Integral of x times a stretched exponential

For a fixed $n>0$, the integral $\int_0^\infty x\,e^{-x^n}\,dx$ can be written as $\tfrac{a}{n}\,\Gamma\!\left(\tfrac bn\right)$ with integers $a$ and $b$. What is $a+b$?

$3$. Substituting $u=x^n$, so $x=u^{1/n}$ and $dx=\tfrac1n u^{1/n-1}du$, turns the integral into $\tfrac1n\int_0^\infty u^{2/n-1}e^{-u}\,du=\tfrac1n\Gamma\!\left(\tfrac2n\right)$. Thus $a=1$, $b=2$.