# Prove $\int_0^{\infty} \frac{x^2}{\cosh^2 (x^2)} dx=\frac{\sqrt{2}-2}{4} \sqrt{\pi}~ \zeta \left( \frac{1}{2} \right)$

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

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

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.

• Maybe $\Gamma (s) \zeta (s)=\int_0^{\infty}\frac{u^{s-1}}{e^u-1}du$ helps. Jun 9 '16 at 22:01
• It probably does, forgot about this, thank you Jun 9 '16 at 22:03

## 3 Answers

$$I=\frac{1}{2\sqrt{2}}\int_{0}^{+\infty}\frac{\sqrt{u}\,du}{1+\cosh(u)}=\frac{1}{\sqrt{2}}\int_{1}^{+\infty}\frac{\sqrt{\log v}}{(v+1)^2}\,dv=\frac{1}{\sqrt{2}}\int_{0}^{1}\frac{\sqrt{-\log v}}{(1+v)^2}\,dv \tag{1}$$ 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:

$$I = \frac{1}{\sqrt{2}}\sum_{n\geq 0}(-1)^n (n+1)\frac{\sqrt{\pi}}{2(1+n)^{3/2}} = \color{red}{\frac{\sqrt{\pi}}{2\sqrt{2}}\cdot\eta\left(\frac{1}{2}\right)}\tag{3}$$

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.

• Can you help me understand what substitution you made to get$$\frac 1{2\sqrt2}\int\limits_0^{\infty}du\,\frac {\sqrt u}{\cosh u+1}=\frac 1{\sqrt2}\int\limits_1^{\infty}dv\,\frac {\sqrt{\log v}}{(v+1)^2}$$? Jun 26 '18 at 23:26
• @FrankW.: $u=\log v$. Jun 27 '18 at 12:18

Hint:

Consider the parametric integral

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

Hence, your integral is simply

$$\int_0^{\infty} \frac{x^2}{\cosh^2x^2}\ dx = -2\ \frac{\partial}{\partial b}\left[\int_0^{ \infty}\frac{t^{a-1}}{1+e^{bt}} \ dt\right]_{a=\frac{1}{2}\ ,\ b=2}$$

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


\begin{align} &\bbox[10px,#ffd]{\int_{0}^{\infty}{x^{2} \over \cosh^{2}\pars{x^{2}}}\,\dd x}\,\,\ \stackrel{x\ \to\ x^{1/2}}{=}\,\,\,\ 2\int_{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}\,\dd\pars{{1 \over \expo{2x} + 1}} = \half\int_{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{\Gamma\pars{\half}\ =\ \root{\pi}}} \\[5mm] = &\ {\root{2} \over 4}\,\root{\pi}\sum_{n = 1}^{\infty}{\pars{-1}^{n + 1} \over n^{1/2}} \end{align}

With the identity $$\ds{\sum_{n = 1}^{\infty}{\pars{-1}^{n + 1} \over n^{s}} = \pars{1 - 2^{1 - s}}\zeta\pars{s}}$$: \begin{align} \bbox[#ffd,10px]{\int_{0}^{\infty}{x^{2} \over \cosh^{2}\pars{x^{2}}}\,\dd x} & = {\root{2} \over 4}\,\root{\pi}\pars{1 - 2^{1 - 1/2}}\zeta\pars{\half} \\[5mm] & = \bbox[10px,border:1px groove navy]{{\root{2} - 2 \over 4}\,\root{\pi}\zeta\pars{\half}} \approx 0.3791 \end{align}