I need help with the integral: $$\int_{-\infty}^\infty\frac{x}{\sinh(x)-1}dx,$$ and I (unfortunately) have to use contour integration techniques. I know how to do the integral $$\int_{-\infty}^\infty\frac{1}{\sinh(x)}dx,$$ so using a similar strategy, I tried integrating $$f(z):=\frac{z}{\sinh(z)-1}$$ around a box of width $2R$ and height $\pi $ centered at the origin. However, $\frac{1}{\sinh(x)-1}$ has poles at $z_n=\ln(2\pm\sqrt{2})+2\pi i n,$ for $n\in\mathbb{N}$, and so we must make $\epsilon$ bumps around $\log(2\pm\sqrt{2})$. However, around these bumps,
$$\int_{C_\epsilon}f(z)dz=\int_0^\pi\frac{2\log(1\pm\sqrt{2})+2\epsilon e^{-i\theta}}{\log(1\pm \sqrt{2})[e^{\epsilon e^{i\theta}}-e^{-\epsilon e^{-\theta}}]-1}\cdot -i\epsilon e^{-i\theta}d\theta$$
Does this simplify? I'm not sure how to tackle this. I also thought about making a substitution $x\mapsto \ln(x)$ at the very beginning, but the integration bounds that I get confuse me (I get from $-\infty +i\pi$ to $\infty$).
Best Answer
DETERMINING CONVERGENCE
We are asked to evaluate the integral $I$ given by
$$I=\int_{-\infty}^\infty \frac{x}{\sinh(x)-1}\,dx\tag1$$
We denote the denominator of the integrand in $(1)$ by $g(x)=\sinh(x)-1$. For $x\in \mathbb{R}$, it is easy to show that $g(x)$ has a single root $x_0-=\log(1+\sqrt 2)$.
Then, from the prosthaphaeresis identity
$$\begin{align} g(x)&=\sinh(x)-1\\\\ &=\sinh(x)-\sinh(x_0)\\\\ &=2\cosh\left(\frac{x+x_0}{2}\right)\sinh\left(\frac{x-x_0}{2}\right) \end{align}$$
we find that $g(x)=O\left(x-x_0\right)$ as $x\to x_0$.
CAUCHY-PRINCIPAL VALUE
However, the Cauchy Principal Value of $(1)$ does exist and is expressed as
$$\begin{align}\text{PV}\left(\int_{-\infty}^\infty \frac{x}{\sinh(x)-1}\,dx\right)&=\lim_{\varepsilon\to 0^+}\left(\int_{-\infty}^{x_0-\varepsilon} \frac{x}{\sinh(x)-1}\,dx\\+\int_{x_0+\varepsilon}^\infty \frac{x}{\sinh(x)-1}\,dx\right)\tag2 \end{align}$$
In the next section, we use contour integration to evaluate $(2)$.
EVALUATION OF THE CAUCHY PRINCIPAL VALUE
Let $f(z)=\frac{z^2}{\sinh(z)-1}$, $z\in \mathbb{C}$. The poles of $f(z)$ are simple and located at $z_n=x_0+i2n\pi$ and $z'_n=-x_0+i(2n+1)\pi$.
Let $J$ be the integral
$$J=\oint_C f(z)\,dz$$
where $C$ is the contour comprised of the six line segments $(i)$ from $-R$ to $z_0-\varepsilon$, $(ii)$ from $z_0+\varepsilon$ to $R$, $(iii)$ from $R$ to $R+i2\pi$, $(iv)$ from $R+in\pi$ to $z_1+\varepsilon$, $(v)$ from $z_1-\varepsilon$ to $-R+i2\pi$, and $(vi)$ from $-R+i2\pi$ to $-R$ and the two semicircular arcs $(i)$ $z_0+\varepsilon e^{i\phi}$, from $\phi=\pi$ to $\phi=0$ and (ii) $z_1+\varepsilon e^{i\phi}$, from $\phi=2\pi$ to $\phi=\pi$.
APPLYING THE RESIDUE THEOREM
The contour $C$ encloses only the simple pole at $z'_0=-x_0+i\pi$. Therefore, the reside theorem guarantees that for $R>|z_0'|$
$$\begin{align} \oint_C f(z)\,dz&=2\pi i \text{Res}\left(f(z), z=z'_0\right)\\\\ &=2\pi i \lim_{z\to z'_0}\left(\frac{z^2(z-z'_0)}{\sinh(z)-1}\right)\\\\ &=2\pi i \lim_{z\to z'_0}\frac{z^2}{\cosh(z)}\\\\ &=2\pi i \frac{(z'_0)^2}{\cosh(z'_0)}\\\\ &=2\pi i \frac{(x_0+i\pi)^2}{-\sqrt 2}\\\\ &=-\frac{i\pi}{\sqrt 2}(2(x_0+i\pi)^2)\tag3 \end{align}$$
EXPRESSING THE INTEGRAL OVER $C$
We also have as $R\to \infty$ and $\varepsilon \to 0^+$
$$\begin{align} \lim_{R\to\infty\\\varepsilon\to 0^+}\oint_C f(z)\,dz&= \text{PV}\left(\int_{-\infty}^{\infty}\frac{x^2}{\sinh(x)-1}\,dx\right)-\text{PV}\left(\int_{-\infty}^{\infty}\frac{(x+i2\pi)^2}{\sinh(x)-1}\,dx\right)\\\\ &-i\pi \frac{z_0^2}{\cosh(z_0)}-i\pi\frac{z_1^2}{\cosh(z_1)}\\\\ &=-i4\pi \text{PV}\left(\int_{-\infty}^{\infty}\frac{x}{\sinh(x)-1}\,dx\right)\\\\ &+4\pi^2\text{PV}\left(\int_{-\infty}^{\infty}\frac{1}{\sinh(x)-1}\,dx\right)\\\\ &-\frac{i\pi }{\sqrt 2}(z_0^2+(z_0+i2\pi)^2)\tag4 \end{align}$$
CONCLUSION
Equating $(3)$ and $(4)$ we find that
$$\bbox[5px,border:2px solid #C0A000]{\text{PV}\left(\int_{-\infty}^\infty \frac{x}{\sinh(x)-1}\,dx\right)=\frac{\pi^2}{2\sqrt 2}}$$