Past exam of the mathematics course of the University of Cambridge / 2012 / iii / Paper 76 / 3 / b

Interpret the Fourier transform as a distributional Fourier transform. Exponentially regularize the Heaviside step function:As $a\downarrow 0$, the first term tends to $\pi \delta (k)$ in the tempered distribution sense, since its integral is $\pi$ and it concentrates at zero. The second tends to $-i\mathrm{PV}(1/k)$. ThereforeThe principal-value convention is essential: the printed $1/(ik)$ is not an ordinary integrable function at zero. This normalization also satisfies $ik\hat{H}=1$, the Fourier transform of a derivative version of $H^{\prime }=\delta$.

The real-space expression $1/x$ means the principal-value reciprocal distribution. Apply even exponential damping. Oddness givesDifferentiate the integral with respect to $k$; its derivative is ${\int }_0^{\infty }{e}^{-ax}\cos (kx)dx=a/(a^2+k^2)$. Its value at $k=0$ is zero, so the integral equals $\arctan (k/a)$. Taking $a\downarrow 0$ givesThe limit is a tempered distribution limit: the regularized transforms are uniformly bounded and converge away from zero, allowing dominated convergence against a Schwartz function. This Fourier transform of a principal-value reciprocal argument also establishes the needed Dirichlet integral ${\int }_0^{\infty }\sin u/udu=\pi /2$ in its Abel interpretation.

The locally integrable logarithm has distributional derivative $(\log |x|{)}^{\prime }=\mathrm{PV}(1/x)$. By the Fourier transform of a derivative and the previous result,For $k\ne 0$, this gives $-\pi /|k|$. At zero the expression must be extended as a Hadamard finite-part integral, not treated as a locally integrable function. For example, define the finite-part inverse-absolute-value distribution byThen $kT=\mathrm{sgn}k$. Different reference cutoffs change $T$ only by a multiple of the Dirac delta distribution. The general solution of the transform equation is consequentlyIndeed multiplication by $k$ annihilates only a delta term among distributions supported at zero; delta derivatives would generate additional nonzero terms. This supplies the precise interpretation of the paper's formula, and the convention-dependent constant $C$ need not be found.

For a locally integrable periodic function, its Fourier series coefficients are $c_m={\int }_0^{1}f(x)e^{-2\pi imx}dx$. Window the function to one period by $w(x)=H(x)-H(x-1)$. With the stated Fourier transform convention,There is no extra factor of $2\pi$: the Fourier coefficient already integrates over a period of length one.

For a completely arbitrary periodic distribution, multiplication by a discontinuous window need not be defined, for example at a derivative of a delta supported on a period boundary. The sharp-window assertion therefore presupposes a windowing convention or enough regularity at the endpoints. The smooth-window Fourier coefficients of a periodic distribution use a smooth compactly supported partition-of-unity window $\chi$ with ${\sum }_{j\in \mathbb{Z}}\chi (x+j)=1$; then $c_m=\widehat{\chi f}(2\pi m)$ in the distribution sense. The logarithmic example below is locally integrable, so the printed sharp window is legitimate and no ambiguity affects it.

For $f(x)=\log |x-1/2|$, put $u=x-1/2$. Symmetry gives, for integer $m\ne 0$,An integration by parts has no endpoint term: $u\log u\to 0$ at zero and $\sin (\pi m)=0$ at the upper endpoint. It follows thatwhere the sine integral is $\mathrm{Si}(z)={\int }_0^z\sin t/tdt$. Its limiting value $\pi /2$ follows from the Abel-regularized calculation in part (ii); integration by parts of its tail gives $\mathrm{Si}(\pi m)=\pi /2-(-1{)}^m/(\pi m)+O(m^{-3})$ for positive integer $m$. Hence the Fourier coefficients of a periodic logarithmic singularity haveThe first term is the requested leading behaviour. Its alternating sign records the singularity at half a period, and its $m^{-1}$ decay records the logarithmic cusp. The $m^{-2}$ correction comes from the derivative mismatch across the periodically joined endpoints. Also $c_0=-1-\log 2$ and $c_{-m}=c_m$, consistent with the real symmetric profile.