# Convergence and value of improper integral

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!

If we replace $x^a$ with $z$ we just find the definition of the $\Gamma$ function:
$$\int_{0}^{+\infty}e^{-x^a}\,dx = \frac{1}{a}\int_{0}^{+\infty}z^{\frac{1}{a}-1}e^{-z}\,dz = \frac{1}{a}\,\Gamma\left(\frac{1}{a}\right)=\Gamma\left(1+\frac{1}{a}\right).$$
• Okay, that sounds quite reasonable, now that I see it. However, I don't quite get one step: where does the $z^{1/a -1}$ in the second integral come from? By substituting $z = x^a$, wouldn't it be $\frac{dz}{dx} = a z^{a - 1} <=> dx = \frac{1}{az^{a-1}}$, which is unequal to $\frac{1}{a} z^{1/a - 1}$? Or where's my mistake? – moran Apr 16 '15 at 19:41
• @moran: the $\frac{1}{a}$ factor is just outside the middle integral, and $x=z^{\frac{1}{a}}$ leads to $dx = \frac{1}{a}z^{\frac{1}{a}-1}\,dz$. – Jack D'Aurizio Apr 16 '15 at 19:53