Past exam of the mathematics course of the University of Cambridge / 2004 / iii / Paper 76 / 1 / b

Take the usual Heaviside step function, $H(x)=1$ for $x>0$ and $0$ for $x<0$. Its Fourier transform must be interpreted as a tempered distribution. For $a>0$, exponential damping gives an ordinary integral:Against any Schwartz function $\phi$, the first term is an approximate identity of total mass $\pi$: changing variables $k=au$ gives the limit $\pi \phi (0)$. For the second term, pair positive and negative $k$ near zero. The difference $\phi (k)-\phi (-k)=O(k)$ removes the singularity and gives the Cauchy principal value of $1/k$. The tails converge by dominated convergence. Since the damped functions also approach $H$ as tempered distributions, continuity of the Fourier transform yields the Fourier transform of the Heaviside step function:The printed plus sign is inconsistent with the printed $e^{-ikx}$ convention for this $H$. In fact $H^{\prime }=\delta$ requires $ik\hat{H}=1$; the printed expression would give $-1$. Its plus sign would instead be correct for $H(-x)$.

To find the Fourier transform of a principal-value reciprocal, exponential damping of its odd part givesThe integral vanishes at $k=0$ and its $k$-derivative is $a/(a^2+k^2)$, so it equals $\arctan (k/a)$. Taking the distributional convergence limit givesAn ordinary improper integral of $1/x$ across zero is not a substitute for the specified Cauchy principal value.

The locally integrable function $\log |x|$ defines a tempered distribution. Its distributional derivative is $\mathrm{PV}(1/x)$: perform integration by parts outside $(-a,a)$, and observe that the boundary contribution $\log a[\phi (a)-\phi (-a)]$ tends to zero. By the Fourier transform of a derivative and the preceding result,Division by $|k|$ needs a finite-part inverse-absolute-value distribution, since $1/|k|$ is not locally integrable at zero. One possible normalization isIt satisfies $k\mathrm{Pf}(1/|k|)=\mathrm{sgn}k$. The general solution of $kT=0$ is a multiple of the Dirac delta distribution: a test function vanishing at zero can be written $k\psi (k)$, so $T$ depends only on its value at zero. Consequently the Fourier transform of the absolute logarithm isChanging the finite-part normalization changes $C$. The printed reciprocal expression is meaningful with this interpretation, not as an ordinary function at $k=0$.

The barred integral in the PDF denotes a Cauchy principal value. With $p(x)=\mathrm{PV}(1/x)$ its left side is the convolution $p\ast f$, whose Fourier transform is $-i\pi \mathrm{sgn}(k)\hat{f}(k)$. The right side is $-1$ on $(-1,1)$ and zero outside; either direct integration or the Fourier transform shift rule givesThus, away from the irrelevant single point $k=0$ in the $L^2$ formulation,By the Fourier transform of the absolute logarithm, the difference $\log |x-1|-\log |x+1|$ has transform $2i\pi \sin k/|k|$; its two delta terms cancel. ThereforeIts endpoint logarithms are locally square integrable, and $f(x)=2/({\pi }^{2}x)+O(x^{-3})$ at infinity, so it belongs to L2 space. The Hilbert transform multiplier is invertible almost everywhere on this space, proving uniqueness there. Without a decay or function-space condition, one may also add a constant under the symmetric principal-value convention. If the reversed step function were used to retain the printed sign in part (i), this answer would acquire the opposite sign.

Write $K_0(z)=-F(z)\log z+G(z)$ locally, with analytic $F,G$ and $F(0)=1$. The coefficient of $\log z$ in the Modified Bessel differential equation requires $z^2{F}^{\prime \prime }+z{F}^{\prime }-z^{2}F=0$. If $F=\sum a_m{z}^m$, its recurrence is $m^2{a}_m=a_{m-2}$, with $a_1=0$. HenceThe analytic part is even: an odd-power term in $G$ would produce an uncancelled odd-power term under the same differential operator, and the recurrence removes all such terms. Insert a smooth even cutoff equal to one near zero. The smooth part and the exponentially decaying tails give no algebraic large-wavenumber terms after repeated integration by parts. Therefore the logarithmic singularities determine the requested terms. For $k\ne 0$,Thus the Fourier transform of the modified Bessel function K0 has expansionAn independent check follows from the integral representation of the modified Bessel function of the second kind. Integrating $e^{-|x|\cosh t}$ first and then setting $u=\sinh t$ gives $2{\int }_0^{\infty }du/(u^2+1+k^2)=\pi /\sqrt{1+k^2}$, with exactly these first two terms.