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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.

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We have \begin{equation} \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 \end{equation} 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 \begin{equation} \frac{1}{2}\int_{0}^{\infty} \frac{1}{\cosh^{2} x} \, \frac{dx}{\sqrt{x}} \end{equation} which is evaluated to

\begin{equation} \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) \end{equation}

where we use Hardy's formula to evaluate the negative argument of Dirichlet eta function \begin{equation} \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) \end{equation} and the relation of the Dirichlet eta function and the Riemann zeta function \begin{equation} \eta(s) = \left(1-2^{1-s}\right)\zeta(s) \end{equation}

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  • 2
    $\begingroup$ I just edited that answer because I had for some reason previously deleted a statement about why the formula is valid if $\text{Re}(a) >-1$. $\endgroup$ – Random Variable Nov 8 '14 at 20:12
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$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\dsc}[1]{\displaystyle{\color{#c00000}{#1}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,{\rm Li}_{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ \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 .

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