SImplification of $I=\int_{x=0}^{\infty}x^{n-1}(\alpha-x)^{m}e^{-\mu x}dx.$

Question

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.

Answer 1

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.