Compute $\int_{0}^{+\infty} e^{-y}y^a\mathrm{d} y$ 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.

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.
