Why does $\int_{-\infty}^\infty \operatorname{sech} x \, \mathrm{d}x =\pi\;$?

Question

According to Desmos, $$\int_{-\infty}^\infty \operatorname{sech} x \, \mathrm{d}x = 3.14159265359$$

Why is $\pi$ here in this definite integral?

Answer 1

• That's magic of Gudermannian function, a bridge (other than Jacobi elliptic functions) relates trigonometric to hyperbolic functions:

\begin{align} \operatorname{gd} x &= \int_0^x \operatorname{sech} t \, dt \\ &= \tan^{-1} \sinh x \\ \operatorname{gd}^{-1} t &= \int_0^t \operatorname{sec} x \, dx \\ &= \tanh^{-1} \sin t \\ \end{align}

• Further coincidence:

\begin{align} \operatorname{gd} x &= \tan^{-1} \sinh x \\ &= \sin^{-1} \tanh x \\ \operatorname{gd}^{-1} t &= \tanh^{-1} \sin t \\ &= \sinh^{-1} \tan t \end{align}

• Domain for $t$ is always $\left( -\frac{\pi}{2},\frac{\pi}{2} \right)$

Answer 2

Note that $\operatorname{sech}(x) = \frac{2e^x}{e^{2x}+1}.$ In that case:

$$I= \int_{-\infty}^{\infty} \operatorname{sech}(x) dx = \int_{-\infty}^{\infty} \frac{2e^x}{e^{2x}+1} dx.$$

For the later, $u=e^x \implies du=e^xdx$,

$$I = 2\int_0^{\infty} \frac{1}{u^2+1} du = 2 \left(\lim_{u \to \infty} \arctan(u) - 0 \right) = \pi.$$

Answer 3

Just as you do it with trigonometric functions, in the case of hyperbolic functions, you can use the corresponding tangent half-angle substitution $$x=2 \tanh ^{-1}(t)\implies dx=\frac{2}{1-t^2}\,dt$$ $$I=\int \text{sech}(x)\,dx=\int\frac{2}{t^2+1}\,dt=2 \tan ^{-1}(t)=2 \tan ^{-1}\left(\tanh \left(\frac{x}{2}\right)\right)$$ Similarly $$J=\int \text{csch}(x)\,dx=\log \left(\tanh \left(\frac{x}{2}\right)\right)$$