Is there closed formula for $\int_{-\infty}^{0}x^n e^{-\alpha x}dx$? I'm trying to find a closed formula for $\int_{-\infty}^{0}x^n e^{-\alpha x}dx$. I'm familiar with the integral $\int_{0}^{\infty}x^n e^{-\alpha x}dx=\frac{n!}{\alpha^{n+1}}$.
I tried to replace $-u=x$ and $-du=dx$ but got stuck because I know that the integral is true only for $\alpha >0$ and $n\in\mathbb{N}$.
Is there a closed formula for that integral? Maybe only for the even numbers of $n\in\mathbb{N}$?
 A: The case $\alpha<0$ reduces to the gamma integral up to constant factors. Thus, we shall assume that $\alpha\geqslant0$. Observe that the integrand makes sense only for $n\in\{pq^{-1}:p\in\mathbb{N},\:q\text { is an odd integer}\}$. We show that the integral diverges for all $n$. To see this, make the change of variable $t = -x$. This transforms the integral into
$$I_n:=(-1)^n\int_{0}^{+\infty} dt\:t^ne^{\alpha t}$$
For $\alpha=0$, the integral clearly diverges for all $n$. Henceforth, we shall assume that $\alpha>0$. For any $n$, observe that the integrand grows without bound, i.e.
$$\lim_{t\to+\infty}\:t^ne^{\alpha t}=+\infty$$
Choose $M>0$ arbitrarily. There exists $t_0 > 0$ such that $t^ne^{\alpha t}>M$ for all $t\geqslant t_0$. Therefore, for any $n\in\mathbb{Z}$, we have
$$\int_0^{+\infty}dt\:t^ne^{\alpha t}=\int_0^{t_0}dt\:t^ne^{\alpha t}+\lim_{x\to+\infty}\int_{t_0}^xdt\:t^ne^{\alpha t}=\int_0^{t_0}dt\:t^ne^{\alpha t}+\lim_{x\to+\infty}M(x-t_0) $$
The first term on the RHS -- the integral, is positive irrespective of whether it converges or diverges. Thus, $I_n$ diverges for all $n$.
