Definite Integral to Infinity of Cosine of a x over Hyperbolic Cosine of b x

Theorem

 * $\displaystyle \int_0^\infty \frac {\cos a x} {\cosh b x} \rd x = \frac \pi {2 b} \sech \frac {a \pi} {2 b}$

where:
 * $a$ and $b$ are positive real numbers
 * $\sech$ denotes the hyperbolic secant function.

Proof
We have, as $b, n > 0$:

We similarly have:

So:

By Mittag-Leffler Expansion for Hyperbolic Secant Function:


 * $\displaystyle \pi \map \sech {\pi z} = 4 \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {2 n + 1} {\paren {2 n + 1} + 4 z^2}$

Setting $z = \dfrac a {2 b}$ gives:


 * $\displaystyle \pi \map \sech {\frac {a \pi} {2 b} } = 4 \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {2 n + 1} {\paren {2 n + 1} + 4 \paren {\frac a {2 b} }^2}$

We therefore have: