\[ I \da \int_\RR {e^{ax} \over \cosh(x) }\dx, \qquad \abs{\Re(a)}<1. \] Show that \[ I=\pi\sec\qty{a\pi\over2}. \]
Solution
Set \[ f(z)=\frac{e^{az}}{\cosh z}. \] The poles of \(f\) occur at \[ z=\qty{k+\frac12}\pi i, \qquad k\in\ZZ. \] Integrate around the positively oriented rectangle with vertices \(-R,R,R+i\pi,-R+i\pi\).
Since \[ \cosh(z+i\pi)=-\cosh z, \qquad e^{a(z+i\pi)}=e^{i\pi a}e^{az}, \] the upper horizontal edge, oriented from \(R+i\pi\) to \(-R+i\pi\), contributes \[ \int_R^{-R}f(x+i\pi)\,dx =e^{i\pi a}\int_{-R}^R f(x)\,dx. \] Therefore the two horizontal edges contribute \[ \qty{1+e^{i\pi a}}I_R, \qquad I_R=\int_{-R}^R\frac{e^{ax}}{\cosh x}\,dx. \]
For \(0\leq y\leq\pi\), \[ \abs{\cosh(R+iy)}^2=\sinh^2R+\cos^2y\geq\sinh^2R. \] Thus, with a constant \(C_a\) independent of \(R\) and \(y\), \[ \abs{f(R+iy)}\leq C_a\frac{e^{R\Re a}}{\sinh R}, \qquad \abs{f(-R+iy)}\leq C_a\frac{e^{-R\Re a}}{\sinh R}. \] The vertical-edge integrals therefore tend to zero when \(-1<\Re a<1\).
The rectangle contains only the pole \(z_0=i\pi/2\). It is simple, and \[ \Res_{z=z_0}f(z) =\frac{e^{az_0}}{\sinh z_0} =\frac{e^{i\pi a/2}}{i} =-i e^{i\pi a/2}. \] Letting \(R\to\infty\) in the residue theorem gives \[ \qty{1+e^{i\pi a}}I =2\pi e^{i\pi a/2}. \] Since \[ 1+e^{i\pi a}=2e^{i\pi a/2}\cos\qty{\pi a\over2}, \] we obtain \[ I=\pi\sec\qty{\pi a\over2}. \]