Use residues to evaluate $$ \int_0^\infty \frac{\cosh(ax)}{\cosh(x)}\,\mathrm{d}x $$ where $|a|<1$.
Try considering the integral of the form $$ \int_C \frac{\exp(az)}{\cosh(z)}\,\mathrm dz, $$ where $C$ is the contour given by $y=0,\, y=\pi,\, x=-R,\, x=R$.
This doesn't use residues until we use $$ \sum_{k\in\mathbb{Z}}\frac{(-1)^k}{z+k}=\pi\csc(\pi z) $$ which can be proven using residues.
We just expand things in powers of $e^x$: $$ \begin{align} &\int_0^\infty\frac{\cosh(ax)}{\cosh(x)}\,\mathrm{d}x\\ &=\int_0^\infty e^{(a-1)x}\frac{1+e^{-2ax}}{1+e^{-2x}}\,\mathrm{d}x\\ &=\int_0^\infty\left(e^{(a-1)x}-e^{(a-3)x}+e^{(a-5)x}-\dots\right)\,\mathrm{d}x\\ &+\int_0^\infty\left(e^{(-a-1)x}-e^{(-a-3)x}+e^{(-a-5)x}-\dots\right)\,\mathrm{d}x\\ &=\frac1{1-a}-\frac1{3-a}+\frac1{5-a}-\dots\\ &+\frac1{1+a}-\frac1{3+a}+\frac1{5+a}-\dots\\ &=\frac1{a+1}-\frac1{a+3}+\frac1{a+5}-\dots\\ &-\frac1{a-1}+\frac1{a-3}-\frac1{a-5}-\dots\\ &=\frac12\left(\dots+\frac1{\frac{a+1}2-2}-\frac1{\frac{a+1}2-1}+\frac1{\frac{a+1}2}-\frac1{\frac{a+1}2+1}+\frac1{\frac{a+1}2+2}-\dots\right)\\ &=\frac12\pi\csc\left(\pi\frac{a+1}2\right)\\ &=\frac\pi2\sec\left(\frac\pi2a\right) \end{align} $$
Denoting the desired integral by $$I=\int_0^{\infty}\!\!\mathrm{d}x\,\frac{\cosh(ax)}{\cosh x},$$ we may extend the integral to $-\infty<x<\infty$, since the integrand is a symmetric function, i.e. $$\begin{align*}I&=\frac{1}{2}\int_{-\infty}^{\infty}\!\!\mathrm{d}x\,\frac{\cosh(ax)}{\cosh x}\\&=\frac{1}{4}\int_{-\infty}^{\infty}\!\!\mathrm{d}x\,\left(e^{ax}+e^{-ax}\right)\,\mathrm{sech}\,x\\&=\frac{f(a)+f(-a)}{4},\qquad(A)\end{align*}$$ where we have defined $$\displaystyle f(z):=\int_{-\infty}^{\infty}\!\!\mathrm{d}x\,e^{zx}\,\mathrm{sech}\,x.$$ Note, as $x\to\infty$, the hyperbolic secant behaves as $e^{-x}$, hence the integrand $\sim e^{-(1-z)x}$. The integral converges iff $z<1$. Likewise, as $x\to-\infty$, the integrand goes like $e^{(z+1)x}$, hence convergence of the integral requires $z>-1$. Thus $f(z)$ is well-defined for all $|z|<1$, which is also the permitted range for the parameter $a$. That said, let’s return to the evaluation of $f(z)$. $$f(z)=2\int_{-\infty}^{\infty}\!\!\mathrm{d}x\,\frac{e^{zx}}{e^x+e^{-x}}=2\int_{-\infty}^{\infty}\!\!\mathrm{d}x\,\frac{e^{(z-1)x}}{1+e^{-2x}}.$$ Substituting $$\displaystyle 1+e^{-2x}=\frac{1}{u}\Rightarrow\mathrm{d}x=\frac{\mathrm{d}u}{2u^2}e^{2x}=\frac{\mathrm{d}u}{2u(1-u)},$$ and changing the limits of integration appropriately, we arrive at $$\begin{align*}f(z)&=\int_0^1\!\!\mathrm{d}u\,\frac{u^{\frac{z-1}{2}}}{(1-u)^{\frac{z+1}{2}}}\\&=\int_0^1\!\!\mathrm{d}u~u^{\frac{1+z}{2}-1}(1-u)^{\frac{1-z}{2}-1}\\&=\mathrm{B}\!\left(\frac{1+z}{2},\frac{1-z}{2}\right),\end{align*}$$ where $\mathrm{B}(x,y)$ is the beta function (https://en.wikipedia.org/wiki/Beta_function). Expressing the beta function in terms of the gamma function (https://en.wikipedia.org/wiki/Gamma_function), via $$\mathrm{B}(x,y)=\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)},$$ and noting that $\Gamma(1)=1$, we can further simplify our result: $$\begin{align*}f(z)&=\Gamma\left(\frac{1+z}{2}\right)\Gamma\left(\frac{1-z}{2}\right)\\&=\Gamma\left(\frac{1+z}{2}\right)\Gamma\left(1-\frac{1+z}{2}\right)\\&=\frac{\pi}{\sin\left(\frac{\pi}{2}+\frac{\pi z}{2}\right)}=\frac{\pi}{\cos\left(\frac{\pi z}{2}\right)}.\end{align*}$$ In the third step above we used Euler’s well know reflection formula (https://en.wikipedia.org/wiki/Reflection_formula) $$\Gamma(x)\Gamma(1-x)=\frac{\pi}{\sin(\pi x)}.$$ Having evaluated $f(z)$ in closed form, we are merely a substitution away from the desired result (cf Eq. (A)). Thus, $$\boxed{\int_0^{\infty}\!\!\mathrm{d}x\,\frac{\cosh(ax)}{\cosh x}=\frac{\pi}{2\cos\left(\frac{\pi a}{2}\right)}}$$
Cheers!
As the original query that asked to use residues has not been answered completely I will contribute some ideas.
Suppose $a$ is a rational number $p/q$ where $p<q$ and $p-q$ is odd. Use a rectangular contour that consists of four segments: $\Gamma_0$ along the real axis from $-R$ to $R$, $\Gamma_1$ parallel to the imaginary axis to $R + \pi i q$, $\Gamma_2$ parallel to the real axis but in the opposite direction to $-R + \pi i q$ and finally, $\Gamma_3$ parallel to the imaginary axis to $-R$ on the real axis.
Now set $$f(z) = \frac{e^{az}+e^{-az}}{e^z+e^{-z}}$$ so that we are looking for $$\frac{1}{2} \int_{-\infty}^\infty f(z) dz$$ and integrate $f(z)$ along $\Gamma_0 - \Gamma_1 - \Gamma_2 - \Gamma_3$. Examine each segment in turn as $R$ goes to infinity. Clearly the integral along $\Gamma_0$ is simply the integral we are looking for. The contributions of $\Gamma_1$ and $\Gamma_3$ vanish in the limit. Along $\Gamma_2$ we have $x= t + \pi i q$, getting $$ \int_\infty^{-\infty} \frac{e^{\frac{p}{q}t + \pi i p}+e^{-\frac{p}{q}t - \pi i p}}{e^{t+ \pi i q}+e^{-t- \pi i q}} dt = - (-1)^{p-q} \int_{-\infty}^\infty \frac{e^{\frac{p}{q}t}+e^{-\frac{p}{q}t}}{e^{t}+e^{-t}} dt =\int_{-\infty}^\infty \frac{e^{\frac{p}{q}t}+e^{-\frac{p}{q}t}}{e^{t}+e^{-t}} dt. $$ The last equality is because $p-q$ is odd.
To conclude we need to compute the poles and residues inside our contour. The poles are at $$\rho_k = \frac{1}{2}\pi i + \pi i k$$ and the residues are $$\lim_{z\to \rho_k} \frac{(z-\rho_k) (e^{az} + e^{-az})}{e^z + e^{-z}} = \lim_{z\to \rho_k} \frac{(z-\rho_k) (a e^{az} -a e^{-az}) + e^{az} + e^{-az}}{e^z - e^{-z}}.$$ But $$ \lim_{z\to \rho_k} \frac{1}{e^z - e^{-z}} = \frac{1}{i e^{\pi i k} - (-i) e^{-\pi i k}} = \frac{1}{i e^{\pi i k} + i e^{-\pi i k}} = \frac{e^{\pi i k}}{i (1+1)} = \frac{(-1)^k}{2i}$$ so that finally $$\operatorname{Res}_{z=\rho_k} f(z) = \frac{(-1)^k}{2i} \left( e^{a\rho_k} + e^{-a\rho_k}\right).$$ With $J$ being the integral we are looking for and $I$ the integral along $\Gamma_0$ we have $$ J = \frac{1}{2} I = \frac{1}{4} 2 I = \frac{1}{4} 2\pi i \sum_{k=0}^{q-1} \operatorname{Res}_{z=\rho_k} f(z)$$ The conclusion is that $$ J = \frac{1}{2} \pi i \sum_{k=0}^{q-1} \operatorname{Res}_{z=\rho_k} f(z) = \frac{\pi}{4} \sum_{k=0}^{q-1} (-1)^k \left( e^{a\rho_k} + e^{-a\rho_k}\right) = \frac{\pi}{2} \sum_{k=0}^{q-1} (-1)^k \cosh(a\rho_k).$$ where we have used the fact that $1/2 + k < q$ implies that $k$ runs up to $q-1.$
Edit. Use the following bound to see that the integral along $\Gamma_1$ vanishes (set $z= R + it$ with $0\le t \le \pi q$): $$ \left| \int_{\Gamma_1} f(z) dz \right| = \left| \int_0^{\pi q} \frac{e^{aR + ait} + e^{-aR -ait}}{e^{R+it} + e^{-R-it}} i dt \right| \le \int_0^{\pi q} \frac{e^{aR} + e^{-aR}}{e^{R} - e^{-R}} dt = \pi q e^{-(1-a) R} \frac{1-e^{-2aR}}{1-e^{-2R}}$$ Now certainly we have $$\lim_{R\to\infty}\frac{1-e^{-2aR}}{1-e^{-2R}} = 1$$ so that the integral is $\theta(e^{-(1-a) R})$ which goes to zero as $R$ goes to infinity. The integral along $\Gamma_3$ is done the same way.
There is some additional simplification that can be done which I'll do in a new answer because my browser is not coping well with those large formulas where speed is concerned.
We have $$ \frac{\pi}{2} \sum_{k=0}^{q-1} (-1)^k \cosh(a\rho_k) = \frac{\pi}{4} \sum_{k=0}^{q-1}(-1)^k e^{a\rho_k} + \frac{\pi}{4} \sum_{k=0}^{q-1}(-1)^k e^{-a\rho_k}$$
The first sum is $$ \sum_{k=0}^{q-1} (-1)^k e^{a\rho_k} = \sum_{k=0}^{q-1} (-1)^k e^{a i \pi /2} e^{a \pi i k} = e^{a i \pi/2} \frac{1-(-e^{a \pi i})^q}{1 + e^{a \pi i}} $$ which is $$ e^{a i \pi/2} \frac{1-(-1)^q e^{p \pi i}}{1 + e^{a \pi i}} = e^{a i \pi/2} \frac{1-(-1)^{p+q}}{1 + e^{a \pi i}} = e^{a i \pi/2} \frac{2}{1 + e^{a \pi i}} = \frac{1}{\cos (a\pi/2)}$$
The second sum is $$ \sum_{k=0}^{q-1} (-1)^k e^{-a\rho_k} = \sum_{k=0}^{q-1} (-1)^k e^{-a i \pi /2} e^{-a \pi i k} = e^{-a i \pi/2} \frac{1-(-e^{-a \pi i})^q}{1 + e^{-a \pi i}} $$ which is $$ e^{-a i \pi/2} \frac{1-(-1)^q e^{-p \pi i}}{1 + e^{-a \pi i}} = e^{-a i \pi/2} \frac{1-(-1)^{p+q}}{1 + e^{-a \pi i}} = e^{-a i \pi/2} \frac{2}{1 + e^{-a \pi i}} = \frac{1}{\cos (a\pi/2)}$$
It follows that the original sum and the integral is $$J = \frac{\pi}{2} \frac{1}{\cos(a \pi/2)}.$$