Show that the integral $$\int_0^{\infty } \frac{x^2 \tanh \left(x^2\right)}{\cosh \left(x^2\right)} \ dx$$ evaluates to: $$\frac{1}{2} \sqrt{\pi } \beta \left(\frac{1}{2}\right)$$ The original question is wrong and now this question is trivial. Please see Jack's answer to know more about Hurwitz if someone need to.

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    $\begingroup$ 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 $e^{-x}$. $\endgroup$ – Simply Beautiful Art Sep 1 '19 at 2:54
  • $\begingroup$ 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) $\endgroup$ – Lee David Chung Lin Sep 1 '19 at 3:44
  • $\begingroup$ @SimplyBeautifulArt Indeed, you're right, the book I mentioned has a typo. $\endgroup$ – 雄凤山 Sep 9 '19 at 9:18

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.

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    $\begingroup$ nice work (+1). I spent some time on this problem before you pointed out the typo. I think sometimes we have to verify the problem before we start working on it as typos are pretty common in math. $\endgroup$ – Ali Shather Sep 1 '19 at 15:21
  • $\begingroup$ Thank you for pointing out the typo in G&R. Also, a neat solution. $\endgroup$ – 雄凤山 Sep 9 '19 at 9:17

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