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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.

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Take a look at en.wikipedia.org/wiki/Gamma_function –  Eric Naslund Sep 11 '12 at 4:19
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That looks like a gamma function. –  Tunococ Sep 11 '12 at 4:20
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@Tunococ: 11 seconds :) –  Eric Naslund Sep 11 '12 at 4:20
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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
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up vote -1 down vote accepted

Wolfram says $\int_0^\infty e^{-y}y^a dy = \Gamma(a+1)$, for $Re(a) > -1$

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And where is your claim? –  Sasha Sep 11 '12 at 5:19
    
nowhere, im just sharing a information –  Integral Sep 12 '12 at 0:47
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