$e^{-x^{\alpha}}$ is Lebesgue-integrable on $[0,\infty)$ for $\alpha>0$ 
Prove that for $\alpha>0$, the function $e^{-x^{\alpha}}$ is $\mu$-integrable on $[0,\infty)$, where $\mu$ is the one-dimensional Lesbesgue measure.

Obviously we want to make an upper bound for the function on this interval and use the dominated convergence theorem, I tried the following: $e^{x^{\alpha}} \leq 1$ for $x\leq 1$ because we have $e^{−x^{\alpha}} \le \dots$. In my exercise class I saw that $x^{-2}$ is an integrable function on $(1,\infty)$, so if I can prove that: $e^{−x^{\alpha}} \leq Mx^{−2}$ for some suitable constant $M = M_\alpha < \infty$, I can easily use the dominated convergence theorem. The main problem is that I don't see how this inequality is actually true. Any hints on proving this?
Kees
 A: For fixed $k \in \mathbb{N}$ we have
$$e^y = \sum_{n \geq 0} \frac{y^n}{n!} \geq \frac{y^k}{k!} \qquad \text{for all $y>0$}.$$
In particular, if $y = x^{\alpha}$ for some $x > 0$, then 
$$e^{x^{\alpha}}\geq \frac{x^{k \alpha }}{k!},$$
i.e.
$$e^{-x^{\alpha}} \leq \frac{k!}{x^{k \alpha}}.$$
Choose $k= k(\alpha)$ sufficiently large such that $k\alpha \geq 2$, then this shows
$$e^{-x^{\alpha}} \leq \frac{k!}{x^2}.$$
Since you already know that $x \mapsto \frac{1}{x^2}$ is integrable on $(1,\infty)$ and $e^{-x^{\alpha}} \leq 1$ for $x \in [0,1]$, this proves that $x \mapsto e^{-x^{\alpha}}$ is integrable on $[0,\infty)$.
A: A quick proof: the change of variable $t = x^\alpha$ transforms your integral into $\int \limits _0 ^\infty \Bbb e ^{-t} \frac 1 \alpha t ^{\frac 1 \alpha -1} \Bbb d t$, and if you know about the Gamma function you will recognize the value of this integral to be $\frac 1 \alpha \Gamma (\frac 1 \alpha)$ which is finite, hence your integral is convergent.
An alternative proof is by noting that $\int \limits _0 ^\infty \Bbb e ^{- x^\alpha} \Bbb d x = \sum \limits _{n=0} ^\infty \int \limits _n ^{n+1} \Bbb e ^{- x^\alpha} \Bbb d x$ and $\Bbb e ^{- x^\alpha} \le \Bbb e ^{- n^\alpha}$ on each interval $[n,n+1]$, therefore your integral will be finite if you manage to prove that the series $\sum \limits _{n=0} ^\infty \Bbb e ^{- n^\alpha}$ converges.
In order to study the convergence of this series, we shall use the logarithm test. Since I haven't been able to find it online (despite it being one of the most powerful convergence tests), here is its statement: assume $x_n > 0 \ \forall n$; if $\lim \dfrac {\log \frac 1 {x_n}} {\log n} \left\{ \begin{matrix} >1 \\ <1 \end{matrix} \right.$, then $\sum x_n \left\{ \begin{matrix} < \infty \\ = \infty \end{matrix} \right.$.
Computing the limit above we get $\lim \dfrac {\log \Bbb e ^{n ^\alpha}} {\log n} = \lim \dfrac {n ^\alpha} {\log n} = \infty > 1$, so the studied series is convergent, therefore the integral will be too (it cannot be $-\infty$ because the integrand is positive).
Note that for $\alpha \ge 1$ you can try an alternate solution: on $[0,1]$ your function is continuous, therefore bounded (the interval is compact), so it is Riemann-integrable. On $(1, \infty)$ you have that $x^\alpha \ge x$, so that $\Bbb e ^{-x^\alpha} \le \Bbb e ^{-x}$ and this last one is integrable. Therefore your function will be integrable. As you can see, the difficulties arise when $\alpha \in (0, 1)$.
A: I assume the domain is $x \ge 0$ since otherwise $e^{x^\alpha}$ isn't defined in general.
If $t < 1$ then $\{x \mid e^{x^\alpha} > t\} = [0,\infty)$.
If $t \ge 1$ then $\{x \mid e^{x^\alpha} > t\} = \{x \mid x^\alpha > \log t\} = ( (\log t)^{1/\alpha},\infty)$.
In all cases $\{x \mid  e^{x^\alpha} > t\}$ is measurable.
Note: the only fact that was really needed was that $e^{x^\alpha}$ is increasing.
A: Use the dominated convergence theorem:
Let $p>1$ be given, then
$$\lim_{x\to\infty}{e^{-x^\alpha}\over x^{-p}}=0.$$
Hence there exists $N>0$ such that for all $x>N$ we have
$$e^{-x^{\alpha}}<x^{-p}$$
Then the function

$$f(x) = \begin{cases} 1 & 0\le x< N \\ x^{-p} & x\ge N\end{cases}$$

is integrable, so by the Lebesgue dominated convergence theorem, $e^{-x^\alpha}$ is integrable.
