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?
 A: 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.
A: 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.
A: 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.
