Timeline for An inverse Laplace transform involving Error function

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Jun 4, 2012 at 18:33	comment	added	Anand		Dear Zen Harper, thanks a lot. It works. After applying inverse Laplace transform of the two functions, I applied convolution, where I can use Lebesgue's dominated convergence theorem to find the right limiting result. Thanks a lot! :-)
Jun 4, 2012 at 9:14	comment	added	Zen Harper		Sorry, I meant "convolve" rather than "multiply", of course (Laplace transforms change convolution into multiplication).
Jun 4, 2012 at 9:05	comment	added	Zen Harper		If you know the inverse transform of $g$, then you can get $g_a$ by a convolution, since $g_a(z)= (1-2 / \sqrt{z})^{-1} g(z)$, so you only need to find the inverse transform of $(1-2 / \sqrt{z})^{-1}$, which is an easy power series in $1/ \sqrt{z}$, then multiply. There's no guarantee that the resulting expression is particularly nice or useful, though.
Jun 2, 2012 at 21:10	history	edited	Anand	CC BY-SA 3.0	edited title; edited title
Jun 2, 2012 at 21:01	history	edited	Anand	CC BY-SA 3.0	added 87 characters in body
Jun 2, 2012 at 20:59	comment	added	Anand		Dear Professor Juan, I tried the hints you gave. It is not very easy. The pattern of the derivatives of $z^{-1/2}e^{az}\text{erfc}(\sqrt{az})$ is a little complicated.
Jun 2, 2012 at 20:49	history	edited	Anand	CC BY-SA 3.0	added 2 characters in body
Jun 2, 2012 at 20:43	history	edited	Anand	CC BY-SA 3.0	added 432 characters in body; added 3 characters in body; added 14 characters in body; added 2 characters in body
Jun 2, 2012 at 20:39	comment	added	Anand		Dear Professor Igor Rivin, I add my motivation in my post above.
Jun 2, 2012 at 20:06	comment	added	Igor Rivin		What exactly is it you want? A closed form? Asymptotics? Something you can evaluate numerically?
Jun 2, 2012 at 20:05	comment	added	Anand		Dear Professor Juan, I think it works by expanding in $z^{-1/2}$. Thanks a lot. I will try.
Jun 2, 2012 at 19:46	comment	added	juan		Perhaps you may expand $(\sqrt{z}-2)^{-1}$ in powers of $z$ then multiply by $e^{az} \text{erfc}(az)$ and apply Laplace transform to each term. The resulting terms would be derivatives one of the others. Possibly obtain a not converging series but maybe with some asymptotic properties. Or apply the same idea but expanding in powers of $z^{-1/2}$. The parameter a can be almost eliminated. This will simplify the computations.
Jun 2, 2012 at 18:11	history	edited	Anand	CC BY-SA 3.0	added 1 characters in body
Jun 2, 2012 at 17:42	history	asked	Anand	CC BY-SA 3.0