Help with Complex integration I have to calculate the following integral
$$\int_{-\infty}^{\infty} \frac{\cos(x)}{e^x+e^{-x}} dx$$
Anyone can give me an idea about what complex function or what path I should choose to calculate the integral?
 A: By following the Daniel Fischer's suggestion, consider the integration path $\gamma=\partial D$ (with counter-clockwise orientation) where $D$ that is the rectangle with vertices in $R,R+i\pi,-R+i\pi,-R$. Since the zero set of $e^x+e^{-x}=2\cosh x$ is $\frac{\pi}{2}i+\pi i\mathbb{Z}$, by the residue theorem:
$$\int_{\gamma}\frac{\cos z}{e^z+e^{-z}}dz = 2\pi i\cdot\operatorname{Res}\left(\frac{\cos z}{e^z+e^{-z}},z=\frac{\pi}{2}i\right)=\pi \cosh\frac{\pi}{2}.$$
Since $\cos(z+\pi i)=\cos(z)\cosh(\pi)-i\sin(z)\sinh(\pi)$ and $\cosh(z+\pi i)=-\cosh(z)$, the contribute given by integrating $\frac{\cos z}{e^z+e^{-z}}$ along the horizontal sides the rectangle equals:
$$(1+\cosh\pi)\int_{-R}^{R}\frac{\cos z}{e^z+e^{-z}}dz,$$
because $\sin(z)$ is an odd function while $\cos(z)$ and $\cosh(z)$ are even functions. 
When $|\Re z|=R$ we have:
$$|\cos z|\leq\sqrt{\cosh^2(|\Im z|)+\sinh^2(|\Im z|)}\leq \cosh(\Im z)$$
$$|2\cosh z|\geq 2\sinh(|\Re z|),$$
hence the contribute given by the vertical sides of the rectangle is negligible when $R\to +\infty$, and:
$$\int_{-\infty}^{+\infty}\frac{\cos z}{e^z+e^{-z}}\,dz = \frac{\pi}{2\cosh(\pi/2)}=\frac{\pi}{e^{\pi/2}+e^{-\pi/2}}.$$
A: $\newcommand{\+}{^{\dagger}}
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\begin{align}&\color{#c00000}{%
\int_{-\infty}^{\infty}{\cos\pars{x} \over \expo{x} + \expo{-x}}\,\dd x}
=\Re\ \overbrace{%
\int_{-\infty}^{\infty}{\expo{\ic x} \over \expo{x} + \expo{-x}}\,\dd x}
^{\ds{\mbox{Set}\ \expo{x} \equiv t\ \imp\ x = \ln\pars{t}}}\ =\
\Re\int_{0}^{\infty}{t^{\ic} \over t + 1/t}\,{\dd t \over t}
\\[3mm]&=\color{#00f}{\Re\int_{0}^{\infty}{t^{\ic} \over t^{2} + 1}\,\dd t}
=\Re\bracks{2\pi\ic\,{\pars{\expo{\pi\ic/2}}^{\ic} \over 2\ic}
+2\pi\ic\,{\pars{\expo{3\pi\ic/2}}^{\ic} \over -2\ic}
-\int_{\infty}^{0}{t^{\ic}\pars{\expo{2\pi\ic}}^{\ic} \over t^{2} + 1}\,\dd t}
\\[3mm]&=\pi\expo{-\pi/2} - \pi\expo{-3\pi/2}
+\expo{-2\pi}\,\color{#00f}{\Re\int_{0}^{\infty}{t^{\ic} \over t^{2} + 1}\,\dd t}
\end{align}

From here we can get an expression for
  $\ds{\color{#c00000}{%
\int_{-\infty}^{\infty}{\cos\pars{x} \over \expo{x} + \expo{-x}}\,\dd x}
=\color{#00f}{\Re\int_{0}^{\infty}{t^{\ic} \over t^{2} + 1}\,\dd t}}\,$:
  \begin{align}&\color{#66f}{\large%
\int_{-\infty}^{\infty}{\cos\pars{x} \over \expo{x} + \expo{-x}}\,\dd x}
={\pi\expo{-\pi/2} - \pi\expo{-3\pi/2} \over 1 - \expo{-2\pi}}
=\pi\,{\expo{\pi/2} - \expo{-\pi/2} \over \expo{\pi} - \expo{-\pi}}
=\pi\,{\sinh\pars{\pi/2} \over \sinh\pars{\pi}}
\\[3mm]&=\pi\,{\sinh\pars{\pi/2} \over 2\sinh\pars{\pi/2}\cosh\pars{\pi/2}}
=\color{#66f}{\large\half\,\pi\,\sech\pars{\pi \over 2}}
\approx {\tt 0.6260}
\end{align}

We used the contour 
There are two simple poles at $\ds{i = \expo{\pi\ic/2}}$ and at
$\ds{-i = \expo{3\pi\ic/2}}$ since the branch cut of $\ds{t^{\ic}}$ is given by:
$$
t^{\ic} \equiv \exp\pars{\ic\ln\pars{\verts{t}} - {\rm Arg}\pars{t}}\,,\qquad
t \not= 0\,,\quad 0 < {\rm Arg}\pars{t} < 2\pi
$$
A: Hint: You should find the solutions of 
$$e^z+e^{-z}=0 ~~(*), $$
with $z=x+iy$. As 
$$e^z+e^{-z}=(e^x+e^{-x})\cos y+i(e^{x}-e^{-x})\sin y,  $$
$(*)$ gives
$$\cos y = \sin y = 0$$ or $$\cos y= 0 \cap x = 0. $$
In summary $e^z+e^{-z}=0 \Leftrightarrow x= 0,~ y= \frac{\pi}{2}+k\pi, $ with $k\in\mathbb Z$: the poles of the function $f(z)=\frac{\cos z}{e^z+e^{-z}}$ lie on the imaginary axis.
Now you can use @DanielFischer comment.
A: You can use this way to do. Clearly
\begin{eqnarray}
I&=&\int_{-\infty}^{\infty} \frac{\cos(x)}{e^x+e^{-x}}dx=2\int_{0}^{\infty} \frac{\cos(x)}{e^x+e^{-x}}dx\\
&=&2\int_{0}^{\infty} \frac{e^{-x}\cos(x)}{1+e^{-2x}}dx=2\text{Re}\int_{0}^{\infty} \frac{e^{-x}e^{-ix}}{1+e^{-2x}}dx\\
&=&2\text{Re}\int_{0}^{\infty} \frac{e^{-(1+i)x}}{1+e^{-2x}}dx=2\text{Re}\int_{0}^{\infty} \sum_{n=0}^\infty e^{-(1+i)x}(-1)^ne^{-2nx}\\
&=&2\text{Re}\int_{0}^{\infty} \sum_{n=0}^\infty(-1)^n e^{-(2n+1+i)x}dx\\
&=&2\text{Re}\int_{0}^{\infty} \sum_{n=0}^\infty(-1)^n e^{-(2n+1+i)x}dx\\
&=&2\text{Re}\sum_{n=0}^\infty(-1)^n \frac{1}{2n+1+i}=2\sum_{n=0}^\infty(-1)^n \frac{2n+1}{(2n+1)^2+1}\\
&=&\sum_{n=0}^\infty(-1)^n \frac{2n+1}{2n^2+2n+1}=\frac{1}{2}\sum_{n=-\infty}^\infty(-1)^n \frac{2n+1}{2n^2+2n+1}\\
&=&\frac{\pi\sinh\frac{\pi}{2}}{\sinh\pi}=\frac{\pi}{2\cosh\frac{\pi}{2}}.
\end{eqnarray}
Here we used
$$ \sum_{n=-\infty}^\infty(-1)^n f(n)=-\pi \sum_{k=1}^m\text{Res}(\frac{f(z)}{\sin(\pi z)},a_k) $$
where $a_1,a_2,\cdots,a_m$ are poles of $f(z)$.
