Prove $\int_0^\infty\operatorname{sech} x\,dx=\pi/2$, and deduce $\int_0^1\operatorname{sech}^{-1}x\,dx$

Question

Prove $$\int_0^\infty\operatorname{sech} x\,dx=\pi/2$$ and deduce $$\int_0^1\operatorname{sech}^{-1}x\,dx$$

I can prove the first statement (see below), but I was unable to deduce the value of the second integral.

Answer 1

Note that by setting $u = \operatorname{sech}^{-1} x$ we have that \begin{align} \int^1_0 \operatorname{sech}^{-1} x\ dx = \int^\infty_0 u \tanh u \operatorname{sech} u\ du \end{align} then by integration by parts you should be done.

Answer 2

When it comes to the second part of your question:

\begin{equation} I = \int f^{-1}(t)\:dt \end{equation}

Let $x = f^{-1}(t)$ or $t = f(x)$, then

\begin{equation} I = \int x \cdot f'(x)\:dx \end{equation}

Integrate by parts:

\begin{align} v'(x) &= f'(x) & u(x) &= x \\ v(x) &= f(x) & u'(x) &= 1 \end{align}

Thus,

\begin{align} I &= \int x \cdot f'(x)\:dx = x \cdot f(x) - \int f(x) \cdot 1 \:dx \\ &= x \cdot f(x) - F(x) \end{align}

Where $F'(x) = f(x)$. As $x = f^{-1}(t)$ we have

\begin{align} I &= \int f^{-1}(t)\:dt = x \cdot f(x) - F(x) = t \cdot f^{-1}(t) - F\left(f^{-1}(t)\right) \end{align}

So in this case $x = \operatorname{arcsech}(t) \rightarrow f(x) = t$.

\begin{equation} F'(x) = \operatorname{sech}(x) \rightarrow F(x) = 2\arctan\left(\tanh\left(\frac{x}{2}\right) \right) \end{equation}

Hence,

\begin{equation} F\left(\operatorname{arcsech}(x)\right) = 2\arctan\left(\tanh\left(\frac{\operatorname{arcsech}(x)}{2}\right) \right) \end{equation}

And thus:

\begin{equation} \int \operatorname{arcsech}(t) \:dx = t\cdot \operatorname{arcsech}(t) - 2\arctan\left(\tanh\left(\frac{\operatorname{arcsech}(x)}{t}\right) \right) + C \end{equation}

Where $C$ is the constant of integration.