Past exam of the short options paper shared by Prelims and Parts A and B of the physics course of the University of Oxford / 2017 / S1

Complex differentiability gives $u_x=v_y$, $u_y=-v_x$. Differentiating once more, for continuously differentiable first derivatives with the needed second derivatives, yields $u_{xx}+u_{yy}=v_{yx}-v_{xy}=0$ and $v_{xx}+v_{yy}=-u_{yx}+u_{xy}=0$. Thus both components satisfy Laplace's equation. The result is local on the domain of holomorphy; it makes no assertion across a branch cut.

Choose a continuous argument $\theta$ on a simply connected domain avoiding zero, and put $r=\sqrt{x^2+y^2}$. With $u=x\log r-y\theta$, direct differentiation givesHere ${\theta }_x=-y/r^2$, ${\theta }_y=x/r^2$; these identities specify the branch needed in the written arctangent. Integrating the Cauchy-Riemann equations givesThe literal principal $\arctan (y/x)$ works only separately on suitable half-planes. It cannot stand for a single continuous argument across the imaginary axis.

For the upper half-plane use $0<\theta <\pi$, and interpret the source's arctangent as this argument. Then $V=(V_0/\pi )\mathrm{Im}(z\mathrm{Log}z)$ is harmonic, so $\mathbf{E}=-\nabla V$ has zero curl and divergence where there is no charge. On the positive boundary $\theta \to 0$ and $y\log r\to 0$; on the negative boundary $\theta \to \pi$. Hence the two boundary values are respectively zero and $V_{0}x$.

Branch qualification: the principal real arctangent alone gives zero on both sides of $y=0$, contradicting the stated negative-axis boundary condition, for example at $x=-1$. The continuous upper-half-plane argument repairs this notation; it is not an optional additive constant.

Along a level curve, $V_{x}dx+V_{y}dy=0$, with $V_x=V_0\theta /\pi$ and $V_y=V_0(1+\log r)/\pi$. ThereforeIn the intended upper half-plane $0<\theta <\pi$, the slope is positive for $0<r<e^{-1}$, vertical at $r=e^{-1}$, and negative for $r>e^{-1}$. At the positive real boundary $\theta =0$ the slope is zero except at the stationary point $(e^{-1},0)$, where both derivatives vanish and the quotient is indeterminate. More generally, outside the chosen half-plane its sign is the sign of $-\theta (1+\log r)$, with the argument branch held fixed. A convenient parametric description of the curves to sketch is $V\pi /V_0=r(\sin \theta \log r+\theta \cos \theta )$.

A branch point obstructs a single-valued analytic continuation around a closed circuit. Continuing $\log z=\log |z|+i\theta$ once around zero increases its value by $2\pi i$. A branch cut removes a curve joining branch points or extending to infinity so that a continuous argument can be chosen on the remainder. Zero and infinity are logarithmic branch points; the positive-axis cut used here selects $0<\arg z<2\pi$. The cut is a choice of representation, not a line of poles of the multivalued logarithm.

The large circle contributes $O((\log R{)}^n/R)$ and the small circle contributes $O(ϵ|\log ϵ{|}^n)$, so both disappear. The enclosed simple poles are $i$ and $-i$, with logarithms $i\pi /2$ and $3i\pi /2$. Their residues are $(i\pi /2{)}^n/(2i)$ and $-(3i\pi /2{)}^n/(2i)$. The positive orientation of the keyhole therefore givesThe upper bank runs outward and the lower bank inward. Their logarithms differ by $2\pi i$; keeping that orientation is essential for the sign.

Taking the difference of the two bank integrals at $n=2$ givesThe $I_j$ are real and convergent. Comparing real and imaginary parts proves $\boxed{I_0=\pi /2,I_1=0}$. Independently, $x\mapsto 1/x$ changes the sign of each odd logarithmic moment and preserves the measure, so every odd $I_j$ vanishes.

At $n=3$ the bank difference is $-6\pi i{I}_2+12{\pi }^2{I}_1+8{\pi }^{3}i{I}_0$, while the residue result is $13i{\pi }^4/4$. Substitution of $I_0,I_1$ yields $I_2={\pi }^3/8$.

At $n=5$, the odd moments vanish and the remaining relation isThusAs an independent generating-function check, $F(s)={\int }_0^{\infty }{x}^s/(1+x^2)dx=\pi /[2\cos (\pi s/2)]$ for $|\mathrm{Re}s|<1$. Differentiating twice and four times at zero gives the same moments.

For a meromorphic function with no poles on a positively oriented closed contour, the contour integral equals $2\pi i$ times the sum of enclosed residues, weighted by winding number for a general contour. At a simple pole $a$, the residue is ${\lim }_{z\to a}(z-a)f(z)$. Analyticity on the contour and the absence of other interior singularities are necessary hypotheses.

Put $z=e^{i\theta }$, so $d\theta =dz/(iz)$, $\cos \theta =(z+z^{-1})/2$ and $\sin \theta =(z-z^{-1})/(2i)$. Rational expressions in these trigonometric functions become rational contour integrands on the unit circle. The method is especially useful when the resulting denominator has a small number of identifiable poles. A pole on the integration circle requires a separately specified principal value or indentation; the residue theorem does not automatically assign an ordinary integral there.

First remove the removable singularities introduced by tangent and secant: the integrand is $\cos \theta /(3+\cos \theta +2\sin \theta )$. The denominator stays positive since $3>\sqrt{5}$. With $z=e^{i\theta }$ the contour integrand isIts interior poles are zero and $r=-1/(1-2i)$; the other quadratic root is $-5/(1-2i)$ and lies outside. The residues are $1/[i(1+2i)]$ and $(r^2+1)/(4ir)$, whose sum is $i/10$. ConsequentlyAssigning arbitrary finite values at the isolated removable points does not change the integral.

Split the sine into exponentials. The $e^{iz}$ term closes in the upper half-plane and picks up the pole $ia$, giving $2\pi i{e}^{-a}$; the $e^{-iz}$ term closes clockwise below and encloses no pole. Jordan's lemma applies to these separate contours. HenceTo pass to $a\downarrow 0$ without an invalid dominated-convergence argument on an infinite interval, pair positive and negative $x$ first: the imaginary part is odd and the real part is $2{\int }_0^{\infty }x\sin x/(x^2+a^2)dx$. On a bounded interval the limit is harmless, while integration by parts bounds the oscillatory tail uniformly for $0<a\le 1$. This gives the convergent improper Dirichlet integral $\boxed{{\int }_{-\infty }^{\infty }\sin x/xdx=\pi }$.