Contour integral of $\int_{-\infty}^\infty \frac{e^{2\pi x / 3}}{\cosh{\pi x}}dx$ I'm having a little trouble figuring this one out. So far I've got
$$I = \int_{-\infty}^\infty \frac{e^{2\pi x / 3}}{\cosh{\pi x}}dx$$
Let 
$$I' = \oint_C \frac{e^{2\pi z / 3}}{\cosh{\pi z}}dz$$
Where the contour $C$ is a rectangle extending from $-R$ to $R$ in the limit of $R \rightarrow \infty$ and of height 1, which gives $$\int_{-R}^R \frac{e^{2\pi x / 3}}{\cosh{\pi x}}dx + \int_{R}^{-R} \frac{e^{2\pi x / 3 + i}}{\cosh{(\pi x + i)}}dx$$
$$=I - \int_{-R}^R \frac{e^{2\pi x / 3 + i}}{\cosh{(\pi x + i)}}dx$$
$$=\text{Res}(z = \frac{i}{z})$$
I'm having trouble solving the problem from here, as I'm not sure how to handle the separation of the denominator $\cosh{(\pi x + i)}$. Any help is appreciated!
 A: Note that $\cosh(\pi x+i)$ should read $\cosh(\pi x+i\pi)$.  
Then note that $\cosh(\pi x\pm i\pi)=\cosh(\pi x)\cos(\pm \pi)+i\sinh(\pi x)\sin(\pm \pi)=-\cosh(\pi x)$.
Therefore, if we take the contour $C$ to be the rectangle with vertices $-R-i$, $R-i$, $R+i$, and $-R+i$, then we can write
$$\begin{align}
\oint_C \frac{e^{2\pi z/3}}{\cosh(\pi z)}\,dz&=\int_{-R}^R \frac{e^{-i2\pi/3}e^{2\pi x/3}}{-\cosh(\pi x)}\,dx+\int_{R}^{-R} \frac{e^{i2\pi/3}e^{2\pi x/3}}{-\cosh(\pi x)}\,dx\\\\
&+\int_{-1}^1 \frac{e^{2\pi (R+iy)/3}}{\cosh(R+iy)}\,i\,dy+\int_{1}^{-1}\frac{e^{2\pi (-R+iy)/3}}{\cosh(-R+iy)}\,i\,dy\tag 1
\end{align}$$
Applying the residue theorem to $(1)$ yields
$$\begin{align}
\oint_C \frac{e^{2\pi z/3}}{\cosh(\pi z)}\,dz&=2\pi i \text{Res}\left(\frac{e^{2\pi z/3}}{\cosh(\pi z)}, z=\pm i/2\right)\\\\
&=2\pi i \left(\frac{e^{i\pi/3}}{\pi \sinh(i\pi/2)}+\frac{e^{-i\pi/3}}{\pi \sinh(-i\pi/2)}\right)\\\\
&=i2\sqrt 3\tag 2
\end{align}$$
As $R\to \infty$, the third and fourth integrals on the right-hand side of $(1)$ approach $0$.  Hence, after letting $R\to \infty$ and setting $(1)$ and $(2)$ equal, we find that 
$$\int_{-\infty}^\infty \frac{e^{2\pi x/3}}{\cosh(\pi x)}\,dx=2$$
