Prove $\exp(-|x|^\alpha)$ is integrable Prove the function
$$x\mapsto\exp(-|x|^\alpha)$$
is $L_p$ integrable $\forall\alpha>0$ (Lebesgue integrable in $\mathbb{R}^p)$
My attempt:
If $t < 1$ then $\{x \mid e^{x^\alpha} > t\} = [0,\infty)$.
If $t \ge 1$ then $\{x \mid e^{x^\alpha} > t\} = \{x \mid x^\alpha > \log t\} = ( (\log t)^{1/\alpha},\infty)$.
Here, I'm stuck. Can someone help me?
 A: Instead, use composition trick: Let $F(w)=\exp(w)$, $G(t)=-t^{\alpha}$, $H(x)=|x|$ and consider $F\circ G\circ H$, all these are continuous.
A: Hints:
Since the function is continuous, it's measurable. To see that it's integrable, switch to polar coordinates:
$$\int\limits_{\mathbb{R}^n}e^{-|x|^\alpha}\, dx=\int\limits_{S^{n-1}}\int\limits_0^\infty e^{-r^\alpha}r^{n-1}\, drd\sigma(\omega)
=\text{vol}(S^{n-1})\int\limits_0^\infty e^{-r^\alpha}r^{n-1}\, dr,$$ where $\sigma$ is the surface measure on $S^{n-1}.$ So, it suffices to show that $$\int\limits_0^\infty e^{-r^\alpha}r^{n-1}\, dr<\infty.$$ Can you see that this integral is finite?
EDIT: Here is a sketch of an argument showing that this integral is finite:
Since, for any fixed $\alpha>0,$ $e^{-r^\alpha}r^{n+1}\rightarrow 0$ as $r\rightarrow\infty$, there exists $N$ so that for all $r\geq N,$ we have that $$e^{-r^\alpha}r^{n+1}<1.$$ Hence,
\begin{align*}\int\limits_0^\infty e^{-r^\alpha}r^{n-1}\, dr&=\int\limits_0^Ne^{-r^\alpha}r^{n-1}\, dr+\int\limits_N^\infty e^{-r^\alpha}r^{n-1}\, dr\\
&<\int\limits_0^Ne^{-r^\alpha}r^{n-1}\, dr+\int\limits_N^\infty \frac{1}{r^2}\, dr\\
&<\infty
\end{align*}
