Help with the integral $\int_{0}^{\infty}\frac{y^{2}e^{y}}{e^{sy}+e^{-sy}-2}dy$

Question

I want to do the integral : $$I(s)=\int_{0}^{\infty}\frac{y^{2}e^{y}}{e^{sy}+e^{-sy}-2} \, \mathrm{d}y$$ $s$ being a complex parameter. I tried expanding the dominator of the integrand, but this way we lose the symmetry $I(s)=I(-s)$. I tried converting the integral into a contour integral, but I don't know how to close the contour.

Answer 1

For the integral \begin{align} I(s) = \int_{0}^{\infty} \frac{y^{2} \, e^{y} \, dy}{e^{sy} - 2 + e^{-sy}} \end{align} it is evident that when $s \to -s$ the result yields $I(-s) = I(s)$. For the evaluation of the integral consider the following. By making use of \begin{align} \frac{1}{(1-x)^{2}} = \sum_{n=0}^{\infty} (n+1) \, x^{n} \end{align} then \begin{align} I(s) &= \int_{0}^{\infty} \frac{y^{2} \, e^{y} \, dy}{(e^{sy/2} - e^{-sy/2})^{2}} \\ &= \int_{0}^{\infty} \frac{y^{2} \, e^{-(s-1)y} \, dy}{(1 - e^{-sy})^{2}} \\ &= \sum_{n=0}^{\infty} (n+1) \, \int_{0}^{\infty} e^{-(sn + s-1)y} \, y^{2} \, dy \\ &= \sum_{n=0}^{\infty} \frac{n+1}{(sn + s -1)^{3}} \, \int_{0}^{\infty} e^{-u} \, u^{2} \, du \\ &= \frac{2}{s^{3}} \, \sum_{n=0}^{\infty} \frac{n+1}{(n+p)^{3}} \hspace{10mm} p = 1 - \frac{1}{s} \\ &= \frac{2}{s^{3}} \left[ \sum_{n=0}^{\infty} \frac{1}{(n+p)^{2}} + \frac{1}{s} \sum_{n=0}^{\infty} \frac{1}{(n+p)^{3}} \right] \\ &= \frac{1}{s^{4}} \left[ 2s \, \psi^{(1)}\left(1 - \frac{1}{s}\right) - \psi^{(2)}\left(1 - \frac{1}{s}\right) \right] \end{align} where $\psi^{(m)}(x)$ are $m^{th}$ derivative of the digamma function.

Answer 2

According to Mathematica:

$$I(s) = \frac{2 s \, \psi_{1}\left(-\frac{1}{s}\right)-\psi_{2}\left(-\frac{1}{s}\right)}{s^4}, \quad \mathrm{Re}(s) > 1, \quad \psi(z) = \Gamma'(z) / \Gamma(z), \quad \psi_n = \psi^{(n)}$$