Using ML inequality

Question

I'm trying to work out the following integral:

$$\int_{-\infty}^\infty \frac{1}{\cosh(x)} dx $$

by complex analysis methods and as a result of the contour I chose (a rectangle) one of the things I need to show is that the integral:

$$\int_{0}^{\pi} \frac{idy}{\cosh(L+iy)} $$ goes to $0$ as L approaches $\infty$ using the ML inequality.

Could someone show me how to do this? :)

Answer 1

$$\left\|2\cosh(L+iy)\right\|^2 = (e^{L+iy}+e^{-L-iy})\cdot(e^{L-iy}+e^{-L+iy})=e^{2L}+e^{-2L}+2\cos(y) $$ gives: $$\left\|\cosh(L+iy)\right\| \geq \sinh(L) $$ hence the integral $$ \int_{0}^{\pi}\frac{i}{\cosh(L+iy)}\,dy $$ is bounded by $$ \frac{\pi}{\sinh(L)} $$ in absolute value. That gives: $$ \int_{-\infty}^{+\infty}\frac{dz}{\cosh z}+\int_{\pi i+\infty}^{\pi i-\infty}\frac{dz}{\cosh z} = 2\pi i\,\text{Res}\left(\frac{1}{\cosh z},z=\frac{\pi i}{2}\right) = 2\pi $$ but the LHS is just twice the original integral since $\cosh(z+\pi i)=-\cosh(z)$, so:

$$ \int_{\mathbb{R}}\frac{dx}{\cosh x}=\color{red}{\pi}.$$

Answer 2

$$ \dfrac1{\cosh (x)} = \dfrac{2e^x}{e^{2x}+ 1} $$

Use the substitution $ y = e^x$, we get $ \dfrac{dy}{dx} = e^x = y $.

$$ \int_{-\infty}^\infty \dfrac1{\cosh(x)} \, dx = \int_0^\infty \dfrac{2}{y^2 + 1} \, dy = 2 \arctan (y)\bigg ]_0^{\infty} =\boxed \pi .$$