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.

Do you mean $\int_0^\infty e^{-x^\alpha},\mathrm{d}x$? – robjohn Nov 13 '17 at 08:23 — robjohn, Nov 13 '17 at 08:23

score 2 · Answer 1 · answered Nov 13 '17 at 08:21

2

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

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

answered Nov 13 '17 at 08:21

Robert Israel

448,999

score 2 · Answer 2 · answered Nov 13 '17 at 08:28

2

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

answered Nov 13 '17 at 08:28

robjohn

345,667

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

2 Answers2