I ran into this problem where I needed to use the following integral equality in my physics textbook.

$$\int_0^\infty x^ne^{-\alpha x} \, dx=\frac{n!}{\alpha^{n+1}}$$

where $n$ is a positive integer and $\alpha$ is a positive constant.

I was just wondering how one arrives at this equality.

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    $\begingroup$ Use integration by parts or look for an answer in some Calculus Text. $\endgroup$ Commented Jul 24, 2020 at 23:54
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    $\begingroup$ First substitute $y=\alpha x$ and then recognize the Gamma function $\endgroup$ Commented Jul 24, 2020 at 23:54
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    $\begingroup$ Do it by induction . $\endgroup$ Commented Jul 24, 2020 at 23:55
  • $\begingroup$ A similar question: math.stackexchange.com/questions/2378736/… $\endgroup$ Commented Jul 25, 2020 at 0:05
  • $\begingroup$ @MichaelHardy I'm a bit new to StackExchange in terms of posting. $\endgroup$ Commented Jul 25, 2020 at 0:15

3 Answers 3



$$I_n=\int_{0}^{\infty }x^ne^{-\alpha x}dx =-\frac1a \int_{0}^{\infty }x^n d(e^{-\alpha x})=\frac n aI_{n-1}$$ $$I_0= \int_{0}^{\infty }e^{-\alpha x}dx=\frac1a$$ Thus, $$I_n=\frac{n!}{\alpha^{n}}I_0=\frac{n!}{\alpha^{n+1}}$$


By subbing $x=-\ln(y)$ we have

$$\int_0^\infty x^n e^{-ax}dx=(-1)^n \int_0^1 y^{a-1}\ln^n(y)dy$$

Assuming $n$ is a positive integer, we have

$$\int_0^\infty x^n e^{-ax}dx=(-1)^n \frac{\partial^n}{\partial^n a}\int_0^1 y^{a-1}dy=(-1)^n \frac{\partial^n}{\partial^n a}\cdot\frac1a=\frac{n!}{a^{n+1}}.$$

The last result follows from:

$$\frac{\partial}{\partial a}\cdot\frac1a=-\frac1{a^2}$$

$$\frac{\partial^2}{\partial^2 a}\cdot\frac1a=\frac{2}{a^3}$$

$$\frac{\partial^3}{\partial^3 a}\cdot\frac1a=-\frac{2\cdot3}{a^4}$$

So, in general we have

$$\frac{\partial^n}{\partial^n a}\cdot\frac1a=(-1)^n\frac{n!}{a^{n+1}}$$


\begin{align} & \int_0^\infty x^ne^{-\alpha x} \, dx \\[8pt] = {} & \frac 1 {\alpha^{n+1}} \int_0^\infty (\alpha x)^n e^{-\alpha x} \, (\alpha\, dx) \\[8pt] = {} & \frac 1 {\alpha^{n+1}} \int_0^\infty y^n \big( e^{-y} \, dy\big) \\[8pt] = {} & \frac 1 {\alpha^{n+1}} \int u\, dv \\[8pt] & \text{where } u = y^n \text{ and } dv = e^{-y}\, dy. \end{align} From here you integrate by parts: $\displaystyle \int u\, dv = uv - \int v\,du.$

In the $uv$ term, the value when $y=0$ will be $0$ except when $n=0.$

The value when $y\to\infty$ can be found via L'Hopital's rule to be $0.$

You should end up with $\displaystyle \frac 1 {\alpha^{n+1}} \cdot n \int_0^\infty y^{n-1} e^{-y} \, dy.$

So doing the same thing again will give you $\displaystyle \frac 1 {\alpha^{n+1}} \cdot n(n-1) \int_0^\infty y^{n-2} e^{-y} \, dy.$

And again, and you get: $\displaystyle \frac 1 {\alpha^{n+1}} \cdot n(n-1)(n-2) \int_0^\infty y^{n-3} e^{-y} \, dy.$

And so on, so you just need to recognize the pattern. Or to put it another way, use mathematical induction on $n.$

  • $\begingroup$ Thanks for the clear and thorough answer. I find it astounding how you all just know how to manipulate integrals and just see through them essentially. Any advice or resources on how to become better at evaluating and proving difficult integrals? (Not just this one in particular, I've seen some much more difficult integral evaluations on this site.) $\endgroup$ Commented Jul 25, 2020 at 0:29
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    $\begingroup$ @CalebWilliamsUIC : Partly it's a matter of experience. The fact that $n!=\int_0^\infty y^n e^{-y} \, dy$ is a somewhat standard exercise. And partly it's a matter of thinking about things rather than just trying to get the answer. With these sorts of elementary integrals, maybe the book titled Street-Fighting Math can help. At a slightly (not enormously) more advanced level, there a book titled Inside Intereseting Integrals. And there are exercises in integration that arise in probability courses, and some that arise in theory-of-statistics courses that don't arise in probability courses. $\endgroup$ Commented Jul 25, 2020 at 0:37

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