# For what values of $\alpha$ does this integral converge?

For what real values of $\alpha$ does this integral converge?

$$\int_0^\infty {x^\alpha e^{-x} dx}$$

And what's the usual approach to this kind of problem?

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en.wikipedia.org/wiki/Gamma_function –  Jack Nov 25 '12 at 21:18
@Jack: But my textbook says it's only convergent when $\alpha>-1$, which differs with what Wikipedia says ($\alpha\not\in\mathbb{Z}_{(-\infty,-1]}$). –  Voldemort Nov 25 '12 at 21:27

## 3 Answers

Note: $$\frac{1}{e} \int_0^1 {x^\alpha dx}<\int_0^1 {x^\alpha e^{-x} dx} <\int_0^1{x^\alpha dx}$$ Thus, $\int_0^1 {x^\alpha e^{-x} dx}$ converges iff $\int_0^1{x^\alpha dx}$ does which happens for $\alpha>-1$. $\int_1^\infty {x^\alpha e^{-x}dx} < \int_1^\infty {x^\alpha ( x^N/N!)^{-1} dx}$ (where $N>\alpha +2$) does always converge.

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+1. Neat. ${}{}$ –  user17762 Nov 25 '12 at 21:34
But why in the Gamma Function article wikipedia says that it can also be defined at all points except $\mathbb{Z}_{(-\infty,-1]}$? –  Voldemort Nov 25 '12 at 21:37
Wikipedia says that you can continue the function defined by this to an analytical function defined everywhere except on the negative integers but not that the integral itself converges everywhere. –  Dominik Nov 25 '12 at 21:41

Note that for $\alpha > -1$, there exists $X(\alpha)$ such that $x^{\alpha} e^{-x} \leq e^{-x/2}$ for all $x > X(\alpha)$.

Hence, $$\int_0^{\infty} x^{\alpha} e^{-x} dx \leq \int_0^{X(\alpha)} x^{\alpha} e^{-x} dx + \int_{X(\alpha)}^{\infty} e^{-x/2} dx = \underbrace{\int_0^{X(\alpha)} x^{\alpha} e^{-x} dx}_{< \infty} + 2e^{-X(\alpha)/2}$$

For $\alpha \leq -1$, the integral diverges, which can be shown as follows. Let $\alpha = -1-\beta$ $$\int_0^{\infty} x^{\alpha} e^{-x} dx = \underbrace{\int_0^{\epsilon} \dfrac{e^{-x}}{x^{1+\beta}} dx}_{I} + \underbrace{\int_{\epsilon}^{\infty} \dfrac{e^{-x}}{x^{1+\beta}} dx}_{II}$$ Note that $$II \leq \displaystyle \int_{\epsilon}^{\infty} \dfrac1{x^{1+\beta}} dx < \infty$$ Note that $$I \geq \int_0^{\epsilon} \dfrac{e^{-\epsilon}}{x^{1+\beta}} dx = \infty$$ Hence, the integral converges for $\alpha > 1$ and diverges for $\alpha \leq -1$.

The integral you are interested in, wherever it converges, is called the Gamma function. The Gamma function is defined for $\alpha \leq -1$ as well through analytic continuation of the above integral. However, the Gamma function still has poles (blows to $\infty$) at negative integers.

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I would split the interval into $[0,1] \cup (1,\infty)$.

In $(1,\infty)$, $e^x$ dominates $x^{\alpha}$ as $x \to \infty$, so $\int_1^\infty {x^\alpha e^{-x} dx}$ is convergent for all $\alpha$. (To make this precise: given $\alpha \in \mathbb R$, we can find $x_0$ such that $x^{\alpha} < e^{x/2}$ for all $x \ge x_0$. So $0 \le x^{\alpha}e^{-x} \le e^{-x/2}$ for all $x \ge x_0$, and the integral converges because the integral of $e^{-x/2}$ converges.)

And in $[0,1]$, $0.3 < e^{-x} \le 1$, so the convergence of the integral is determined only by the convergence of $\int_0^1 {x^\alpha dx}$, which you can work out for yourself.

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