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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. 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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    @Shiyu: Yes, it is correct. For the comparison, I picked the smallest *integer* $k$ that does the job, for no special reason. For sure we want to use a $k\gt 1$.2012-09-11

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I am formalizing the comments:

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

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

Convergence:

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.