Convergence and value of improper integral

Question

Show, that the integral $\int_0^\infty e^{-x^a}dx$ exists for all $a > 0$, and show that it's value is $\frac{1}{a}\Gamma(\frac{1}{a})$ where $\Gamma(x)$ is the gamma function.

I've tried substituting $-x^a$ aswell as rewriting the integral as $\int_0^\infty e^{-e^{a lnx}}dx$, but both attempts didn't lead anywhere so far. Thanks in advance!

Answer 1

If we replace $x^a$ with $z$ we just find the definition of the $\Gamma$ function:

$$ \int_{0}^{+\infty}e^{-x^a}\,dx = \frac{1}{a}\int_{0}^{+\infty}z^{\frac{1}{a}-1}e^{-z}\,dz = \frac{1}{a}\,\Gamma\left(\frac{1}{a}\right)=\Gamma\left(1+\frac{1}{a}\right).$$