Integration contour for improper integral $\int^\infty_{-\infty} \frac{dx}{(e^x + 1)(1 + e^{-x-z})}$. A while ago, I read through a book in solid state physics (Ziman's Electrons and Photons) and, as an algebraic step in the derivation of a formula, the author uses

$$
\int_{-\infty}^{\infty}\frac{d\eta}{\{\exp\eta + 1\}\{1 + \exp[-(\eta + z)]\}} = \frac{z}{1 - \exp(-z)}
$$

without proof, merely stating the integral is "elementary". Today I used this equation to derive an analytical result in a paper, so I would like to find proof that this evaluation of the integral is correct. By the form of the integrand and the infinite limits of the integral, I guessed that the result should be obtainable with contour integration, but I am struggling to find an appropriate integration contour.
Any suggestions for which contour of integration I should take or even if contour integration is the correct idea to evaluate this integral?
 A: You don't need to use contour integration, write
\begin{align}
\int\frac{1}{(e^\eta+1)(1+e^{-\eta-z})}d\eta
&= \int\frac{e^\eta}{(e^\eta+1)(e^\eta+e^{-z})}d\eta \\
&= \frac{e^z}{e^z-1}\int\frac{1}{e^\eta+1}d\eta - \frac{1}{e^z-1}\int\frac{1}{e^\eta+e^{-z}}d\eta \\
&= \frac{e^z}{e^z-1}\ln\dfrac{e^\eta+e^{-z}}{e^\eta+1}
\end{align}
take limit with integral bounds and find the result $\dfrac{z}{1-e^{-z}}$.
A: The integral is indeed elementary.
For initial notational ease, let $a = e^{-z}$ and we will consider the indefinite integral
$$I = \int \frac{dx}{(e^x + 1)(1 + a e^{-x})}.$$
Observing that
$$\frac{1}{(e^x + 1)(1 + a e^{-x})} = \frac{1}{1 -a} \left [\frac{1}{e^x + 1} - \frac{a e^{-x}}{1 + a e^{-x}} \right ],$$
the integral can be rewritten as
\begin{align*}
I &= \frac{1}{1 - a} \int \frac{dx}{e^x + 1} - \frac{1}{1 - a} \int \frac{ae^{-x}}{1 + a e^{-x}} \, dx\\
&= \frac{1}{1 - a} \int \frac{e^{-x}}{1 + e^{-x}} - \frac{1}{1 - a} \int \frac{ae^{-x}}{1 + a e^{-x}} \, dx.
\end{align*}
Setting $u = e^{-x}, du = - e^{-x} \, dx$ yields
\begin{align*}
I &= -\frac{1}{1 - a} \int \frac{du}{1 + u} + \frac{1}{1 - a} \int \frac{a}{1 + au} \, du\\
&= -\frac{1}{1 - a} \ln (1 + u) + \frac{1}{1 - a} \ln (1 + au) + C\\
&= -\frac{1}{1 - a} \ln (1 + e^{-x}) + \frac{1}{1 - a} \ln (1 + ae^{-x}) + C.
\end{align*}
Rearranging, after setting $a = e^{-z}$, the expression for the integral can be brought into the following form
$$I = \frac{1}{1 - e^{-z}} \ln \left (\frac{1 + e^{x + z}}{1 + e^x} \right ) + C.$$
Now for your improper integral, if we note that
$$\lim_{x \to \infty} \ln \left (\frac{1 + e^{x + z}}{1 + e^x} \right ) = z,$$
and
$$\lim_{x \to -\infty} \ln \left (\frac{1 + e^{x + z}}{1 + e^x} \right ) = 0,$$
then
$$\int^\infty_{-\infty} \frac{dx}{(e^x + 1)(1 + e^{-x-z})} = \frac{z}{1 - \exp(-z)},$$
as required.
