= Solution
Because $\operatorname{Re}a>0$, 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.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2018/ii/paper-1-laplace-contours.png]
{title=Contour choices for the <Laplace transform>}
For $a=1$, $f(t)=1$ is entire, so path independence makes either contour give, up to orientation,
$$
\int_{-i}^{i}e^{zt}\,dt
=\frac{e^{iz}-e^{-iz}}{z}
=\frac{2i\sin z}{z}.
$$
Hence a nonzero constant multiple gives the solution
$$
\boxed{\ w_1(z)=\frac{\sin z}{z}\ }.
$$
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