Evaluating: $\int_{-\infty}^{\infty}{\frac{1}{\cosh(kx)}dx}$

Question

How can you integrate: $$\int_{-\infty}^{\infty}{\dfrac{1}{\cosh(kx)}dx}$$

I know that: $$\int_{0}^{\infty}{\dfrac{1}{\cosh(x)}dx}=\int_{0}^{\infty}{\dfrac{2}{e^x+e^{-x}}dx}=\int_{0}^{\infty}{\dfrac{2e^{-x}}{1+e^{-2x}}dx}=$$ $$=2\int_{0}^{\infty}{(1-1/3+1/5-1/7+...)dx}=2\pi/4=\pi/2$$

But i don´t know how to integrate: $$\int_{-\infty}^{\infty}{\dfrac{1}{\cosh(kx)}dx}$$

Answer 1

In order to collect all the contributions from @marty cohen and @Stephen Montgomery-Smith, i will write the answer:

$$\int_{0}^{\infty}{\dfrac{1}{\cosh(x)}dx}=\pi /2$$

Let $u=kx\implies du=kdx\implies dx=\dfrac{1}{k}du$

Substituting: $$\int_{-\infty}^{\infty}{\dfrac{1}{\cosh(kx)}dx}=\dfrac{1}{k}\int_{-\infty}^{\infty}{\dfrac{1}{\cosh(u)}du}$$

Since $\cosh(u) =\cosh(-u)\implies \int_{-\infty}^{\infty}\dfrac{1}{\cosh(u)}du =2\int_{0}^{\infty}\dfrac{1}{\cosh(u)}du$

So: $$\int_{-\infty}^{\infty}{\dfrac{1}{\cosh(kx)}dx}=2\int_{0}^{\infty}\dfrac{1}{\cosh(u)}du=\dfrac{\pi}{k}$$

Answer 2

Since $\cosh(x) =\cosh(-x) $, $\int_{-\infty}^{\infty} =2\int_{0}^{\infty} $.

$\int_{-\infty}^{\infty} \frac{dx}{\cosh(kx)} =\frac1{k}\int_{-\infty}^{\infty} \frac{kdx}{\cosh(kx)} =\frac1{k}\int_{-\infty}^{\infty} \frac{dx}{\cosh(x)} $