Past exam of the mathematics course of the University of Cambridge / 2013 / iii / Paper 67 / 4 / b

Use the standard Fourier transform on the whole real axis. The printed lower limit is missing a minus sign. For this integrable even function, its only obstruction to smoothness is the origin, so its high-frequency algebraic terms come from its local cusp:Localize near zero with a smooth cutoff. The smooth polynomial terms, including $|x{|}^6=x^6$, contribute no algebraic cusp term; their localized transforms decay faster than any prescribed inverse power. The supplied half-line formula, interpreted away from $k=0$ in an Abel-regularized or tempered distribution sense, gives the Fourier transform of an algebraic cuspFor $p=3$ the cosine is $1$, and for $p=9$ it is $-1$. Both cusp coefficients in the local expansion are $-1$. HenceThe second coefficient is $29!=725760$. The absence of a $|k{|}^{-7}$ term follows from the smooth even power $x^6$, not from neglecting an available correction.

One can verify the signs by integration by parts on $2\mathrm{Re}{\int }_0^{\infty }{e}^{-ikx}(1+x^3{)}^{-1}dx$. Its endpoint derivatives first have nonzero relevant values $f^{(3)}(0)=-3!$ and $f^{(9)}(0)=-9!$, which contribute twice those values divided by $(ik{)}^4$ and $(ik{)}^{10}$. Higher derivatives are integrable on the half-line, justifying the displayed remainder after further integrations. Equivalently these coefficients are the Fourier decay from a derivative jump at the origin.

The ordinary Fourier transform does not exist. There is a real simple pole at $x=-1$, withThe two separate improper integrals diverge logarithmically. Thus no ordinary large-$k$ expansion is defined under the stated integral convention. A Cauchy principal value or a contour prescription would be additional data, and must be stated rather than silently introduced.

For completeness, the symmetric Cauchy principal value has a precise answer. The real pole contributes the principal-value Fourier transform of a real poleThe other poles are $z_{\pm }=(1\pm i\sqrt{3})/2$ with residues $r_{\pm }=1/(3z_{\pm }^2)=(-1\mp i\sqrt{3})/6$. For $k>0$, close the contour integration in the lower half-plane, with clockwise orientation. The nonreal pole contributes $-2\pi i{r}_{-}{e}^{-ik{z}_{-}}$, while the real principal-value pole supplies the half-residue above. For $k<0$, the upper-half-plane contour gives the conjugate result. With $s=\mathrm{sgn}(k)$, the exact principal-value formula for $k\ne 0$ isThese are its two nonzero asymptotic contributions: a nondecaying oscillatory real-pole term and an exponentially small complex-pole term. There is no second nonzero inverse-power term. This formula is qualified by the principal-value choice; it is not the ordinary transform requested in the statement.

Other prescriptions change the leading term. For example, replacing the real pole by its upper or lower boundary value changes the distribution by $\mp i\pi \delta (x+1)/3$, and hence changes the transform by $\mp i\pi e^{ik}/3$. This demonstrates why a pole prescription is essential. The smooth Taylor series at the origin alone would miss the decisive real-pole contribution.