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Does anyone know how to compute analytically the following integral:

$$\int\limits_{0}^{+\infty}\dfrac{x^2\mathrm{d}x}{e^{x}-1}$$

It should be equal to $2\zeta(3)$ according to Maple. I tried the following using the binomial theorem for negative integer exponents:

$$I = \int\limits^{+\infty}_{0} e^{-x}(1-e^{-x})^{-1}x^2\mathrm{d}x = \int\limits^{+\infty}_{0}\left[\sum_{k=0}^{+\infty}(-1)^k(-1)^ke^{-(k+1)x}\right]x^2\mathrm{d}x=\int\limits^{+\infty}_{0}\left[\sum_{k=0}^{+\infty}(-1)^{2k}e^{-(k+1)x}\right]x^2\mathrm{d}x$$

After another change of variables, $y=(k+1)x$:

$$I = \sum_{k=0}^{+\infty}(-1)^{2k} \frac{1}{(k+1)^3}\int\limits_0^{+\infty} y^2e^{-y}\mathrm{d}y$$

The keen eye might recognize $\int\limits_0^{+\infty} y^2e^{-y}\mathrm{d}y$ as the gamma function, $\Gamma(3)=(3-1)!=2$. This, together with a slight nudge to the bottom limit of the summation we can rewrite things as:

$$I = \Gamma(3)\sum_{k=1}^{+\infty} \dfrac{(-1)^{2k}}{k^3}$$

And i see immediately (since the beginning in fact...) an infinite sum that makes me troubles and i can't get rid of. I tried to found if i did any trivial error but i'm focusing since to many hours to found it. That's why I need an external view to point me out my obvious error.

Thanks in advance for your help

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  • $\begingroup$ Instead of using tricks to emulate displayed math mode (\limits and \dfrac), simply use displayed math model by using $$ instead of $. Also try to have a descriptive title, and avoid shorthands like "thx", since you're not paying by the letter here. $\endgroup$
    – Asaf Karagila
    Commented Nov 13, 2020 at 9:27
  • $\begingroup$ Just to be sure, do you know how to justify that $\int_0^\infty x^2\sum_{n\ge 1} e^{-nx}dx=\sum_{n\ge 1}\int_0^\infty x^2 e^{-nx}dx$ ? It follows from that $\lim_{N\to \infty}\int_0^\infty x^2\frac{e^{-Nx}}{e^x-1}dx=0$. $\endgroup$
    – reuns
    Commented Nov 13, 2020 at 11:05

1 Answer 1

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You're actually almost there! Note that

$$(-1)^{2k} = \Big( (-1)^2 \Big)^k = 1$$

Thus,

$$\sum_{k=1}^\infty \frac{(-1)^{2k}}{k^3} = \sum_{k=1}^\infty \frac{1}{k^3} = \zeta(3)$$

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  • $\begingroup$ Thx. Indeed that was something obvious as i suspected it. Thx a lot! $\endgroup$ Commented Nov 13, 2020 at 9:00

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