n+4)\pi x}\right)\;dx \\ &= 4\sum_{n=0}^{\infty}\binom{n+5}5 \left[\int_0^{\infty}x^{s-1}e^{-(n+2)\pi x}\;dx - \int_0^{\infty}x^{s-1}e^{-(n+4)\pi x}\;dx\right] \\ \end{align*}$$ = 4\sum_{n=0}^{\infty}\binom{n+5}5 \left[\frac1{((n+2)\pi)^s}\int_0^{\infty}((n+2)\pi x)^{s-1}e^{-(n+2)\pi x}\;(n+2)\pi\;dx - \frac1{((n+4)\pi)^s}\int_0^{\infty}((n+4)\pi x)^{s-1}e^{-(n+4)\pi x}\;(n+4)\pi\;dx\right] $$\begin{align*} &= 4\sum_{n=0}^{\infty}\binom{n+5}5 \left[\frac1{((n+2)\pi)^s}\int_0^{\infty}u^{s-1}e^{-u}\;du - \frac1{((n+4)\pi)^s}\int_0^{\infty}u^{s-1}e^{-u}\;du\right] \\ &= \frac{4\Gamma(s)}{\pi^s}\sum_{n=0}^{\infty}\binom{n+5}5 \left[\frac1{(n+2)^s} - \frac1{(n+4)^s}\right] \\ &= \frac{4\Gamma(s)}{\pi^s}\sum_{n=1}^{\infty}\left[\binom{n+3}5-\binom{n+1}5\right]\frac1{n^s} \\ &= \frac{4\Gamma(s)}{\pi^s}\sum_{n=1}^{\infty}\frac{(n+3)(n+2)(n+1)n(n-1) - (n+1)n(n-1)(n-2)(n-3)}{120n^s} \\ &= \frac{\Gamma(s)}{30\pi^s}\sum_{n=1}^{\infty}\frac{(n+1)n(n-1)((n+2)(n+3) - (n-2)(n-3))}{n^s} \\ &= \frac{\Gamma(s)}{30\pi^s}\sum_{n=1}^{\infty}\frac{(n+1)n(n-1)(10n)}{n^s} \\ &= \frac{\Gamma(s)}{3\pi^s}\sum_{n=1}^{\infty}\frac{n^4-n^2}{n^s} \\ &= \frac{\Gamma(s)}{3\pi^s}\left(\zeta(s-4) - \zeta(s-2)\right)\;\blacksquare \end{align*}