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I tried using certain substitutions like $u=ax^b$ but that lead to $\displaystyle\frac{1}{a^{\frac{1}{b}}b^n}\int _0^{\infty }e^{-u}\:\ln ^n\left(\frac{u}{a}\right)u^{\frac{1}{b}-1}du\:$ i tried to use special functions to evaluate this but that $\ln ^n\left(\frac{u}{a}\right)$ is very annoying, i'd appreciate any help.

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  • $\begingroup$ What about $u=\ln(x) $? $\endgroup$ – EDX Jul 2 '20 at 3:37
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You can start using the following identity, $$\int _0^{\infty }x^m\:e^{-ax^b}\:dx=\frac{\Gamma \left(\frac{m+1}{b}\right)}{b\:a^{\frac{m+1}{b}}}$$ You can now differentiate both sides $n$ times with respect to m and then set it to $0$, $$\int _0^{\infty }x^m\:\ln ^n\left(x\right)\:e^{-ax^b}\:dx=\frac{\partial ^n}{\partial m^n}\frac{\Gamma \left(\frac{m+1}{b}\right)}{b\:a^{\frac{m+1}{b}}}$$ $$\boxed{\int _0^{\infty }\ln ^n\left(x\right)\:e^{-ax^b}\:dx=\lim _{m\to 0}\frac{\partial ^n}{\partial m^n}\frac{\Gamma \left(\frac{m+1}{b}\right)}{b\:a^{\frac{m+1}{b}}}}$$

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  • $\begingroup$ Is it obvious that you can differentiate under the integral sign here? I don't remember what conditions are necessary for that to work. $\endgroup$ – Robert Shore Jul 2 '20 at 4:26

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