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

With the stated Fourier transform convention, the exponential transform is a shifted Dirac delta distribution:Writing the sine as $(e^{i{\omega }_{0}t}-e^{-i{\omega }_{0}t})/(2i)$ then givesThese are tempered distributions, since the time-domain functions are not integrable on the full real line. The second formula also correctly vanishes when ${\omega }_0=0$.

By the definition of convolution and Fourier inversion,The second line uses $\nu =\omega -\eta$. The factor $2\pi$ comes from the inverse-transform normalization. This is the product form of the convolution theorem, initially justified for sufficiently decaying functions and extended when the indicated products and convolutions exist.

The reciprocal in this transform must mean the principal-value reciprocal distribution. Its inverse transform isusing the given Dirichlet integral and the sign of $t$. Also ${\mathcal{F}}^{-1}[\delta ](t)=1/(2\pi )$. The inverse of the proposed expression is consequently $(iA/2)\mathrm{sgn}t+B/(2\pi )$. Comparison with $H(t)=[1+\mathrm{sgn}t]/2$ for $t\ne 0$ givesThe value assigned at the single point $t=0$ does not affect the distribution. This is the Fourier transform of the Heaviside step function; treating $1/\omega$ as an ordinary integral would miss the required prescription.

Multiply $H$ by the exponential form of the sine and use the modulation property of the Fourier transform:Inserting part (iii) givesThe reciprocal terms are real principal-value distributions and the delta terms are imaginary. Equivalently, exponential damping $H(t)e^{-ϵt}$ gives ${\omega }_0/[(ϵ+i\omega {)}^2+{\omega }_0^2]$, whose distributional limit is the boxed expression. The shifted-difference formula remains valid at ${\omega }_0=0$, where all terms cancel.