Show that the improper integral $\int_{1}^{\infty} x^p e^{-x} dx$ converges for every value of $p \in \mathbb{R}$

Question

Ok from reading the question, "for every value of $p \in \mathbb{R}$" tells me that there might be some type of discontinuity at some value of $p$. Also from looking at the integrand I kind of get the hint that it might be a comparison test.Looking at the integrand of $e^{-x}$ it converges. However I'm not sure how to move further.

Answer 1

Hint Show that for each $p$ you have $$\lim_{x \to \infty} \frac{ x^p e^{-x}}{\frac{1}{x^2}}=0$$

Deduce that there exists some $A$ such that, for all $x >A$ you have $$0 \leq x^p e^{-x} \leq \frac{1}{x^2}$$

Use the comparison test. If you covered the limit comparison test for integrals you can skip one step above.

Hint 2 You can calculate $$\lim_{x \to \infty} \frac{ x^p e^{-x}}{\frac{1}{x^2}}=\lim_{x \to \infty}\frac{x^{p+2}}{e^x}$$ by applying L'H until the degree in the numerator becomes negative (i.e. $\lfloor p \rfloor +3$ times).

A much faster solution is by oberving that, as long as $p+2 \neq 0$ we have
$$\frac{x^{p+2}}{e^x}= \left( \frac{x}{e^{x/(p+2)}}\right)^{p+2}$$ and the limit inside can be calculated by applying LH once.

If $p+2=0$ then your limit is trivial: $$\lim_{x \to \infty} \frac{ x^p e^{-x}}{\frac{1}{x^2}}=\lim_{x \to \infty}\frac{1}{e^x}$$

Answer 2

Choose $m\in \Bbb Z^+$ with $m> p+2.$ For $x\ge 1$ we have $$e^x=\sum_{n=0}^{\infty}x^n/n!>x^m/m!\ge x^{p+2}/m!>0.$$ So $0<x^pe^{-x}<x^p(m!/x^{p+2})=m!/x^2.$

Therefore $$0\le \lim_{r\to \infty}\sup_{\ge r}\int_r^sx^pe^{-x}dx\le \lim_{r\to \infty}\sup_{s\ge r}\int_r^s(m!/x^2)dx=0.$$