integration - Evaluating $int_{0}^{infty} frac{cos {ax}}{cosh{x}}dx$

$\int_{0}^{\infty} \frac{\cos {ax}}{\cosh{x}}dx$

$a$ is a real number.
Hint: Consider a suitable rectangular contour

I know that $\frac{1}{\cosh z}$ has simple poles when $z=\frac{(2n-1)\pi}{2}i$.

What should I do next?
I would appreciate any help. Thank you.

Answer

First off, let us notice that the function $\cos (ax) / \cosh (x)$ does not change under $x \mapsto -x$. So,

$$\int\limits_0^\infty \frac{\cos ax}{\cosh x} \, dx = \frac{1}{2} \int\limits_{-\infty}^\infty \frac{\cos ax} {\cosh x} \, dx.$$

Next, we should take the contour $\gamma_M$ to be like that. Line (1) is a part of real line $[-M, \, M]$, and the upper line (3) is the same line translated by $\pi i$.

The integral over line (3) could be rewritten as

$$\int\limits_{(3)} \frac{\cos ax}{ \cosh x} \, dx = \int\limits_{(1)} \frac{\cos a(x + \pi i)}{\cosh (x + \pi i)} \, dx = (*).$$

Notice, that $\cosh (x + \pi i) = - \cosh x$ and

$$\cos a (x + \pi i) = \cos (ax) \cos ( a \pi i) - \sin (ax) \sin ( a \pi i).$$

Therefore,

$$(*) = -\cos (a \pi i )\int\limits_{(1)} \frac{\cos a x}{\cosh x} \, dx + \sin (a \pi i) \int\limits_{(1)} \frac{\sin a x}{\cosh x} \, dx.$$

Last integral clearly equals zero, as it is the sum of symmetric and antisymmetric function over symmetric interval.

Integrals over lines (2) and (4) go to zero as $M \to \infty$. So, we have now, via residuals theorem

$$2 \pi i \, \mathrm{res} \; \frac{\cos a x}{\cosh x} = \lim\limits_{M \to \infty} \int\limits_{\gamma_M} \frac{\cos a x}{\cosh x} \, dx = (1 - \cos (a \pi i)) \int\limits_{-\infty}^\infty \frac{\cos a x}{\cosh x} \, dx$$