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Dear MOs,

I need to calculate the inverse Laplace transform of the following function

$$ g_a(z) = \frac{e^{a z}\: \text{erfc}(\sqrt{a z})}{\sqrt{z}-2},\quad a>0. $$

I have checked, among many others, the book "Table of Integral Transform, Vol. I". In P.267, Eq. (14) is for

$$ g(z) = \frac{e^{a z}\: \text{erfc}(\sqrt{a z})}{\sqrt{z}} $$

which is almost what I need. Other than this formula, I didn't find the one that I need. I have tried mathematica, which couldn't give an answer. I think the hope to find out the solution is quite small.

EIDT: here is some motivation of the problem.

Suppose the inverse transform gives us a function $f_a(t)$. I want to see the limit $$\lim_{a\rightarrow 0_+} f_a(t)=?$$

Can I simply do this:

$$ \lim_{a \rightarrow 0_+} \mathcal{L}^{-1}\left(g_a\right)(t) \stackrel{?}{=} \mathcal{L}^{-1}\left(\lim_{a\rightarrow 0_+} g_a\right)(t) = \mathcal{L}^{-1}\left(g_0\right)(t) =\frac{1}{\sqrt{\pi t}} + 2 e^{4t} \text{erfc}(-2\sqrt{t}) $$

Are there some Lebesgue's dominated convergence theorems to use in complex analysis?

Thank you very much for any hints!

Anand

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    $\begingroup$ 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. $\endgroup$
    – juan
    Commented Jun 2, 2012 at 19:46
  • $\begingroup$ Dear Professor Juan, I think it works by expanding in $z^{-1/2}$. Thanks a lot. I will try. $\endgroup$
    – Anand
    Commented Jun 2, 2012 at 20:05
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    $\begingroup$ What exactly is it you want? A closed form? Asymptotics? Something you can evaluate numerically? $\endgroup$
    – Igor Rivin
    Commented Jun 2, 2012 at 20:06
  • $\begingroup$ Dear Professor Igor Rivin, I add my motivation in my post above. $\endgroup$
    – Anand
    Commented Jun 2, 2012 at 20:39
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    $\begingroup$ 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. $\endgroup$
    – Zen Harper
    Commented Jun 4, 2012 at 9:05

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