How may one show that $\int_{0}^{\infty}x^{n-1}e^{-x^n}\ln^2(x)\mathrm dx={6\gamma^2+\pi^2\over 6n^3}?$

Question

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 $(2)?$

Answer 1

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 $(1)$.