= Solution
Taking the <Laplace transform> in $t$ and using the zero initial condition gives
$$
p\widetilde T=\kappa\widetilde T_{xx}.
$$
The solution bounded as $x\to\infty$ is
$$
\widetilde T(x,p)=\widetilde T_0(p)
\exp\left(-x\sqrt{\frac p\kappa}\right).
$$
The transformed boundary condition is
$$
\boxed{\widetilde T_0(p)=\frac{\omega}{p^2+\omega^2}},
$$
so this is the <sinusoidally forced heat equation on a half-line>. Integrating in $x$ gives
$$
\boxed{
\widetilde I(p)
=\int_0^\infty\widetilde T(x,p)\,dx
=\frac{\omega\sqrt\kappa}
{\sqrt p\,(p^2+\omega^2)}
}.
$$
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 $p=\pm i\omega$ contribute
$$
\sqrt{\frac{\kappa}{\omega}}
\cos\left(\omega t-\frac{3\pi}{4}\right)
=-\sqrt{\frac{\kappa}{2\omega}}\cos\omega t
+\sqrt{\frac{\kappa}{2\omega}}\sin\omega t.
$$
Hence
$$
\boxed{A=-\sqrt{\frac{\kappa}{2\omega}}},
\qquad
\boxed{B=\sqrt{\frac{\kappa}{2\omega}}}.
$$
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 give
$$
\frac1\pi\int_0^\infty
\frac{\omega\sqrt\kappa\,e^{-st}}
{(s^2+\omega^2)\sqrt s}\,ds.
$$
This 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.
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