Calculating the envelope of a function in the Laplace Domain I have a PDE problem (namely the heat equation for a semi-infinite solid) and I'm trying to solve it using the Laplace Transform. 
$$\frac{1}{\alpha} \frac{\partial T}{\partial t} = \frac{\partial^2 T}{\partial x^2}$$
The conditions are:


*

*$T(x,0)=0$

*$T(0,t)=A\cos(\omega t$)


I can find the solution in the $s$ domain, $F(x,s)$, but I cannot antitransform it to the time domain.
However, I'm not interested in the solution $f(x,t)$ but rather on the envelope of the function, that is, a function $f_e(x)$ that gives the maximum temperature reached at each point of the solid. I kown from the numerical solution that the envelope is a decaying exponential.
Is there any way to compute this envelope mathematically from the Laplace Transform $F(x,s)$?
 A: If you view this question by power series method approach, you will find that it is easier to solve.
Similar to Model for soil temperature underground using the Heat Equation:
Let $T(x,t)=\sum\limits_{n=0}^\infty\dfrac{x^n}{n!}\dfrac{\partial^nT(0,t)}{\partial x^n}$ ,
Then $T(x,t)=\sum\limits_{n=0}^\infty\dfrac{x^{2n}}{(2n)!}\dfrac{\partial^{2n}T(0,t)}{\partial x^{2n}}+\sum\limits_{n=0}^\infty\dfrac{x^{2n+1}}{(2n+1)!}\dfrac{\partial^{2n+1}T(0,t)}{\partial x^{2n+1}}$
$=\sum\limits_{n=0}^\infty\dfrac{x^{2n}}{\alpha^n(2n)!}\dfrac{\partial^nT(0,t)}{\partial t^n}+\sum\limits_{n=0}^\infty\dfrac{x^{2n+1}}{\alpha^n(2n+1)!}\dfrac{\partial^{n+1}T_x(0,t)}{\partial t^n}$
$=\sum\limits_{n=0}^\infty\dfrac{x^{2n}}{\alpha^n(2n)!}\dfrac{\partial^n(A\cos\omega t)}{\partial t^n}+\sum\limits_{n=0}^\infty\dfrac{x^{2n+1}}{\alpha^n(2n+1)!}\dfrac{\partial^{n+1}T_x(0,t)}{\partial t^n}$
$=\sum\limits_{n=0}^\infty\dfrac{(-1)^nA\omega^{2n}x^{4n}\cos\omega t}{\alpha^n(4n)!}-\sum\limits_{n=0}^\infty\dfrac{(-1)^nA\omega^{2n+1}x^{4n+2}\sin\omega t}{\alpha^n(4n+2)!}+\sum\limits_{n=0}^\infty\dfrac{x^{2n+1}}{\alpha^n(2n+1)!}\dfrac{\partial^{n+1}T_x(0,t)}{\partial t^n}$
