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Observing

$$\int_{0}^{\infty}e^{-x}\ln^2(x)\mathrm dx=\gamma^2+{\pi^2\over 6}\tag1$$

The general of $(1)$ would be

$$\int_{0}^{\infty}x^{n-1}e^{-x^n}\ln^2(x)\mathrm dx={6\gamma^2+\pi^2\over 6n^3}\color{red}?\tag2$$

Where $n\ge 1$

My try:

Too complicate, I don't where to begin.

How may one prove $(2)?$

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By the change of variable $$ u=x^n,\qquad du=n\cdot x^{n-1}dx, $$ one gets $$ \begin{align} \int_{0}^{\infty}x^{n-1}e^{-x^n}\ln^2(x)\:\mathrm dx=\frac1{n^2}\int_{0}^{\infty}x^{n-1}e^{-x^n}\ln^2(x^{n})\:\mathrm dx=\frac1{n^3}\int_{0}^{\infty}e^{-u}\ln^2(u)\:\mathrm du \end{align} $$then you can finish with your result $(1)$.

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  • $\begingroup$ thank you, you are too quick!, I thought it was sort of interesting result, but it seems not. $\endgroup$ – gymbvghjkgkjkhgfkl May 12 '17 at 11:50
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    $\begingroup$ @AbuHurayrah France has great mathematicians :P $\endgroup$ – ParaH2 May 12 '17 at 11:56
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    $\begingroup$ yes you are absolutely right, he is one of my favourite mathematician. $\endgroup$ – gymbvghjkgkjkhgfkl May 12 '17 at 12:00
  • $\begingroup$ Abu Hurayrah: Next time, please, repack better the integral you want to modify. I know you can do better than that ;) $\endgroup$ – FDP May 12 '17 at 17:17

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