 Solution

The derivative formula requires additional regularity: continuity alone does not define $f^{\prime }$. Assume $f$ is continuously differentiable, or locally absolutely continuous with its almost-everywhere derivative. Integration by parts on $[0,R]$ givesFor $\mathrm{Re}p>c$, the given exponential bound makes the first term tend to zero. Thus the improper transform exists under these regularity assumptions andFor the primitive, absolute integrability justifies reversing the integration order:Indeed the corresponding absolute double integral is finite when $\mathrm{Re}p>c$.

For real $p>0$, substitution $t=s^2/p$ and the supplied Gaussian integral giveBoth sides are holomorphic for $\mathrm{Re}p>0$, so the identity theorem extends this equality there, taking the square root with positive real part. The integrable singularity at zero causes no difficulty for this improper integral. The primitive rule then gives $\mathcal{L}[2\sqrt{t}]=\sqrt{\pi }{p}^{-3/2}$.

For the conductor, take the Laplace transform in time and write $U(x,p)$. The initial temperature gives $U_{xx}=pU$. Use the usual physical condition that the temperature is bounded far down the conductor; this selects the decaying solutionThe constant applied flux has transform $Q/p$, so $-k{U}_x(0,p)=Q/p$, giving $A(p)=Q/(k{p}^{3/2})$. This is the surface temperature under constant heat flux on a half-line. Using the transform just calculated yieldsHere $k>0$ is the thermal conductivity. The far-field boundedness is the customary physical selection condition; without a spatial growth restriction the written PDE and initial/boundary data alone do not specify that selection. Within the physical solution, the face temperature tends continuously to zero as $t\downarrow 0$.