Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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

share|improve this question
2  
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. –  juan Jun 2 '12 at 19:46
    
Dear Professor Juan, I think it works by expanding in $z^{-1/2}$. Thanks a lot. I will try. –  Anand Jun 2 '12 at 20:05
1  
What exactly is it you want? A closed form? Asymptotics? Something you can evaluate numerically? –  Igor Rivin Jun 2 '12 at 20:06
    
Dear Professor Igor Rivin, I add my motivation in my post above. –  Anand Jun 2 '12 at 20:39
1  
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. –  Zen Harper Jun 4 '12 at 9:05
show 3 more comments

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.