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

Question

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.

Answer 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 $n.$

Answer 2

Note

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

Answer 3

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

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