= Solution
Choose the principal <branch of a multivalued function>, $-\pi<\arg s<\pi$, for $s^{1/2}$. The integrand $G(s)=e^{st}/((s^2+1)\sqrt s)$ has simple poles at $\pm i$ and a <branch cut> along the negative real axis. Its residues are
$$
\operatorname{Res}_{i}G=\frac{e^{it-i\pi/4}}{2i},\qquad \operatorname{Res}_{-i}G=-\frac{e^{-it+i\pi/4}}{2i},
$$
whose sum is $\sin(t-\pi/4)$.
For $t>0$, close the <Bromwich contour> to the left using a rectangle with outer real part $-R$ and horizontal edges at imaginary parts $\pm R$, indenting the cut and zero. On every distant edge, $|s^2+1|\ge|s|^2-1$, $|\sqrt s|=|s|^{1/2}$ and $|e^{st}|\le e^{\gamma t}$. Each edge has length $O(R)$ and $|s|\ge R$, so its <integral> is $O(R^{-3/2})$. The small circle of radius $\varepsilon$ about zero contributes $O(\varepsilon^{1/2})$. The tails of the original vertical <integral> are absolutely bounded by $O(R^{-3/2})$ as well.
On the upper and lower banks at $s=-r$, the square roots are $i\sqrt r$ and $-i\sqrt r$, respectively. With the upper bank traversed toward zero and the lower bank away from zero, their combined <integral> is
$$
-2i\int_\varepsilon^R\frac{e^{-rt}}{(r^2+1)\sqrt r}\,dr.
$$
The cut <integral> converges near zero since $r^{-1/2}$ is integrable, and at infinity by exponential decay. The <residue theorem>, followed by $R\to\infty$ and $\varepsilon\to0$, therefore gives
$$
\boxed{f(t)=\sin(t-\pi/4)+\frac1\pi\int_0^\infty\frac{e^{-rt}}{(r^2+1)\sqrt r}\,dr\quad(t>0).}
$$
For $t<0$, close the <Bromwich contour> to the right. There are no enclosed poles or cuts, and $|e^{st}|\le e^{\gamma t}$ there. The same distant-edge estimates show \b[$f(t)=0$ for $t<0$].
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2009/ib/paper-1-bromwich.png]
{title=Left-closing Bromwich contour with the principal square-root cut, two pole residues, and a small indentation at zero}
For the large-$t$ expansion, $1/(1+r^2)=1-r^2+r^4/(1+r^2)$. Scaling $r=u/t$ and using the given <integral> yields $\int_0^\infty e^{-rt}r^{-1/2}\,dr=\sqrt\pi\,t^{-1/2}$. Integration by parts twice gives the next coefficient $\Gamma(5/2)=3\sqrt\pi/4$. The remainder is bounded by $\int_0^\infty e^{-rt}r^{7/2}\,dr=\Gamma(9/2)t^{-9/2}$. Thus, without needing a globally convergent termwise integration of the local power series,
$$
f(t)=\sin(t-\pi/4)+\frac1{\sqrt{\pi t}}-\frac3{4\sqrt\pi\,t^{5/2}}+O(t^{-9/2}).
$$
In particular, the requested two leading terms are \b[$\sin(t-\pi/4)+(\pi t)^{-1/2}$], interpreted as an additive asymptotic expansion. This is <Bromwich inversion with a square-root branch cut>.
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