Why is $\int_{0}^{\infty} e^{-x}x^kdx = k!$?

Question

I was answering some math exercises, and by accident, I 'discovered' the following equation:

$$ \int_{0}^{\infty} e^{-x}x^kdx = k! $$

for instance, if $k=3$, we have the following (using an online integral calculator, i.e. Wolfram):

$$ \int_{0}^{\infty} e^{-x}x^3dx = 3! = 6 $$

However, I could not figure out how the equation makes sense... Is there a way to analytically transform the integral to $k!$ ?

Answer 1

$$I_0=\int_{0}^{\infty} e^{-at} dt =\frac{1}{a}~~~(1)$$ D.w.r.t. $a$ both sides you get $$I_1=\int_{0}^{\infty}t e^{-at} dt =\frac{1}{a^2}$$ Again differentiate to get $$I_2=\int_{0}^{\infty} t^2 e^{-at} dt =\frac{2}{a^3}$$ Finally , differentiate w.r.t. $a$ $k$-times to get $$I_k=\int_{0}^{k} t^k e^{-at} dt = \frac{k!}{a^{k+1}}$$ Put= $a=1$ to get $I_k=k!$

Answer 2

For a proof by mathematical induction:

Basic step: $$\int_{0}^{\infty }{{{e}^{-x}}}{{x}^{0}}dx=0!\ and\ \int_{0}^{\infty }{{{e}^{-x}}}{{x}^{1}}dx=1!\ $$

Induction step : $$\int_{0}^{\infty }{{{e}^{-x}}}{{x}^{k+1}}dx=\int_{0}^{\infty }{\overbrace{{{x}^{k+1}}}^{u}\overbrace{{{e}^{-x}}dx}^{dv}}=\left. {{x}^{k+1}}{{e}^{-x}} \right|_{\infty }^{0}+\left( k+1 \right)\int_{0}^{\infty }{{{e}^{-x}}{{x}^{k}}dx}=\left( k+1 \right)k!=\left( k+1 \right)!$$

note that $\left. {{x}^{k+1}}{{e}^{-x}} \right|_{\infty }^{0}=0$ by repeated application of L'Hôpital's rule.