Finding definite integral using contour integration

Question

Wanting to find the value of the integral $\int_{0}^{\infty} \dfrac{1}{\cosh (x)} dx$ and know I have to find the residue at $\dfrac{\pi i}{2}$ which I find to be $-i$. So far so good. So then, I know to take the contour from R to -R, $-R to -R$ to $-R-i \pi$ to $-R-i \pi$ to $R + i \pi$ back to R. Now, where I'm getting caught up on is how to proceed. First, I get the wrong expression for the RHS which becomes $2 \pi i * -i= 2\pi$ and on the LHS, I can't work out how to get the definite integral out of the tangle of 4 integrals. Any ideas?

Answer 1

Going counterclockwise around a rectangle with vertices at $z=-R, z=R, z=R+ i \pi$, and $z=-R+ i \pi$,

$$ \int_{-R}^{R} \frac{1}{\cosh x} \ dx + i \int_{0}^{\pi} \frac{1}{\cosh (R+it)} \ dt + \int_{R}^{-R} \frac{1}{\cosh( t + i \pi)} \ dt + i \int_{\pi}^{0} \frac{1}{\cosh (-R + it)} \ dt$$

$$ = 2 \pi i \ \text{Res} \left[\frac{1}{\cosh z}, \frac{i \pi}{2} \right] = 2 \pi$$

But $$ \begin{align} \cosh (t + i \pi) &= \frac{e^{t + i \pi} + e^{-t-i \pi}}{2} \\ &= \frac{-e^{t}-e^{-t}}{2} \\ &= -\frac{e^{t}+e^{-t}}{2} \\ &= -\cosh t \end{align}$$

So combining the first and third integrals,

$$ = 2 \int_{-R}^{R} \frac{1}{\cosh x} \ dx + i \int_{0}^{\pi} \frac{1}{\cosh (R+it)} \ dt - i \int_{0}^{\pi} \frac{1}{\cosh (-R + it)} \ dt = 2 \pi$$

Now we need to show that as $R \to \infty$, the second and third integrals vanish.

Using the ML inequality,

$$ \begin{align} \Big| \int_{0}^{\pi} \frac{1}{\cosh (R +it)} \ dt \Big| &\le \int_{0}^{\pi} \Big|\frac{1}{\cosh (R+it)} \Big| \ dt \\ &= \pi \ \frac{2}{|e^{R+it} +e^{-R-it}|}\\ &\le 2 \pi \frac{1}{|e^{R+it}|-|e^{-R-it}|} \ \ \text{reverse triangle inequality} \\ &= 2 \pi \frac{1}{e^{R}-e^{-R}} \to 0 \ \text{as} \ R \to \infty \end{align}$$

Showing that the the other integral vanishes is similar.

And we have

$$ 2 \int_{-\infty}^{\infty} \frac{1}{\cosh x} \ dx = 2 \pi$$

or

$$ \int_{0}^{\infty} \frac{1}{\cosh x} \ dx = \frac{\pi}{2}$$