The functions $f$ and $g$ have Laplace transforms $\widehat{f}$ and $\widehat{g}$, and satisfy $f(t)=0=g(t)$ for $t<0$. The convolution $h$ of $f$ and $g$ is defined by

$$h(u)=\int_{0}^{u} f(u-v) g(v) d v$$

and has Laplace transform $\widehat{h}$. Prove (the convolution theorem) that $\widehat{h}(p)=\widehat{f}(p) \widehat{g}(p)$.

Given that $\int_{0}^{t}(t-s)^{-1 / 2} s^{-1 / 2} d s=\pi \quad(t>0)$, deduce the Laplace transform of the function $f(t)$, where

$$f(t)=\left\{\begin{array}{l} t^{-1 / 2}, \quad t>0 \\ 0, \quad t \leqslant 0 \end{array}\right.$$