Derive the identity $\int_0^{\infty} e^{x^\alpha} dx = \Gamma(\frac{\alpha+1}{\alpha})$

Question

I'm having a hard time figuring out how to solve this problem. I've tried using $\Gamma(z) = \int_0^\infty t^{z-1} e^{-t} dt$ but I haven't gotten anywhere. I don't know if I should use factorial either.

Any help would be appreciated.

Answer 1

Hint: change of variables $x^\alpha = t$.

EDIT: But I think you mean $\int_0^\infty e^{-x^\alpha}\; dx$.

Answer 2

Hint: in addition to Robert Israel's hint, remember that $$ \frac1\alpha\Gamma\!\left(\frac1\alpha\right)=\Gamma\!\left(\frac{\alpha+1}\alpha\right) $$