 Solution

ID: past-exam-of-the-algebra-1-and-differential-equations-1-course-of-the-university-of-oxford/2014/6/c/solution

- \frac{x ^{2}}{4 t} - iω x = - (\frac{x}{2 t} + i t ω)^{2} - t ω^{2} .

After

z = x / (2 t)

q_{t} (ω) = \frac{e ^{- t ω^{2}}}{π} \int_{- \infty}^{\infty} e^{- (z + i t ω)^{2}} d z .

The supplied shifted Gaussian integral equals

π

for real

t ω

. Hence

q_{t} (ω) = e^{- t ω^{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.

 New to topics? Read the docs here!