For a free particle in one dimension, whose Hamiltonian is
\(\)\Hhat = \frac{\phat^2}{2m}\(\)
compute the propagator \(K(x,t;x',0)\).
We simply compute
$$\begin{align}
\bra{x} e^{-i\phat t/2m\hbar} \ket{x'} ={}& \int_{-\infty}^\infty \bra{x} e^{-i\phat^2 t/2m\hbar} \ket{p}\bra{p} \ket{x'} \d p \\
={}& \int_{-\infty}^\infty e^{-ip^2 t/2m\hbar}\bra{x}\ket{p}\bra{p}\ket{x'} \d p \\
={}& \int_{-\infty}^\infty e^{-ip^2 t/2m\hbar} e^{ip(x-x')/\hbar} \frac{\d p}{2\pi\hbar}
\end{align}$$
Now, we complete the square. This allows us to evaluate the integral exactly
$$\begin{align}
\bra{x} e^{-i\phat t/2m\hbar} \ket{x'} ={}& \int_{-\infty}^\infty \exp\left(-\frac{it}{2m\hbar}\left(p - m\frac{x-x'}{t}\right)^2\right) e^{im(x-x')^2/2m\hbar t} \frac{\d p}{2\pi\hbar} \\
={}& e^{im(x-x')^2/2m\hbar t} \int_{-\infty}^\infty \exp\left(-\frac{it}{2m\hbar}p^2\right) \frac{\d p}{2\pi\hbar} \\
={}& \frac{e^{im(x-x')^2/2m\hbar t}}{2\pi\hbar} \sqrt{\frac{2m\hbar\pi}{it}} \\
={}& e^{im(x-x')^2/2m\hbar t}\sqrt{\frac{m}{2\pi i\hbar t}}
\end{align}$$