Show that $\int_0^{+\infty}x^pe^{-px}\,dx<+\infty$ where $p\ge 1$

Question

Show that without using beta or gamma function

$$I=\int_0^{+\infty}x^pe^{-px}\,dx<{+\infty}$$ where $p\ge 1$.

My try :

$$I=\int_0^{1}x^{p}e^{-px}\,dx+\int_1^{+\infty}x^{p}e^{-px}\,dx$$

First integral converges. But how I prove that the second integral $<+\infty$?

Answer 1

There is a constant $C$ such that $x^{p}e^{-px} \leq Ce^{-px/2}$ for $x>1$. This is because $x^{p}e^{-px/2}$ is a continuous function which tends to $0$ as $x \to \infty$ (which makes it a bounded function). Since $\int_1^{\infty} e^{-px/2}\, dx <\infty$ we are done.

More details: $\lim_{x\to \infty}\frac { x^{p}e^{-px}} {e^{-px/2}} =\lim_{x\to \infty} { x^{p}e^{-px/2}} =0$ since $e^{-px/2} \to 0$ faster than any power of $x$. (This is a standard fact proved using L'Hopital's Rule). Now $\frac { x^{p}e^{-px}} {e^{-px/2}}$ is a continuous function on $[1,\infty)$ which tends to $0$ as $x \to \infty$. Any such function is necessarily bounded. Take $C$ to be a bound for this function.