Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Does the following integral have a finite value? How to compute it? $$\int_{0}^{+\infty} e^{-x^k}\mathrm{d} x$$ where $k$ is given and $0<k<1$. By substituting $x^k=y$ we may obtain an equivalent integral $$\int_{0}^{+\infty} e^{-y}y^a\mathrm{d} y$$ where $a>0$ is given.

share|cite|improve this question
Take a look at –  Eric Naslund Sep 11 '12 at 4:19
That looks like a gamma function. –  Tunococ Sep 11 '12 at 4:20
@Tunococ: 11 seconds :) –  Eric Naslund Sep 11 '12 at 4:20
Existence does not require knowing about the Gamma function. Our integrand behaves nicely near $0$, and in the long run decays (far) more rapidly than $1/x^2$. –  André Nicolas Sep 11 '12 at 4:55
@AndréNicolas: I got your meaning that once the integrand decays faster than $1/x^2$ and then the integral has a finite value. Is it correct in the long run the integrand decays more rapidly than any $1/x^k$ with $k>0$? –  Shiyu Sep 11 '12 at 6:27

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.