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I'm trying to find for which values $r$ the following improper integral converges. $$\int_0^\infty x^re^{-x} dx$$ What I have so far is that $x^r < e^{\frac{1}{2}x}$ for $x \geq a$, which splits the integral into $$\int_0^a x^re^{-x} dx + \int_a^\infty e^{-\frac{1}{2}x}$$ We know the latter interval converges, but I don't know what to do with the first one. (For reference, graphing the functions reveals the answer to be $x > -1$.)

Edit: I would like a proof without the gamma function. Preferably one that uses the comparison test to compare limits.

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    $\begingroup$ This is actually one of the more interesting integrals in mathematics. try plugging in $ r = 1, 2, 3, 4...$ etc. What do you notice? $\endgroup$ – Aniruddh Venkatesan Jan 18 at 3:41
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    $\begingroup$ Hint:$$\int\limits_0^{\infty}\mathrm dz\, x^n e^{-x}=n!=\Gamma(n+1)$$ $\endgroup$ – Frank W. Jan 18 at 3:42
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    $\begingroup$ Did anybody even read the OP's question or do they just want to use the post to mention the gamma function? $\endgroup$ – Dionel Jaime Jan 18 at 3:49
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    $\begingroup$ @AniruddhVenkatesan I know it equals $n!$ for natural numbers, but for this question I'm mainly interested in for which real values of $r$ the interval converges. $\endgroup$ – Chase K Jan 18 at 3:53
  • $\begingroup$ @DionelJaime As I recall, the original question didn't mention not using the gamma function, with this being added afterwards. $\endgroup$ – John Omielan Jan 18 at 3:58
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Note that for all $r \in \mathbb{R}$,

$$\lim_{x \to \infty} \frac{x^r e^{-x}}{x^{-2}} = \lim_{x \to \infty} \frac{x^{r+2}}{e^x} = 0 $$

since the exponential function tends to infinity faster than any polynomial. Hence, by the limit comparison test the integral over $[1, \infty)$ converges since $\int_1^\infty x^{-2}\, dx = 1$.

See if you can finish by finding the condition on $r$ such that the integral over $[0,1]$ converges.

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