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does the following integral converge for all $p$?

$$\int_1^\infty e^{-x}x^p\mathrm dx$$

I believe it converges for all $p$, since exponential decays faster than polynomial. But I cannot seem to form a sequence of inequalities to prove that the integral converges.

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3 Answers 3

up vote 5 down vote accepted

Since exponentials decay faster than powers can grow, $e^{-x/2}x^p$ is bounded by some $M>0$ on $[1,\infty)$, and therefore $$\int_1^\infty e^{-x/2}(e^{-x/2}x^p)dx\leq M\int_1^\infty e^{-x/2}dx<\infty. $$

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Use integration by parts to relate your integral to the one with $p$ replaced with $p-1$. Repeat until the exponent is negative, then compare to the case $p=0$.

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We can also get the result as follows.

Let $n$ be a positive integer such that $n-p > 1$. For any $x\geq 1$, it follows from $e^x > \frac{{x^n }}{{n!}}$ (recall that $e^x = \sum\nolimits_{n = 0}^\infty {\frac{{x^n }}{{n!}}}$) that $$ e^{ - x} x^p < \frac{{n!}}{{x^{n - p} }}. $$ Since $\int_1^\infty {\frac{n!}{{x^{n - p} }}\,dx}$ converges, so does $\int_1^\infty {e^{ - x} x^p \,dx} $.

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