Integral of Function related to Error Function (e.r.f)

Question

Need help solving this question - I've tried solving it multiple times but to no avail:
$$ I=\int_0^\infty e^{-y^3}.\sqrt y~dy $$ My approach has been to substitute: $$ y^3=t\Rightarrow 3.y^2~dy = dt \Rightarrow I= \frac13\int_0^\infty e^{-t}t^{-1/2}~dt $$ I'm lost on solving the above Integral. A similar problem suggested the answer in terms of e.r.f.
Related Question
However, I'm unable to understand how to solve this integral and get a numerical answer such as $\frac{\sqrt\pi}3 $. (Also, I'm curious as to how did $\pi$ appear in the expression? - My understanding of error functions is very limited. Wikipedia doesn't offer much help in explaining it either).

Answer 1

Hint. The gamma function is given by $$ \Gamma(s)=\int_0^\infty t^{s-1}e^{-t}~dt, \quad s>0. $$ Can you apply it here? Do you see the corresponding value for $s$?

Answer 2

Try setting $u=y^{\frac{3}{2}}$. Your integral reduces to $$ \int_0^\infty e^{-y^3}\sqrt{y}dy=\frac{2}{3}\int_0^\infty e^{-u^2}du=\frac{1}{3}\sqrt{\pi} $$