# How can i find $\int _0^{\infty }\ln ^n\left(x\right)\:e^{-ax^b}\:dx$

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.

• What about $u=\ln(x)$? – EDX Jul 2 '20 at 3:37

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 $$0$$, $$\int _0^{\infty }x^m\:\ln ^n\left(x\right)\:e^{-ax^b}\:dx=\frac{\partial ^n}{\partial m^n}\frac{\Gamma \left(\frac{m+1}{b}\right)}{b\:a^{\frac{m+1}{b}}}$$ $$\boxed{\int _0^{\infty }\ln ^n\left(x\right)\:e^{-ax^b}\:dx=\lim _{m\to 0}\frac{\partial ^n}{\partial m^n}\frac{\Gamma \left(\frac{m+1}{b}\right)}{b\:a^{\frac{m+1}{b}}}}$$