# Understanding the inverse Laplace transform of a function with essential singularities

I need to do an inverse Laplace transform of a function with essential singularities for a specific problem. I find it is very similar to an equation J. Noolandi worked out in one of his papers in 1977. The paper is at this page, and I have taken a snapshot of the Appendix page where the math problem and the solution were given (view it here). But I don't quite get some part of the solving process which I need help on deciphering.

If I can understand what he was doing, my specific problem can be solved. So now let's take the problem in the Appendix of Noolandi's paper for example. The function is

$\displaystyle \tilde I(s) = \frac{1-\exp(-a(s)t_0)}{a(s)}$

where $\displaystyle a(s)t_0 = \Sigma_{i=1}^n \frac{s\omega_i}{s+r_i} t_0 = \Sigma_{i=1}^n \frac{sM_i}{s+r_i}$. "$s$" is the variable. $\omega_i$, $r_i$, $t_0$, and $M_i$ are all real numbers greater than zero.

The question is how to get inverse Laplace transform $L^{-1}[\tilde I(s)]$. I think the author used Residue Theorem to evaluate the contour integral $\int_C \tilde I(s) \exp(st) ds$, but the hard part is the singularities are essential singularities, thus the residues can not be got conventionally.

However, it seems that the author constructed a contour and solved it anyway. I don't get Equation A5-A8 in the Appendix, and don't understand how to determine the parameter $R$ and $R_1$ he introduced in.

By the way, the author also said the inverse Laplace transform could also be evaluated using Laplace transform tables. And the result involved a convolution of the modified Bessel function of the first order. I still haven't worked it out either...

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Hi, this question will probably be closed as not being "research level". But I would say: (1) what is the exact function you want to invert? You haven't said in your question; (2) Inverse Laplace transforms don't always exist, you need conditions on the analytic function; (3) The Residue Theorem works for essential singularities also, only the coefficient of 1/z is important. Whether the negative powers of z are finite or infinite in number is irrelevant, since the Laurent series always converges nicely enough to interchange contour integration with the infinite sum. – Zen Harper May 31 '12 at 11:20
Thanks for your reply. (1) Sorry about the ambiguity. What I was trying to say was I needed to understand how to do the inverse Laplace transform of function like $\tilde I(s)$ in that paper. Once I understood that, my real question would be solved consequently. (2) The answer doesn't have to be analytic. I think Noolandi gave the guideline for doing the transform numerically. But the problem is I don't understand his approach on the paper. (3) Yes. But expanding the function in Laurent series and filtering out the z^{-1} term is not easy. I am hoping for a more clear. – user24116 May 31 '12 at 15:36
I've added the Fourier Analysis tag, since I think it is appropriate [although that usually goes without saying, when Laplace transforms are used!] – Zen Harper Jun 1 '12 at 10:41

After taking a quick look at the paper, I agree with Robert Israel.

Almost no detailed justification is given (perhaps not surprising for a Physics journal...), and it seems to be not so easy to justify.

Take the special case $N=1$ for simplicity; so up to constants, we consider (note the use of (A3) in the paper, this constant is subtracted from the original $\tilde{I}$)

$$\tilde{I}(s) = \frac{(1-\exp(-Ks/(s+r))}{s/(s+r)} - (1-\exp(-K))$$

for $K$, $r$ constants. Now consider what happens as $|s| \to \infty$, then we get [assuming, dangerously, that my calculations are error free]

$$\tilde{I}(s) \sim -Kr e^{-K}/(s+r) \sim A/s$$

as $|s| \to \infty$, for some constant $A$. The contour is originally a vertical line, but $1/s$ is not in $L^1$ so the integral is not a standard Lebesgue integral (i.e., is not absolutely convergent). However, it is is $L^2$, i.e. is square integrable, so you can use the $L^2$ theory of the Fourier transform to give a meaning.

Now, if you deform the original vertical line contour over { Re(s) = c } by Cauchy's Theorem (push part of it to the left), you get the integral of $\tilde{I}(s) e^{st}$ over the circle $C$ plus an error term, being the sum of integrals over the contours

C1 = { $L + iy : |y| < R$ },

C2 = { $x \pm i R : L < x < c$ },

C3 = { $c + iy : |y|>R$ }

where $c$ is fixed; now let $L \to -\infty$ and $R \to +\infty$ appropriately to get the error term going to zero.

[More detail: as $R \to \infty$ the integral over C3 goes to 0 uniformly in $t$, by the $L^2$ properties of the Fourier transform; also the integral over C2 goes to zero because $|1/s| < 1/R$ on C2.

The integral over C1 goes to zero as $L \to -\infty$ because $| e^{st} / s | < e^{Lt}$ on C1, which clearly $\to 0$ rapidly, at least for $t>0$. ]

There are still 2 things to be justified in this special case: (i) what about $t<0$, and (ii) how do we know the analytic function $\tilde{I}(s)$ really can be represented as the Laplace transform of something?

I think (ii) follows from a general result in my last paper (about Laplace transform representation theorems, which is in Documenta Mathematica 2010, or on my website); but (i) I am not sure about.

[Strangely enough, the contours C1, C2, C3 are exactly the ones I used in my paper, although I got them from an earlier paper by C.Batty and M.D.Blake; and I suspect the original use of these contours dates back at least 50 years, since similar stuff has been used in Tauberian theory and analytic number theory for a long time].

So, even in this special case with just one singularity, there are a lot of extra details to fill in if you want to make it properly rigorous. Since the original paper is from a Physics journal, I would guess it's not the best place to look for rigour(!) Approach with caution...

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The paper is not using residue theory, he is evaluating the contour integral numerically. I don't see a justification given for deforming the path of integration from a vertical line to a circle, but assuming sufficiently nice properties at infinity that should be OK.

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