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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 – 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
up vote 0 down vote accepted

I am formalizing the comments:

$$\Gamma(a)=\int_0^\infty e^{-y}y^{a-1}\text{ d}y$$

$$\Gamma(a+1)=\int_0^\infty e^{-y}y^a\text{ d}y$$


So we know that $\Gamma(x)$ converges (for at least $x\geqslant0$)

because it converges significantly faster than $x^{-2}$ for arbitrarily large $x$ values.

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