$\int \limits_{0}^{\infty}x^n e^{-ax} dx = \frac{n!}{a^{n+1}}$

Question

This probably comes up a lot with exponential probability distributons:

How to prove that:

$$\int \limits_{0}^{\infty}x^n e^{-ax} dx = \frac{n!}{a^{n+1}}$$

I look in a math table of integrals, I get:

$$\int x^n e^{ax}~dx = \frac{x^n~e^{ax}}{a} - \frac{n}{a} \int x^{n-1}~e^{ax}~dx$$

$$\int x^n e^{ax}~dx = \frac{e^{ax}}{a}\bigg[ x^n - \frac{n~x^{n-1}}{a} + \frac{n~(n-1)~x^{n-2}}{a^2} - \cdots \frac{(-1)^n~ n!}{a^n}\bigg]~~~(\text{n is positive})$$

not exactly the nice easy formula I was looking for...

Search for "gamma function" or just use induction on $n$. Your attempt after "I look in a math table of integrals [...]" is a viable approach, if you want to use induction. — Batominovski, Aug 21 '20 at 01:25
Instead of looking it up in a table, why not do it yourself? It's an easy exercise in integration by parts. Besides, the definite integral from $0$ to $\infty$ is easier than the indefinite integral. — Angina Seng, Aug 21 '20 at 01:48
For $a > 0$, $\int_0^\infty e^{-ax},{\rm d}x = 1/a$. View both sides as functions of $a$ and differentiate both sides $n$ times, where you use differentiation under the integral sign on the left side. After doing that $n$ times you get $\int_0^\infty x^n e^{-ax},{\rm d}x = n!/a^{n+1}$. — KCd, Aug 21 '20 at 02:39
Does this answer your question [after applying a substitution $ax\to x$]? How to come up with the gamma function? — Integrand, Aug 21 '20 at 03:36

score 1 · Answer 1 · edited Aug 21 '20 at 02:40

\begin{align} & \int\limits_0^\infty x^n e^{-ax} \, dx \\[8pt] = {} & \int\limits_0^\infty (ax)^n e^{-ax} (a\,dx) \cdot \frac 1 {a^{n+1}} \\[8pt] = {} & \int\limits_0^\infty t^n e^{-t} \, dt \cdot \frac 1{a^{n+1}} \end{align} That takes care of $\text{“}a\text{.''}$

\begin{align} & \int\limits_0^\infty t^n (e^{-t}\, du) = \overbrace{\int\limits_0^\infty u\,dv = \Big[ uv\Big]_{x\,:=\,0}^\infty - \int\limits_0^\infty v\, du}^\text{integration by parts} \\[10pt] = {} & \Big[ -t^n e^{-t} \Big]_{x\,:=\,0}^\infty -\int\limits_0^\infty -nt^{n-1} e^{-t} \, dt \\[10pt] = {} & \lim_{t\to\infty} \frac{-t^n}{e^t} + n \int\limits_0^\infty t^{n-1} e^{-t} \, dt. \end{align} The limit can be shown to be $0$ via L'Hopital's rule. Now we have \begin{align} \int\limits_0^\infty t^n e^{-t} \, dt & = n \int\limits_0^\infty t^{n-1} e^{-t} \, dt \\[10pt] & = n(n-1) \int\limits_0^\infty t^{n-2} e^{-t} \, dt \\[10pt] & = n(n-1)(n-2) \int\limits_0^\infty t^{n-3} e^{-t} \, dt \quad \text{etc.} \end{align}

$\int \limits_{0}^{\infty}x^n e^{-ax} dx = \frac{n!}{a^{n+1}}$

1 Answers1