Improper integral complex analysis $\int_{-\infty}^\infty \frac{e^{ax} \, dx}{\cosh(x)}$ I tried the following problem but I don't think I got the right answer. I checked it by substituting $a=\frac{1}{2}$ into the integral and putting that through Wolfram Alpha but it didn't match the answer I found when I substituted $a=\frac{1}{2}$ into my final answer. So I would be so grateful if someone could tell me where I went wrong. I'm not sure if there's an easier contour than a rectangle but if there is please don't suggest any because I was required to solve this problem using a rectangular contour. Thank you in advance! :)
Problem:
Evaluate $$\int_{-\infty}^\infty \frac{e^{ax} \, dx}{\cosh(x)}$$ where $-1<\operatorname{Re}(a)<1$ using a rectangular contour integral and residue calculus.
My attempt:
$$\cosh(z)=\cos(iz)=0$$
$$z=\frac{i\pi}{2}-k\pi i,k\in Z$$
Let $L>0$ be a real number, and $C_1, C_2, C_3, C_4$ be the line segments that go from $-L$ to $L$, from $L$ to $L+\pi i$, from $L + \pi i$ to $-L+\pi i$ and from $-L+\pi i$ to $-L$, respectively.  Let $C = C_1 + C_2 + C_3 + C_4$, a  rectangular contour surrounding the singularity $z=\frac{i\pi}{2}$.
$$\oint_C\frac{e^{az} \, dz}{\cosh(z)}=\int_{-L}^L \frac{e^{ax}dx}{\cosh(x)} + \int_0^\pi \frac{e^{a(L+iy)}i \, dy}{\cosh(L+iy)}+\int_L^{-L} \frac{e^{a(x+\pi i)} \, dx}{\cosh(x+\pi i)}+\int_\pi^0 \frac{e^{a(-L+iy)}i \, dy}{\cosh(-L+iy)}$$
Now, 
$$\oint_C\frac{e^{az} \, dz}{\cosh(z)}=2\pi i \operatorname{Res} \left(\frac{e^{az}}{\cosh(z)};\frac{i\pi}{2}\right) =2\pi i \lim_{z \rightarrow \frac{i\pi}{2}}\frac{e^{az}(z-\frac{i\pi}{2})}{\cosh(z)}=-ie^{\frac{ai\pi}{2}}$$ and
$$\int_{-L}^L \frac{e^{ax} \,dx}{\cosh(x)}+\int_L^{-L} \frac{e^{a(x+\pi i)} \, dx}{\cosh(x+\pi i)} = (1+e^{a\pi i})\int_{-L}^L \frac{e^{ax} \, dx}{\cosh(x)}$$ using the fact that $\cosh(x+\pi i)=-\cosh(x)$.
$$\Rightarrow -ie^{\frac{ai\pi}{2}} = (1+e^{a\pi i})\int_{-L}^L \frac{e^{ax} \, dx}{\cosh(x)} + \int_0^\pi \frac{e^{a(L+iy)}i \, dy}{\cosh(L+iy)}+\int_\pi^0 \frac{e^{a(-L+iy)}i \, dy}{\cosh(-L+iy)}$$
Then after showing that $\int_0^\pi \frac{e^{a(L+iy)}i \, dy}{\cosh(L+iy)}$, $\int_\pi^0 \frac{e^{a(-L+iy)}i\,dy}{\cosh(-L+iy)} \rightarrow 0$ as $L\rightarrow \infty$ we find that
$$-ie^{\frac{ai\pi}{2}} = (1+e^{a\pi i})\int_{-\infty}^\infty \frac{e^{ax} \, dx}{\cosh(x)}$$
$$\Rightarrow \int_{-\infty}^\infty \frac{e^{ax} \, dx}{\cosh(x)} = \frac{-i e^{\frac{ai\pi}{2}}}{(1+e^{a\pi i})}$$
 A: Let $I(a)$ be given by the integral 
$$I(a)=\int_{-\infty}^\infty \frac{e^{ax}}{\cosh(x)}\,dx \tag 1$$
To evaluate the integral in $(1)$, we analyze the contour integral
$$J(a)=\oint_C \frac{e^{az}}{\cosh(z)}\,dz$$
where $C$ is the closed rectangular contour with vertices at $-L$, $L$, $L+i\pi$, and $-L+i\pi$.  From the residue theorem, we have
$$\begin{align}
J(a)&=2\pi i \text{Res}\left(\frac{e^{az}}{\cosh(z)}, z=i\pi/2\right)\\\\
&=2\pi i \frac{e^{ia\pi/2}}{\sinh(i\pi/2)}\\\\
&=2\pi e^{ia\pi/2} \tag 2
\end{align}$$
In addition, as $L\to \infty$ the contributions to $J(a)$ from integrating over the vertical strips vanish.  Therefore, we find that 
$$\begin{align}
\lim_{L\to \infty}J(a)&=\int_{-\infty}^{\infty}\frac{e^{ax}}{\cosh(x)}\,dx+\int_{\infty}^{-\infty}\frac{e^{a(x+i\pi)}}{\cosh(x+i\pi)}\\\\
&=(1+e^{ia\pi})\int_{-\infty}^{\infty}\frac{e^{ax}}{\cosh(x)}\,dx \tag 3
\end{align}$$
Equating $(2)$ and $(3)$ reveals
$$(1+e^{ia\pi})I(a)=2\pi e^{ia\pi/2} $$
whence we find
$$\bbox[5px,border:2px solid #C0A000]{I(a)=\frac{\pi}{\cos(\pi a/2)}}$$
