2
$\begingroup$

I was inspired in the shape of the integrals for $\zeta(2)$ in A. Córdoba, Encounters at the interface between Number Theory and Harmonic Analysis, Proceedings of the Segundas Jornadas de Teoría de Números, page 102 (2007) Biblioteca de la Revista Matemática Iberoamericana, to ask to Wolfram Alpha online calculator about integrals of a different kind, to obtain integrals that get zeta values.

Example 1. After a lot of trials with Wolfram Alpha I've found (here was fixed a typo, see the comments from the users, thanks them) $$\int_{-\infty}^\infty\frac{x^2}{-1+\cosh (2x)}dx=\frac{\pi^2}{6}.$$

When I did the change of variables $u=e^{2x}$ I can show (using the change of variables and the evaluation from Wolfram Alpha) that previous integral is evaluated as $$\int_0^\infty\frac{\log^2 u}{-8u+4u^2+4}du=\frac{\pi^2}{6},$$ see it:

integrate (log x)^2/(-8x+4x^2+4) dx from x=0 to x=infinite

Example 2. After a lot experiments, I would like to get $\frac{\pi^4}{90}$ as an integral of previous kind, I say a double integral, and not a multiple of an integral of this kind, some of my attempts were, for example (in comments you can see an ample variety) this $$\int_{-\infty}^\infty\int_{-\infty}^\infty\frac{x^2y^2}{(-1+\cosh (2x))(1+\cosh(y)))}dxdy=\frac{\pi^4}{9},$$ or this $$\int_{-\infty}^\infty\int_{-\infty}^\infty\frac{x^2y^2}{(1+\cosh (2x))(1+\cosh( \sqrt{5}y)))}dxdy=\frac{\pi^4}{90\sqrt{5}}.$$ In comments are more of my attempts to get a similar integral for $\zeta(4)=\frac{\pi^4}{90}$.

These are the codes of previous Example 2:

integrate x^2y^2/((-1+cosh(2x))(1+cosh(y)))dx dy from x=-infty to x=infty, from y=-infty to y=infty

integrate x^2y^2/((1+cosh(2x))(1+cosh(sqrt(5)y)))dx dy from x=-infty to x=infty, from y=-infty to y=infty

I would like to know

Question. Can you provide to me hints to finish the calculation $\int_0^\infty\frac{\log u}{-8u+4u^2+4}du=\frac{\pi^2}{6}$? Can you obtain an integral, not a multiple of an integral of previous kind (see also the comments) to get $\zeta(4)$? Many thanks.

$\endgroup$
  • 1
    $\begingroup$ The integral in your question diverges. $\endgroup$ – Ahmed S. Attaalla Jul 11 '16 at 15:16
  • 1
    $\begingroup$ @user243301 I think there is a typo in your first integral, the result is not $\frac{\pi^2}6$, it's rather $\frac43 \pi^2 \approx 13.15947\cdots$. See my answer below. $\endgroup$ – Olivier Oloa Jul 11 '16 at 16:06
  • 1
    $\begingroup$ Very thanks much, you are rigth, thus I'm sorry @AhmedS.Attaalla There was a typo, that now will fix, the right code was integrate x^2/(-1+cosh(2x)) dx from x=-infinite to x=infinite $\endgroup$ – user243301 Jul 11 '16 at 18:36
  • 1
    $\begingroup$ Very thanks much and I'm sorry @OlivierOloa There was a typo, that now will fix, the right code was integrate x^2/(-1+cosh(2x)) dx from x=-infinite to x=infinite $\endgroup$ – user243301 Jul 11 '16 at 18:36
  • 1
    $\begingroup$ It's better to give the right integral in the title, that's why I've edited it ;) $\endgroup$ – Olivier Oloa Jul 11 '16 at 18:59
5
$\begingroup$

Can you help me to evaluate $\displaystyle \int_{-\infty}^\infty\frac{x^2}{-1+\cosh (2x)}dx$ as $\pi^2/6$?

And can you find a similar integral for $\zeta(4)$?

Hint. One may observe that, for $p\ge1$, one has $$ \begin{align} \int_{-\infty}^\infty\frac{x^{2p}}{-1+\cosh(2x)}dx&=2\int_0^\infty\frac{x^{2p}}{-1+\cosh(2x)}dx \\\\&=2\int_0^\infty\frac{x^{2p}}{-1+\dfrac{e^{2x}+e^{-2x}}2}dx \\\\&=4\int_0^\infty\frac{x^{2p}\:e^{-2x}}{(1-e^{-2x})^2}dx \\\\&=4\sum_{n=1}^\infty n\int_0^\infty x^{2p}e^{-2nx}dx \\\\&=4\:\sum_{n=1}^\infty \frac{\Gamma(2p+1)}{2^{2p+1}n^{2p}} \\\\&= \frac1{2^{2p-1}}\:\Gamma(2p+1)\zeta(2p). \end{align} $$ Then, taking for example $p=1$ and $p=2$, one gets

$$ \begin{align} \int_{-\infty}^\infty\frac{x^2}{-1+\cosh (2x)}dx&=\zeta(2)=\frac{\pi^2 }6 \\\\ \int_{-\infty}^\infty\frac{x^4}{-1+\cosh(2x)}dx&=3\zeta(4)=\frac{\pi^4 }{30}. \end{align} $$

$\endgroup$
  • $\begingroup$ I've modified the question, and I vote up your good answer, very thanks much for your hints that now I will read. I am wait about a week, and after I will accept an answer. Notice that but my mistake and my clumsiness (= torpeza in spanish) the question was modified. $\endgroup$ – user243301 Jul 11 '16 at 18:40
  • $\begingroup$ Also closed your forms are nice Olivier Oloa! $\endgroup$ – user243301 Jul 11 '16 at 18:47
  • $\begingroup$ @user243301 I've edited my answer accordingly. Thanks. $\endgroup$ – Olivier Oloa Jul 11 '16 at 18:55
  • $\begingroup$ You are quick! Very thanks much one more time for your results. $\endgroup$ – user243301 Jul 11 '16 at 18:56
  • 1
    $\begingroup$ @user243301 Since $\cosh 2x-1=\sinh^2x$, one might consider the general form of $$ \int_{-\infty}^{\infty}\frac{x^{a-1}}{\sinh^2bx}\ dx=2\cdot\frac{\partial}{\partial b}\int_{0}^{\infty}\frac{x^{a-1}}{1-e^{bx}}\ dx $$ $\endgroup$ – Anastasiya-Romanova 秀 Jul 12 '16 at 3:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy