Is there a formula for $\int {x^n e^{-x} dx}$

Question

I saw in this question that $$ \int {x^n e^x dx} = \bigg[\sum\limits_{k = 0}^n {( - 1)^{n - k} \frac{{n!}}{{k!}}x^k } \bigg]e^x + C. $$

and I was wondering if we can get some formula like that for

$$ \int {x^n e^{-x} dx} $$

when $n \in \mathbb N$. I already know that $$ \int^{\infty}_0 {x^n e^{-x} dx} = \Gamma(n+1) $$

but I'm looking for a general formula.

Just substitute $-x$ in your formula and account for the extra $(-1)^n$ term that appears. — Nigel Overmars, Sep 23 '16 at 17:32
This integral is known as the Incomplete Gamma Function: $\int x^n e^{-x}{\rm d}x = -\Gamma(n+1,x) + C$. — Winther, Sep 23 '16 at 17:36
@Winther Wrong. The incomplete Gamma is a definite integral. The OP wants an answer without extrema in the integral. — , Sep 23 '16 at 17:37

score 1 · Accepted Answer · answered Sep 23 '16 at 17:33

The general formula is simply $n\to z$ in which $z\in\mathbb{R}$:

$$\int_0^{+\infty}\ t^{z-1}e^{-t}\ \text{d}t = \Gamma(z)$$

Possibly avoid the poles: $z = 0, -1, -2, \ldots$.

If you are talking about a general formula for the indefinite integral, then the series expansion is what you are searching for.

Just expand the exponential in series

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

and you'll get the result:

$$\int x^n e^{-x}\ \text{d}x = \sum_{k = 0}^{+\infty} \frac{(-1)^k}{k!}\int x^{n+k}\ \text{d}x$$

Which is

$$\sum_{k = 0}^{+\infty}\frac{(-1)^k}{k!} \frac{1}{n+k+1}x^{n+k+1}$$

You are confusing the integral with the function $\Gamma(z)$ (which is an analytical continuation of the integral). The latter has poles at negative integers. The former simply does not exist for any $z\leq 0$. — Kibble, Sep 24 '16 at 00:22

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