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Take the branch of the complex logarithm with $-\pi <\arg s<\pi$ and write $s^{-1/2}=\exp [-\log s/2]$. Deform the Bromwich contour into a Hankel contour about the negative real axis. The small circle at zero contributes $O(r^{1/2})$ and vanishes; the large closing arcs vanish by the exponential estimates for $t>0$. On the upper lip $s=-r+i0$, $s^{-1/2}=-i/\sqrt{r}$, while on the lower lip it is $i/\sqrt{r}$. Keeping the lip orientations in the Bromwich inversion with a square-root branch cut givesWith $r=u^2$, the supplied Gaussian integral evaluates this as $2{\pi }^{-1}{\int }_0^{\infty }{e}^{-t{u}^2}du$. HenceThe inverse is singular but locally integrable at zero, which is compatible with existence of its Laplace transform.

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 arewhose 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 $\epsilon$ about zero contributes $O({\epsilon }^{1/2})$. 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 isThe 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 $\epsilon \to 0$, therefore givesFor $t<0$, 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 $f(t)=0$ for $t<0$.

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,In particular, the requested two leading terms are $\sin (t-\pi /4)+(\pi t{)}^{-1/2}$, interpreted as an additive asymptotic expansion. This is Bromwich inversion with a square-root branch cut.

Taking the Laplace transform in $t$ and using the zero initial condition givesThe solution bounded as $x\to \infty$ isThe transformed boundary condition isso this is the sinusoidally forced heat equation on a half-line. Integrating in $x$ gives

Use the principal square root, with a branch cut along the negative real axis. The Bromwich contour is a vertical line to the right of $p=\pm i\omega$ and the branch point $p=0$. For $t>0$, close it in the left half-plane, indent around the cut, and let the large semicircle recede. The simple poles at $p=\pm i\omega$ contributeHence

On the upper and lower sides of the cut, put $p=-s$; the two values of $\sqrt{p}$ are $i\sqrt{s}$ and $-i\sqrt{s}$. Their jump and the opposite contour orientations combine to giveThis is precisely Bromwich inversion with a square-root branch cut: the residues produce the permanent periodic response, while the cut integral is the transient that decays with time.