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

$a$ 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.

Teddy38
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GouldBach
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  • I think you can apply Cauchy's residue theorem here in the limit sense even if there are infinitely many poles. See Cauchy's residue theorem with an infinite number of poles. – Alex Vong Nov 30 '17 at 11:07
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2 Answers2

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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

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

$$\int\limits_{0}^\infty \frac{\cos a x}{\cosh x} \, dx = \pi i \, \mathrm{res} \; \frac{\cos a x}{\cosh x} * \frac{1}{(1 + \cos (a \pi i))}$$

rpr
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Pavel Ievlev
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Let us assume $a>0$. By geometric series we have $$\begin{eqnarray*} \int_{0}^{+\infty}\frac{\cos(ax)}{\cosh(x)}\,dx &=& 2\sum_{n\geq 0}(-1)^n\int_{0}^{+\infty}\cos(ax) e^{-(2n+1)x}\,dx\\&=&2\sum_{n\geq 0}\frac{(2n+1)(-1)^n}{a^2+(2n+1)^2}\end{eqnarray*}$$ and we may notice that $$\operatorname*{Res}_{z=(2n+1)}\left(\frac{\pi}{2}\sec\frac{\pi z}{2}\right)=(-1)^{n+1} $$ so by invoking Herglotz' trick we get: $$ \int_{0}^{+\infty}\frac{\cos(ax)}{\cosh(x)}\,dx = \color{red}{\frac{\pi}{2\cosh\frac{\pi a}{2}}}. $$

Jack D'Aurizio
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