= Solution
Complete the square inside the <Fourier transform> of the <Gaussian heat kernel>:
$$
-\frac{x^2}{4t}-i\omega x
=-\left(\frac{x}{2\sqrt t}+i\sqrt t\,\omega\right)^2-t\omega^2.
$$
After $z=x/(2\sqrt t)$,
$$
\widehat q_t(\omega)=\frac{e^{-t\omega^2}}{\sqrt\pi}
\int_{-\infty}^{\infty}e^{-(z+i\sqrt t\,\omega)^2}\,dz.
$$
The supplied shifted <Gaussian integral> equals $\sqrt\pi$ for real $\sqrt t\,\omega$. Hence
$$
\boxed{\widehat q_t(\omega)=e^{-t\omega^2}.}
$$
At zero this says $\int q_t=1$, so convolution with the heat kernel averages rather than rescales the initial mass. The formula is an instance of the <Fourier transform of a Gaussian>.
Back to article page
