Solution:

$I=\int_{0}^{\infty} \frac{e^{-x^{3}}}{\sqrt{x}} d x$

Put $x^{3}=t$

$\therefore x=t^{\frac{1}{3}}$

$d x=\frac{1}{3} t^{\frac{-2}{3}}$

$\begin{aligned} \therefore I &=\int_{0}^{\infty} e^{-t} \cdot t^{-\frac{1}{6}} \cdot \frac{1}{3} \cdot t^{\frac{-2}{3}} d t \\ & \therefore I=\int_{0}^{\infty} e^{-t} \cdot t^{-\frac{5}{6}} d t \end{aligned}$

$\therefore I=\frac{1}{3} | \frac{\overline{1}}{6}$