Past exam of the mathematics course of the University of Cambridge / 2018 / ii / Paper 1 / 14B (Further Complex Methods)

Insertinto the equation. Since $w^{\prime }={\int }_{\gamma }t{e}^{zt}fdt$ and $w^{\prime \prime }={\int }_{\gamma }{t}^2{e}^{zt}fdt$, use $z{e}^{zt}={\partial }_t{e}^{zt}$ and integration by parts:The integral vanishes whenso, up to a constant and on a consistently chosen branch,The remaining contour condition isThis is the Laplace contour-integral method for a linear differential equation.

Because $\mathrm{Re}a>0$, the endpoint factor $(1+t^2{)}^a$ vanishes at the branch points $t=\pm i$. A finite contour may therefore run from $-i$ to $i$ on either side of a chosen branch cut; the diagram shows contours bowing through the left and right half-planes.

For $a=1$, $f(t)=1$ is entire, so path independence makes either contour give, up to orientation,Hence a nonzero constant multiple gives the solution

For real $z>0$, the exponential decays as $\mathrm{Re}t\to -\infty$, so two allowed contours run from $-\infty$ to $i$ and from $-\infty$ to $-i$. For real $z<0$, it decays as $\mathrm{Re}t\to +\infty$, so reverse the construction: run from $i$ or $-i$ to $+\infty$. These are the infinite contours in the diagram.

For $z>0$ their integrals arefor $z<0$ the corresponding orientation changes only an overall sign. Taking the sum of the two exponentials produces a solution independent of $\sin z/z$:Its nonzero Wronskian with $w_1$ on $z\ne 0$ proves linear independence.