Calculating $ \int_p^\infty x^{n-1} e^{-ax} \, dx$ $$ \int_p^\infty x^{n-1} e^{-ax} \, dx \quad p \in \mathbb{R}, a>0, n \in \mathbb{N}$$ 
Wolfram can't deal with this. How to calculate it?
 A: This can be expressed with the help of the incomplete gamma function given by
$$ \Gamma(n,p) = \int_p^{\infty} t^{n-1}\,e^{-t}\,{\rm d}t.$$
You just need to make a simple change of variable $t=ax$ to have your closed-form expression.
A: WolframAlpha can handle this integral, but there are a couple of tricks


*

*You need the extended time option that's only available through WolframAlpha pro.

*You need to indicate certain assumptions on the parameters.



A: You can use a formula by Hermite. For any polynomial $p(x)$ of degree $m$, we have:
$$\int_y^\infty e^{-x} p(x) dx = e^{-y} \left( p(y) + p'(y) + p''(y) + \cdots + p^{(m)}(y)\right)$$
A heuristic way to derive this formula is to adopt an ansatz that for any polynomial $p(x)$
of degree $m$, the integral of $e^{-z} p(x)$ is again another polynomial of same degree  multiplied by $e^{-x}$:
$$\int_y^\infty e^{-x} p(x) dx = e^{-y} P(y)$$
Differentiate both sides with $y$, and introduce $D$ as a short hand for $\frac{d}{dy}$,
we have:
$$- e^{-y} p(y) = e^{-y} \left( \frac{d}{dy} P(y) - P(y) \right)
\quad\iff\quad  p(y) = (1 - D) P(y)$$
Multiply both sides by $1 + D + D + \cdots D^{m}$, one get:
$$ \left( 1 + D + D^2 + \cdots D^{m}\right)p(y) = (1 - D^{m+1}) P(y) = P(y)$$
Apply this to the integral at hand, we gave:
$$\begin{align}
\int_p^\infty x^{n-1} e^{-ax}dx = & a^{-n}\int_{ap}^\infty t^{n-1} e^{-t}dt\\
= & a^{-n} e^{-ap} \left[(ap)^{n-1} + (n-1)(ap)^{n-2} + (n-1)(n-2)(ap)^{n-3} + \cdots\right]\\
= & \frac{p^{n-1} e^{-ap} }{a} \sum_{k=0}^{n-1} \frac{(n-1)!}{(n-1-k)!}(ap)^{-k}
\end{align}$$
