Integral $\int_{-\infty}^{\infty} \frac{\sinh(x)}{x [a+\cosh(x)]^2}dx$

Question

I have difficulties with calculating the following integral:

$$I(a)=\int_{-\infty}^{\infty} \frac{\sinh(x)}{x [a+\cosh(x)]^2} \mathrm dx~~~~~~~,\text{where } a>1$$

For the case with $a=1$ the solution is $-28 \zeta'(-2)$.

Answer 1

I will do the case $a=1$ in order to give some insight:$$I=\int_{-\infty}^{\infty} \frac{\sinh(x)}{x (1+\cosh(x))^2}dx\overset{x=\ln t}=2\int_0^\infty \frac{t-1}{(t+1)^3}\frac{dt}{\ln t}$$ We can consider the following integral and perform Feynman's trick: $$I(n)=2\int_0^\infty \frac{t^{n-1}-1}{(t+1)^3}\frac{dt}{\ln t}$$ $$\Rightarrow \frac{d}{dn}I(n)=2\int_0^\infty \frac{t^{n-1}}{(t+1)^3}dt=2B(n,3-n)=\Gamma(n)\Gamma(3-n)$$ $$=(2-n)(1-n)\Gamma(n)\Gamma(1-n)=\pi\frac{(1-n)(2-n)}{\sin(\pi n)}$$ $$I(1)=0 \Rightarrow I(2)-I(1)=I =\pi\int_1^2 \frac{(1-n)(2-n)}{\sin(\pi n)}dn$$ $$ \overset{n-1=x}=\boxed{\pi \int_0^1 \frac{x(1-x)}{\sin(\pi x)}dx = \frac{7\zeta(3)}{\pi^2 }=-28\zeta'(-2)}$$ The last integral can be found here.

By the same idea one gets: $$I(a)=\int_{-\infty}^\infty \frac{\sinh x}{x(a+\cosh x))^2}dx=2\int_0^\infty \frac{x^2-1}{(x^2+2ax+1)^2}\frac{dx}{\ln x}$$ Now consider the same parameter take a derivative with respect to it, apply a Mellin transform, and try to carry on.