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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 $-28 \zeta'(-2)$.

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    $\begingroup$ Expand $\frac{\sinh(x)}{(a+\cosh(x))^2} = \sum_{k=1}^\infty c_k(a) e^{-kx}$ then $\int_0^\infty x^{s-1} \frac{\sinh(x)}{(a+\cosh(x))^2} dx = \sum_{k=1}^\infty c_k(a) k^{-s} \Gamma(s)$ and $\int_{-\infty}^\infty x^{s-1} \frac{\sinh(x)}{(a+\cosh(x))^2} dx = 2\lim_{s \to 0} \sum_{k=1}^\infty c_k(a) k^{-s} \Gamma(s)$. Things like $\zeta'(-2)$ appear from the functional equation, or the residue theorem. $\endgroup$ – reuns Jan 29 at 21:14
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    $\begingroup$ What have you tried? Where did you struggle? Please show some effort you made and include it within your post with an edit :) $\endgroup$ – mrtaurho Jan 29 at 21:15
  • $\begingroup$ Actually, I've tried an expansion but a very crude one. The solution by @reuns looks impressive. $\endgroup$ – Gregory Rut Jan 29 at 21:17
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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.

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