# How to find closed-form of $\int_{0}^{+\infty} \operatorname{sech}^2 (x^2)\,dx$

How to find this integral closed form:

$$I=\int_{0}^{+\infty}\operatorname{sech}^2{(x^2)}\,dx$$

where $\operatorname{sech}{(x)}$ is defined as secant of hyperbolic function.

This problem form is very simple and it's interesting problem, but I use computer to help me to find its closed-form and W|A turns its numerical result $$I=\int_{0}^{+\infty}\operatorname{sech}^2{(x^2)}\,dx\approx 0.952781\ldots$$

We have $$\int_{0}^{\infty} \frac{x^{a}}{\cosh^{2} x} \ dx=\frac{2 \Gamma(a+1) \eta(a)}{2^{a}}\qquad,\qquad\mbox{for}\,\,a>-1$$ where $\eta(a)$ is the Dirichlet eta function. Proof can be seen here.

Making substitution $x^2\mapsto x$ and using ${\rm{sech}}^2 x=\dfrac{1}{{\rm{cosh}}^2 x}$, then the considered integral can be rewritten as $$\frac{1}{2}\int_{0}^{\infty} \frac{1}{\cosh^{2} x} \, \frac{dx}{\sqrt{x}}$$ which is evaluated to

$$\sqrt{2}\, \Gamma\left(\frac{1}{2}\right) \eta\left(-\frac{1}{2}\right)=\frac{\left(2\sqrt{2}-1\right)}{2\sqrt{2\pi}}\zeta\left(\frac{3}{2}\right)$$

where we use Hardy's formula to evaluate the negative argument of Dirichlet eta function $$\eta(-s) = \frac{s}{\pi^{1+s}} \frac{2^{1+s}-1}{2^{s}-1} \sin\left({\pi s \over 2}\right) \Gamma(s)\eta(s+1)$$ and the relation of the Dirichlet eta function and the Riemann zeta function $$\eta(s) = \left(1-2^{1-s}\right)\zeta(s)$$

• I just edited that answer because I had for some reason previously deleted a statement about why the formula is valid if $\text{Re}(a) >-1$. – Random Variable Nov 8 '14 at 20:12


We used the well known identity: $${\tanh\pars{z} \over z} =8\sum_{n\ =\ 0}^{\infty}{1 \over 4z^{2} + \bracks{\pars{2n + 1}\pi}^{2}}$$ and the $\ds{\color{#c00000}{\mbox{red result}}}$ was found with one of my previous answers .