# Evaluate $\int_0^{\infty } \frac{x^2 \tanh \left(x^2\right)}{\cosh \left(x^2\right)} \, dx$

How to show $$\int_0^{\infty } \frac{x^2 \tanh \left(x^2\right)}{\cosh \left(x^2\right)} \ dx=\frac{1}{2} \sqrt{\pi } \beta \left(\frac{1}{2}\right)$$ Any help will be appreciated.

• Numerical computation shows a large gap between the integral and the supposed answer. Are you sure you didn't mean to have $\int_0^\infty x^2\operatorname{sech}^2(x)\tanh^2(x)~\mathrm dx$? If you did, then it should boil down to rewriting it in terms of $e^{-x}$ and appropriately series expanding in terms of $e^{-x}$. Sep 1, 2019 at 2:54
• I have the 7th edition (2007). What's printed in the book is indeed as posted. (this doesn't rule out typos from the publisher/authors) Sep 1, 2019 at 3:44

That $$\cosh^2$$ reminds me of one integral representation for the $$\zeta$$ function. Let's start from scratch: for any $$s>0$$,
$$\sum_{n\geq 1}\frac{(-1)^{n+1}}{n^s} = \frac{1}{\Gamma(s)}\int_{0}^{+\infty}\frac{x^{s-1}}{e^x+1}\,dx$$ is a straightforward consequence of the (inverse) Laplace transform. If we apply integration by parts we get that $$\eta(s) = \frac{1}{\Gamma(s+1)}\int_{0}^{+\infty}\frac{x^s e^{x}}{(e^x+1)^2}\,dx = \frac{2^{s-1}}{\Gamma(s+1)}\int_{0}^{+\infty}\frac{x^s}{\cosh^2(x)}\,dx$$ holds for any $$s>-1$$. Similarly, $$\beta(s)=\sum_{n\geq 0}\frac{(-1)^n}{(2n+1)^s}=\frac{1}{2^{s+1}\Gamma(s)}\int_{0}^{+\infty}\frac{z^{s-1}}{\cosh(z/2)}\,dz$$ for any $$s>0$$ leads to $$\beta(s) = \frac{1}{2\Gamma(s+1)}\int_{0}^{+\infty}\frac{z^s\sinh(z) }{\cosh^2(z)}\,dz$$ for any $$s>-1$$. It follows that $$\frac{\sqrt{\pi}}{2}\beta\left(\tfrac{1}{2}\right) = \frac{1}{2}\int_{0}^{+\infty}\frac{\sqrt{z}\sinh(z)}{\cosh^2(z)}\,dz=\int_{0}^{+\infty}\frac{z^2\tanh(z^2)}{\cosh(z^{\color{red}{2}})}\,dz.$$ Long story short: there's a typo.