 Solution

Under the usual existence and boundary hypotheses for the Laplace transform,The first formula requires a finite right limit and sufficient regularity for integration by parts; it is not to be applied blindly to the singular function $t^{-1/2}$. For that function and real $p>0$, substitute $s=\sqrt{pt}$:Use the second transform identity to obtain

Take the time transform of the heat equation, with differentiation and boundary limits interpreted in the standard decaying heat-solution class. The initial value is zero, so $v_{xx}=pv-u(x,0)=pv$. Transforming the constant flux gives $v_x(0,p)=-1/p$. Decay selects the decreasing solution of the transformed ordinary differential equation, givingConsequentlyThe boundary value follows by comparing $v(0,p)=p^{-3/2}$ with the transform of $\sqrt{t}$.

One can also check the decay and transform steps without relying on a merely pointwise limit under an improper integral. Using the complementary error function, the classical heat solution isDifferentiation verifies $u_{xx}=u_t$, $u_x(0,t)=-1$, and $u(x,0)=0$ for $x>0$. Its integral representation gives $0\le u(x,t)\le 2\sqrt{t/\pi }$ and pointwise decay as $x\to \infty$. Thus $e^{-pt}2\sqrt{t/\pi }$ supplies an integrable majorant, proving $v(x,p)\to 0$ by dominated convergence. The same representation gives the required regularity for the transforms. This explicitly verifies that the solution selected by the decaying transformed equation lies in the intended heat-solution class.