I am having trouble proving this ,$\int^{\infty}_0 x^n e^{-cx} dx = \frac{n!}{c^{n+1}}$.

Question

To be more precise, $n$ is a nonnegative integer and c is a positive constant.

$\int^{\infty}_0 x^n e^{-cx} dx = \frac{n!}{c^{n+1}}$

supposedly, but I cannot prove it.

I've tried integration by parts, but can someone give me a clue ?

Answer 1

HINT:

Let $$I_n=\int^{\infty}_0 x^n e^{-cx} dx$$

Using LIATE rule for Integration by Parts,

$$=x^n\int^{\infty}_0e^{-cx} dx-\int^{\infty}_0\left(\frac{d x^n}{dx}\cdot \int e^{-cx} dx\right)dx$$

$$=x^n\frac{e^{-cx}}{-c}\big|_0^{\infty}+\frac nc\int^{\infty}_0x^{n-1}e^{-cx}dx=\frac ncI_{n-1} $$

Now, $$I_0=\int^{\infty}_0 e^{-cx} dx=\frac1c$$