Inverse Laplace transform of a certain expression I am trying to solve the following  problem
$$\rho\,C_{{p}}{\frac {\partial }{\partial t}}T \left( x,t \right) =
\kappa\,{\frac {\partial ^{2}}{\partial {x}^{2}}}T \left( x,t \right) 
$$
$$T(x,0)=T_0$$
$$T(0,t)=T_1e^{-\beta t}$$
$$T \left( \infty ,t \right) =T_0$$
where the initial  condition is for $x >0$.
I am using the laplace transform method.  In the laplace domain I am obtaining the following solution
$$T \left( x \right) ={{\rm e}^{-{\frac {\sqrt {\rho}\sqrt {C_{{p}}}
\sqrt {s}x}{\sqrt {\kappa}}}}} \left( {\frac {T_{{1}}}{s+\beta}}-{
\frac {T_{{0}}}{s}} \right) +{\frac {T_{{0}}}{s}}
$$
I am implementing the procedure both in Mathematica and Maple respectively but none of them is able to compute the inverse laplace transform of the last expression.
My question is: it is possible to obtain a closed form for the solution?
 A: OK, it is possible to find a closed form for this. Mathematica informs me the term proportional to $T_0$ is $\mathcal{L}\left[T_0\mathrm{erf}\left(\frac{x}{2}\sqrt{\frac{\rho C_p}{\kappa t}}\right)\right]$. For the term muliplying $T_1$, we use
$$
\mathcal{L}\left[e^{-\beta t}f(\beta t)\right] = \frac{1}{\beta}F\left(\frac{s}{\beta}+1\right)
$$
Mathematica informs me that
$$
\mathcal{L}^{-1}\left[\frac{\exp\left(-a \sqrt{u - 1}\right)}{u}\right] = \mathrm{Re}\left[e^{i a}\mathrm{erfc}\left(\frac{a + 2iv}{2\sqrt{v}}\right)\right]
$$
So we have
$$
\mathcal{L}^{-1}\left[T_1 \frac{\exp\left(-\sqrt{\frac{\rho C_p}{\kappa}} x\sqrt{s}\right)}{s+\beta}\right]
= T_1 e^{-\beta t}\mathrm{Re}\left[e^{i\alpha x}\mathrm{erfc}\left(\frac{\alpha x + 2i\beta t}{2\sqrt{\beta t}}\right)\right]
$$
where $\alpha = \sqrt{\rho C_p \beta/\kappa}$. I don't know of any way to simplify that real part term, but this should give you your answer.
A: Using Maple I am obtaining the following closed solution

Where "CylinderD" is  the Whittaker's Parabolic Function (https://en.wikipedia.org/wiki/Parabolic_cylinder_function)
The trick is to use
 
Using the identity

it is possible to rewrite the closed form solution in terms of Hermite functions as

From other side, using Mathematica and the same trick we obtain the closed solution in terms of hypergeometric functions as follows

This last solution coincides with the solution given by Maple in terms of parabolic cylinder functions when the following identity is used

The universal closed solution is obtained using Heun functions and it takes the form

Also it is possible to write the closed solution in terms of the MeijerG function as

The closed solution can be obtained using Mathematica with Raspberry pi. It is wonderful.
A: Other closed solution is given by

