What is the general solution for integrals of the form $\int_{0}^{\infty}\frac{x^{2 m}\;\ln^{n}(x) }{e^{\frac{2 p +1}{2}x}} dx$? I have this integral 
$$\int_0^\infty \frac{x^{2 m}\;\ln^n (x)  }{e^{\frac{2 p +1}{2}x}} dx\;\; m,n,p \in \mathbb{N}$$
and I'like to know the general solution.
From WolframAlpha I got some solutions, e.g.
$$\int_0^\infty \frac{x^2\ln(x)}{e^{\frac{3x}{2}}} dx  = \frac{8}{27}\left(3-2\gamma-\ln\left(\frac{9}{4}\right)\right) $$
and
$$\int_0^\infty \frac{x^{2}\ln^{2}(x)}{e^{\frac{3x}{2}}} dx  = \frac{8}{81}\left(6-18\gamma+6\gamma^{2}+\pi^{2}-6 \ln \left(\frac{3}{2}\right)\left(3-2\gamma-\ln\left(\frac{3}{2}\right)\right) \right) $$
etc... but I'd like to know the general form of the solution and the process/method to get there. Note that $\gamma$ is the Euler-Mascheroni constant.
Thanks.
 A: Let's rewrite a little your integral :
Set $a:=2m$, $b:=\frac{2 p +1}{2}$ then
$$\int_0^\infty \frac{x^{2 m}\ln^{n}(x)}{e^{\frac{2 p +1}{2}x}}\, dx=\int_0^\infty x^a\ln^{n}(x)e^{-bx}\, dx$$
$$=\int_0^\infty \left(\frac d{da}\right)^n e^{a\ln(x)}e^{-bx}\, dx$$
$$=\left(\frac d{da}\right)^n \int_0^\infty  x^a e^{-bx}\, dx$$
$$=\left(\frac d{da}\right)^n \left(b^{-a-1}\int_0^\infty t^a e^{-t}\, dt\right)$$
$$=\left(\frac d{da}\right)^n \left(b^{-a-1}\Gamma(a+1)\right)$$
A common trick to compute $\Gamma'(x)$ and derivatives is to use $\psi(x)=\log(\Gamma(x))'= \dfrac{\Gamma'(x)}{\Gamma(x)}$
(with $\psi$ the Digamma function and the derivatives the Polygamma function) so that
$\Gamma'(x)=\psi(x)\Gamma(x)$,
$\Gamma''(x)=\left(\psi'(x)+\psi(x)^2\right)\Gamma(x)$
and so on...
Let's consider some examples of 
$I(n,a,b)=\left(\dfrac d{da}\right)^n \left(b^{-a-1}\Gamma(a+1)\right)$


*

*$n=1$ :
$$I(1,a,b)=b^{-a-1}\left[-\ln(b)+\psi(1+a)\right]\Gamma(a+1)$$
getting a generalization of your first example (case $a=2,b=\frac 32$) :
$$I(1,2,\frac 32)=\left(\frac 32\right)^{-3}\left[-\ln\left(\frac 32\right)+\psi(3)\right]\Gamma(3)$$ 
with $\psi(3)=H_2-\gamma=1+\frac 12-\gamma$.

*$n=2$ :
$$I(2,a,b)=b^{-a-1}\left(\psi'(1+a) + \psi(1+a)^2 - 2\ln(b)\psi(1+a)+\ln(b)^2\right)\Gamma(a+1)$$
$\cdots$
