What about whether or not this integral converges:
$$\int_0^\infty x^\alpha(\text{log}_e(x))^ne^{-x}\text{d}x\:\:\:\:\:\:\:(\star)$$
$\alpha,n,x:=\begin{cases}\alpha\in\mathbb{R^+}\cup\text{(-1, 0]}\\n\in\mathbb{N}\\x\in\mathbb{R^+}\cup\text{{0}}\end{cases}$
Here's what I did:
From $\mathcal{E}$uler's definition$$\:\:\Gamma(\alpha+1):=\int_0^1 (-\text{log}_e(x)\text{)}^{\alpha}\text{d}x$$$\\$Let$\:\:\text{log}_e(x):=\ln(x)\:;\:\mu=-\ln(x)\to\text{d}x=-\text{d}\mu\cdot e^{-\mu}$
Hence after some simplifications,$$\Gamma(\alpha+1)=\int_0^\infty\mu^\alpha\:e^{-\mu}\:\text{d}\mu$$
Now, under $\mathcal{L}$eibniz rule
$$\Gamma^{(1)}(\alpha+1):=\Gamma^{\large '}(\alpha+1)=\frac{\text{d}}{\text{d}\alpha}\Gamma(\alpha+1)=\frac{\text{d}}{\text{d}\alpha}\int_{0}^{\infty}\mu^\alpha\:e^{-\mu}\:\text{d}\mu=\int_{0}^{\infty}\frac{\text{d}}{\text{d}\alpha}\mu^\alpha\:e^{-\mu}\:\text{d}\mu$$
$$=\int_{0}^{\infty}\mu^\alpha\:e^{-\mu}\ln(\mu)\:\text{d}\mu\:,\:\:\:\:\:\:\mu>0 $$
Assume$$\Gamma^{(n-1)}(\alpha+1)=\int_{0}^{\infty}\mu^\alpha\:e^{-\mu}\ln^{n-1}(\mu)\:\text{d}\mu\:,\:\:\:\:\:\:\mu>0;\:n\in\mathbb{N}$$Then$$\frac{\text{d}}{\text{d}\alpha}\Gamma^{(n-1)}(\alpha+1)=\int_{0}^{\infty}\frac{\text{d}}{\text{d}\alpha}\mu^\alpha\:e^{-\mu}\ln^{n-1}(\mu)\:\text{d}\mu=\int_{0}^{\infty}\mu^\alpha\:e^{-\mu}\ln^{n}(\mu)\text{d}\mu=\Gamma^{(n)}(\alpha+1)$$
Hence$\:(\star):=\Gamma^{(n)}(\alpha+1)$
We can see that if$\:\:\mu\in(0,1],\:\mu^\alpha\:e^{-\mu}\ln^{n}(\mu)\le\mu^\alpha e^{-\mu (n+1)}\le e^{\alpha-\mu (n+1)}\:$
Thus,$$e^{\alpha}\int_0^1 e^{-\mu (n+1)}\text{d}\mu<\infty$$
If $\:\:\mu\in(1,\infty),\:\mu^\alpha\:e^{-\mu}\ln^{n}(\mu)\le \ln^n(\mu)e^{-\mu/2}\le \mathcal C< 1$
Thus,$$\mathcal C\int_1^\infty \text{d}\mu<\infty$$
We conclude that$\:\:\Gamma^{(n)}(\alpha+1)\:$converges in$\:\mathbb{R}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\square$
Now if everything's fine, how can I find the value of the integral?