Krdytkyu

Wolfram Alpha evaluates this integral numerically as

$$int_0^{infty} frac{x^2}{cosh^2 (x^2)} dx=0.379064 dots$$

Its value is apparently

$$frac{sqrt{2}-2}{4} sqrt{pi}~ zeta left( frac{1}{2} right)=0.37906401072dots$$

How would you solve this integral?

Obviously, we can make a substitution $t=x^2$

begin{align}
int_0^{infty} frac{x^2}{cosh^2 (x^2)} dx&=frac{1}{2} int_0^{infty} frac{sqrt{t}}{cosh^2 (t)} dt\[10pt]
&=int_0^{infty} frac{sqrt{t}}{cosh (2t)+1} dt\[10pt]
&=frac{1}{2 sqrt{2}}int_0^{infty} frac{sqrt{u}}{cosh (u)+1} du
end{align}

We could use geometric series since $cosh (u) geq 1$, but I don't know how it will help.

$$I=frac{1}{2sqrt{2}}int_{0}^{+infty}frac{sqrt{u},du}{1+cosh(u)}=frac{1}{sqrt{2}}int_{1}^{+infty}frac{sqrt{log v}}{(v+1)^2},dv=frac{1}{sqrt{2}}int_{0}^{1}frac{sqrt{-log v}}{(1+v)^2},dv tag{1}$$
but since
$$ int_{0}^{1}v^k sqrt{-log v},dv = frac{sqrt{pi}}{2(1+k)^{3/2}} tag{2}$$
by expanding $frac{1}{(1+v)^2}$ as a Taylor series we get:

$$ I = frac{1}{sqrt{2}}sum_{ngeq 0}(-1)^n (n+1)frac{sqrt{pi}}{2(1+n)^{3/2}} = color{red}{frac{sqrt{pi}}{2sqrt{2}}cdotetaleft(frac{1}{2}right)}tag{3}$$

and the claim follows from the well-known:
$$ eta(s) = (1-2^{1-s}),zeta(s)tag{4} $$
that gives an analytic continuation for the $zeta$ function.

Hint:

Consider the parametric integral

begin{equation}
I(a) = int_0^infty frac{t^{a-1}}{cosh^{2} t} dt=4 int_{0}^{infty} frac{t^{a-1}}{(e^{t}+e^{-t})^{2}} dt = 4 int_{0}^{infty}frac{t^{a-1} e^{-2t}}{(1+e^{-2t})^{2}} dt
end{equation}

Hence, your integral is simply

begin{equation}
int_0^{infty} frac{x^2}{cosh^2x^2} dx = -2 frac{partial}{partial b}left[int_0^{ infty}frac{t^{a-1}}{1+e^{bt}} dtright]_{a=frac{1}{2} , b=2}
end{equation}

I believe you can evaluate the last expression by your own.

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newcommand{dd}{mathrm{d}}
newcommand{ds}[1]{displaystyle{#1}}
newcommand{expo}[1]{,mathrm{e}^{#1},}
newcommand{half}{{1 over 2}}
newcommand{ic}{mathrm{i}}
newcommand{iff}{Leftrightarrow}
newcommand{imp}{Longrightarrow}
newcommand{ol}[1]{overline{#1}}
newcommand{pars}[1]{left(, #1 ,right)}
newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
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begin{align}
&bbox[10px,#ffd]{int_{0}^{infty}{x^{2} over cosh^{2}pars{x^{2}}},dd x},,
stackrel{x to x^{1/2}}{=},,,
2int_{0}^{infty}{x^{1/2}expo{2x} over pars{expo{2x} + 1}^{2}},dd x
\[5mm] = &
-int_{x = 0}^{x to infty}x^{1/2},ddpars{{1 over expo{2x} + 1}} =
halfint_{0}^{infty}{x^{-1/2}expo{-2x} over 1 + expo{-2x}},dd x
\[5mm] stackrel{2x to x}{=},,,&
{root{2} over 4}int_{0}^{infty}{x^{-1/2}expo{-x} over 1 + expo{-x}},dd x
\[5mm] = &
{root{2} over 4}sum_{n = 0}^{infty}pars{-1}^{n}
int_{0}^{infty}x^{-1/2}expo{-pars{n + 1}x},dd x
\[3mm] stackrel{pars{n + 1}x to x}{=},,,&
{root{2} over 4}sum_{n = 0}^{infty}
{pars{-1}^{n} over pars{n + 1}^{1/2}} overbrace{%
int_{0}^{infty}x^{-1/2}expo{-x},dd x}^{ds{Gammapars{half} = root{pi}}}
\[5mm] = &
{root{2} over 4},root{pi}sum_{n = 1}^{infty}{pars{-1}^{n + 1} over n^{1/2}}
end{align}