Solution:

Let $\quad I=\int_{0}^{\infty} \frac{e^{-x^{2}}}{\sqrt{x}} d x$ Put $\quad x^{2}=t \quad \Rightarrow \quad x=\sqrt{t} \quad \Rightarrow \sqrt{x}=t^{1 / 4}$

Differentiate w.r.t x ,

$\begin{array}{ll}{\therefore \mathrm{dx}=\frac{1}{2 \sqrt{t}} d t} & {\text { lim } \rightarrow[0, \infty]} \\ {\therefore \mathrm{I}=} {\int_{0}^{\infty} \frac{e^{-t}}{1} \frac{t^{-3 / 4}}{2} d t}\end{array}$

But we know that,

$\int_{0}^{\infty} e^{-t} \cdot t^{n-1} d t= \operatorname{gamma}(n)$

$\therefore \mathrm{I}=\frac{1}{2} \mathrm{r}\left(\frac{1}{4}\right)---by \ definition \ of \ gamma \ function$