 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2021/paper-2/5/a/iii/solution

The reciprocal in this transform must mean the principal-value reciprocal distribution. Its inverse transform is

F^{- 1} (PV \frac{1}{ω}) (t) = \frac{i}{π} \int_{0}^{\infty} \frac{sin ( ω t )}{ω} d ω = \frac{i}{2} sgn t,

using the given Dirichlet integral and the sign of

t

. Also

F^{- 1} [δ] (t) = 1/ (2 π)

. The inverse of the proposed expression is consequently

(i A /2) sgn t + B / (2 π)

. Comparison with

H (t) = [1 + sgn t] /2

for

t \neq = 0

gives

A = - i, B = π, H (ω) = - i PV \frac{1}{ω} + π δ (ω) .

The value assigned at the single point

t = 0

does not affect the distribution. This is the Fourier transform of the Heaviside step function; treating

1/ ω

as an ordinary integral would miss the required prescription.

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