I'm trying to find $$ \int_0^\infty\ln(x)\,x^2e^{-x}\,\mathrm{d}x $$ Could anyone help explain this to me? I'm also interested in changing the $e^{-x}$ to an $e^{-ax}$. Thank you.
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This is most easily handled using the Gamma function $$ \Gamma(x)=\int_0^\infty t^{x-1}e^{-t}\,\mathrm{d}x\tag{1} $$ Note that if we take the derivative, we get $$ \Gamma'(x)=\int_0^\infty\log(t)\,t^{x-1}e^{-t}\,\mathrm{d}x\tag{2} $$ So your question is asking for $\Gamma'(3)$.
In this answer, it is shown that $$ \Gamma'(n+1)=n!\,(-\gamma+H_n)\tag{3} $$ Where $H_n$ is a Harmonic Number and $\gamma$ is the Euler-Mascheroni Constant.
Note that with the change of variables $t\mapsto at$, we get $$ \begin{align} \Gamma(x) &=\int_0^\infty\log(at)\,(at)^{x-1}e^{-at}\,\mathrm{d}at\\ &=a^x\int_0^\infty(\log(a)+\log(t))\,t^{x-1}e^{-at}\,\mathrm{d}t\tag{4} \end{align} $$ Using the fact that $\Gamma(n+1)=n!$ and $(1)$-$(4)$, it is not difficult to compute $$ \int_0^\infty\log(t)\,t^{x-1}e^{-at}\,\mathrm{d}t\tag{5} $$
