Past exam of the mathematics course of the University of Cambridge / 2017 / ii / Paper 2 / 29E (Asymptotic Methods)

Keep $\nu$ fixed as $x\to +\infty$. On both vertical lines $\sin (\pm \pi +iy)=-i\sinh y$. The right side is traversed upward and the left downward, so their combined contribution isSet $v=\sinh y$. The integral becomes ${\int }_0^{\infty }{e}^{-xv}a(v)dv$, where $a(v)=e^{-\nu \mathrm{arsinh}v}/\sqrt{1+v^2}$, $a(0)=1$ and $a^{\prime }(0)=-\nu$. Successive integration by parts, with exponentially vanishing terms at infinity, givesHence the leading vertical contribution is $\boxed{V(x)=-\sin (\pi \nu )/(\pi x)+O(x^{-2})}$. It is exactly zero for integer $\nu$. These estimates hold for fixed real $\nu$; they are not uniform when the order grows with $x$.

The horizontal contribution is $B(x)=(2\pi {)}^{-1}{\int }_{-\pi }^{\pi }{e}^{-ix\sin z}{e}^{i\nu z}dz$. The two nondegenerate stationary points of the phase $-\sin z$ are $\pi /2$ and $-\pi /2$, with second derivatives $1$ and $-1$. The stationary phase method gives their contributionswith remainder $O(x^{-3/2})$ for smooth fixed amplitudes localized near the stationary points. On the remaining portions, use integration by parts. The endpoint term isCutoff endpoints cancel between pieces; a second integration gives the other nonstationary contributions $O(x^{-2})$. Thus this endpoint term cancels the leading vertical term. AltogetherThis stationary phase endpoint cancellation in a half-strip contour is necessary: omitting it would leave an erroneous $x^{-1}$ remainder for noninteger $\nu$.