SImplification of $I=\int_{x=0}^{\infty}x^{n-1}(\alpha-x)^{m}e^{-\mu x}dx.$ Let $n$ and $m$ be positive integers, $\mu$ be real positive and $\alpha$ positive real number.
I would like to compute the following integral if there is close formula 
$$
I=\int_{x=0}^{\infty}x^{n-1}(\alpha-x)^{m}e^{-\mu x}dx.
$$ 
Thanks.
 A: Here we will address your integral:
\begin{equation}
I_{n,m}(\alpha, \mu) = \int_0^\infty x^{n - 1}\left(\alpha - x\right)^m e^{-\mu x}\:dx
\end{equation}
We first identify that:
\begin{equation}
 \frac{d^{n - 1}}{d\mu^{n - 1}} \left[e^{-\mu x}\right] = (-1)^{n - 1}x^{n - 1}e^{-\mu x} \Longrightarrow x^{n - 1}e^{-\mu x} = \frac{1}{(-1)^{n - 1}}\frac{d^{n - 1}}{d\mu^{n - 1}}\left[e^{-\mu x}\right]  = (-1)^{n - 1}\frac{d^{n - 1}}{d\mu^{n - 1}}\left[e^{-\mu x}\right]
\end{equation}
Thus, 
\begin{equation}
I_{n,m}(\alpha, \mu) = \int_0^\infty x^{n - 1}\left(\alpha - x\right)^m e^{-\mu x}\:dx = \int_0^\infty \left(\alpha - x\right)^m (-1)^{n - 1}\frac{d^{n - 1}}{d\mu^{n - 1}}\left[e^{-\mu x}\right] \:dx
\end{equation}
By Leibniz's Integral Rule, this becomes:
\begin{equation}
I_{n,m}(\alpha, \mu) = (-1)^{n - 1}\frac{d^{n - 1}}{d\mu^{n - 1}}\left[\underbrace{\int_0^\infty \left(\alpha - x\right)^m e^{-\mu x}\:dx}_{J_m(\alpha, \mu)}\right]
\end{equation}
We now resolve $J_m(\alpha, \mu)$ and begin by expanding the $(\alpha - x)^m$ using the Binomial Theorem:
\begin{equation}
\left(\alpha - x\right)^m = \sum_{j = 0}^{m} {m \choose j}\alpha^j (-x)^{m - j}= \sum_{j = 0}^{m} {m \choose j}\alpha^j (-1)^{m - j}x^{m - j}
\end{equation}
Thus our $J_m(\alpha, \mu)$ becomes:
\begin{align}
J_m(\alpha, \mu) &= \int_0^\infty \left(\alpha - x\right)^m e^{-\mu x}\:dx = \int_0^\infty \left(\sum_{j = 0}^{m} {m \choose j}\alpha^j (-1)^{m - j} x^{m - j}\right) e^{-\mu x}\:dx \nonumber \\
&= \sum_{j = 0}^{m} {m \choose j}\alpha^j(-1)^{m - j} \int_0^\infty x^{m - j} e^{-\mu x}\:dx
\end{align}
We now let $t = \mu x$:
\begin{align}
J_m(\alpha, \mu) &=  \sum_{j = 0}^{m} {m \choose j}\alpha^j(-1)^{m - j} \int_0^\infty \left(\frac{t}{\mu}\right)^{m - j} e^{-t}\cdot \frac{1}{\mu}\:dt = \sum_{j = 0}^{m} {m \choose j}\frac{\alpha^j(-1)^{m - j}}{\mu^{m - j + 1}} \int_0^\infty t^{m - j} e^{-t}\:dt \nonumber \\
&= \sum_{j = 0}^{m} {m \choose j}\frac{\alpha^j(-1)^{m - j}}{\mu^{m - j + 1}} \Gamma\left(m - j + 1 \right)
\end{align}
Returning to our integral we arrive at:
\begin{align}
I_{n,m}(\alpha, \mu) &= (-1)^{n - 1}\frac{d^{n - 1}}{d\mu^{n - 1}}\left[ 
\sum_{j = 0}^{m} {m \choose j}\frac{\alpha^j(-1)^{m - j}}{\mu^{m - j + 1}} \Gamma\left(m - j + 1 \right)\right] \nonumber \\
&= \sum_{j = 0}^{m} {m \choose j}\alpha^j(-1)^{m - j} \Gamma\left(m - j + 1 \right)\frac{d^{n - 1}}{d\mu^{n - 1}}\left[ \frac{1}{\mu^{m - j + 1}}\right] \\ &=\sum_{j = 0}^{m} {m \choose j}\alpha^j(-1)^{m +n- j-1} \Gamma\left(m - j + 1 \right)\left[ \frac{(m-j+1)\cdots (m+n-1-j)}{\mu^{m +n- j }}\right]
\end{align}
Hope this is of use. 
