TITLE: How to find closed-form of $\int_{0}^{+\infty} \operatorname{sech}^2 (x^2)\,dx$
QUESTION [11 upvotes]: 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 $$I=\int_{0}^{+\infty}\operatorname{sech}^2{(x^2)}\,dx\approx 0.952781\ldots$$
Thank you for your help.

REPLY [4 votes]: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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\begin{align}&\color{#66f}{\large%
\int_{0}^{\infty}\sech^{2}\pars{x^{2}}\,\dd x}
=\int_{x\ =\ 0}^{x\ \to\ \infty}{\dd\tanh\pars{x^{2}} \over 2x}
=\half\int_{0}^{\infty}{\tanh\pars{x^{2}} \over x^{2}}\,\dd x
\\[5mm]&=4\sum_{n\ =\ 0}^{\infty}\ \overbrace{%
\int_{0}^{\infty}{\dd x \over 4x^{4} + \bracks{\pars{2n + 1}\pi}^{2}}}
^{\ds{\color{#c00000}{1 \over 4\root{\pi}\pars{2n + 1}^{3/2}}}}\ =\
{1 \over \root{\pi}}\sum_{n\ =\ 0}^{\infty}{1 \over \pars{2n + 1}^{3/2}}
\\[5mm]&={1 \over \root{\pi}}\bracks{%
\sum_{n\ =\ 1}^{\infty}{1 \over n^{3/2}}-
\sum_{n\ =\ 1}^{\infty}{1 \over \pars{2n}^{3/2}}}
={1 \over \root{\pi}}\pars{%
\sum_{n\ =\ 1}^{\infty}{1 \over n^{3/2}}-
2^{-3/2}\sum_{n\ =\ 1}^{\infty}{1 \over n^{3/2}}}
\\[5mm]&={1 - 2^{-3/2} \over \root{\pi}}\sum_{n\ =\ 1}^{\infty}{1 \over n^{3/2}}
=\color{#66f}{\large{1 - 2^{-3/2} \over \root{\pi}}\,\zeta\pars{3 \over 2}}
\approx {\tt 0.9528}
\end{align}

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 .