put ,

$log\frac{1}{x}=t \hspace{0.5cm} =\gt\hspace{0.5cm} \frac{1}{x}=e^t \hspace{0.5cm} =\gt\hspace{0.5cm} x=e^{-t} \hspace{0.5cm} =\gt\hspace{0.5cm} dx=-e^{-t}dt$

when $x=0, t=\infty ,x=1,t=0$

$\int_{\infty}^{0} \frac{-e^{-t}dt}{\sqrt{e^{-t}}t^{1/2}}=\int_{0}^{\infty}e^{-t/2}t^{-1/2}dt$

Put,$\frac{t}{2}=u\hspace{0.5cm} =\gt\hspace{0.5cm} t=2u\hspace{0.5cm} =\gt\hspace{0.5cm} dt=2du$

$\int_{0}^{\infty}e^{-u}(2u)^{-1/2}2du=\frac{2}{\sqrt{2}} \int_{0}^{\infty} e^{-u}u^{-1/2}du$

${3.5cm}= \sqrt{2} \sqrt{1/2} =\sqrt{2\pi} $