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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 $0$\int_{0}^{+\infty} e^{-y}y^a\mathrm{d} y$$ where $a>0$ is given.

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    Take a look at http://en.wikipedia.org/wiki/Gamma_function2012-09-11
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    That looks like a [gamma function](http://en.wikipedia.org/wiki/Gamma_function).2012-09-11
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    @Tunococ: 11 seconds :)2012-09-11
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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 $1/x^2$.2012-09-11
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    @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 $k>0$?2012-09-11
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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.