calculus - Evaluate: $int_{0}^{infty}left(x^2-3x+1right)e^{-x}ln^3(x) dx$

$$I=\large \int_{0}^{\infty}\left(x^2-3x+1\right)e^{-x}\ln^3(x)\mathrm dx$$

$$e^{-x}=\sum_{n=0}^{\infty}\frac{(-x)^n}{n!}$$

$$I=\large \sum_{n=0}^{\infty}\frac{(-1)^n}{n!}\int_{0}^{\infty}\left(x^2-3x+1\right)x^n\ln^3(x)\mathrm dx$$

$$J=\int \left(x^2-3x+1\right)x^n\ln^3(x)\mathrm dx$$

We can evaluate $J$ by integration by parts but problem, the limits does not work.

How to evaluate integral $I?$

Answer

Here's a way that only requires the identity $\Gamma'(1) = - \gamma$ , since all derivatives of higher order cancel:
\begin{align}
I &=\int \limits_0^\infty (x^2 - 3x+1) \ln^3 (x) \mathrm{e}^{-x} \, \mathrm{d} x \\
&= \left[\frac{\mathrm{d}^2}{\mathrm{d} t^2} + 3 \frac{\mathrm{d}}{\mathrm{d}t}+1 \right] \int \limits_0^\infty \ln^3 (x) \mathrm{e}^{- t x} \, \mathrm{d} x ~\Bigg\vert_{t=1}\\
&= \left[\frac{\mathrm{d}^2}{\mathrm{d} t^2} + 3 \frac{\mathrm{d}}{\mathrm{d}t}+1 \right] \frac{1}{t} \int \limits_0^\infty \left[\ln^3 (y) - 3 \ln^2(y) \ln(t) + 3 \ln(y) \ln^2(t) - \ln^3 (t)\right] \mathrm{e}^{-y} \, \mathrm{d} y ~\Bigg\vert_{t=1}\\
&= \left[\frac{\mathrm{d}^2}{\mathrm{d} t^2} + 3 \frac{\mathrm{d}}{\mathrm{d}t}+1 \right] \frac{1}{t} \left[\Gamma'''(1) - 3 \Gamma''(1) \ln(t) + 3 \Gamma'(1) \ln^2(t) - \ln^3 (t)\right] ~\Bigg\vert_{t=1} \\

&= 2 \Gamma'''(1) + 6 \Gamma''(1) + 3 \Gamma''(1) + 6 \Gamma'(1) - 3 \Gamma'''(1) - 9 \Gamma''(1) + \Gamma'''(1) \\
&= 6 \Gamma'(1)\\
&= - 6 \gamma \, .
\end{align}

In fact, integration by parts yields the following generalisation:
\begin{align}
\gamma &= \int \limits_0^\infty (-\ln (x)) \mathrm{e}^{-x} \, \mathrm{d} x
= \int \limits_0^\infty \frac{-\ln (x)}{x} x \mathrm{e}^{-x} \, \mathrm{d} x \\
&= \int \limits_0^\infty (-\ln (x))^2 \frac{1-x}{2} \mathrm{e}^{-x} \, \mathrm{d} x \\

&= \int \limits_0^\infty (-\ln (x))^3 \frac{x^2 - 3x +1}{6} \mathrm{e}^{-x} \, \mathrm{d} x \\
&= \dots \, \\
&= \int \limits_0^\infty (-\ln (x))^{n+1} \frac{p_n (x)}{(n+1)!} \mathrm{e}^{-x} \, \mathrm{d} x \, .
\end{align}
The polynomials $p_n$ are defined recursively by $p_0(x) = 1$ and
$$p_n (x) = \mathrm{e}^{x} \frac{\mathrm{d}}{\mathrm{d}x} \left(x p_{n-1} (x) \mathrm{e}^{-x}\right) \, , \, n \in \mathbb{N} \, ,$$
for $x \in \mathbb{R}$ . The exponential generating function
$$ \sum \limits_{n=0}^\infty \frac{p_n(x)}{n!} t^n = \mathrm{e}^{t+x(1-\mathrm{e}^t)}$$
can actually be computed from a PDE and it turns out that the polynomials are given by
$$p_n(x) = \frac{B_{n+1}(-x)}{-x} \, , \, x \in \mathbb{R} \, , \, n \in \mathbb{N}_0 \, , $$

where $(B_k)_{k \in \mathbb{N}_0}$ are the Bell or Touchard polynomials.