# How do I estimate/bound the error in an inverse Laplace transform?

Suppose I have a Laplace transform,

$$F(s) = \int_0^{\infty}dx\ f(x)e^{-s x} \ .$$

I know that

$$F(s) \approx e^{A/(4s)}$$

(for $s$ real) where $A$ is very large, and I want to estimate $f(x)$. I suspect that $f(x) \approx e^{\sqrt{A x}}$, because then the integral can be done by the saddlepoint method and agrees with $F(s)$. But can you suggest a method or reference to make this sort of statement precise? (Or is it wrong?)

In other words, can I reliably take the inverse Laplace transform of a function $F(s)$ that is known only approximately?

This question is motivated by a physics calculation, where $F(s)$ is the free energy at temperature $T=1/s$ and I am trying to infer something rigorous about the density of states $f(x)$.

• What does your formula $F(s)\approx s^{A/(4s)}$ mean? Is this for small $s$? For large $s$? Please explain. – Alexandre Eremenko Nov 3 '13 at 2:32

## 2 Answers

A very sobering warning (with some relevant estimates) about inverting the Laplace transform was given by Charlie Epstein and John Schotland here. (the paper appeared in the SIAM review in 2008).

With $\hat{}$ standing for the Fourier transform, you have, at least formally, $$F(s)=\int H(x) f(x) e^{-2i\pi\frac{sx}{2iπ}}dx=\widehat{fH}(\frac{s}{2iπ})$$ and still formally, this entails $$f(x) H(x)=\int_\mathbb R F(2iπ s)e^{2iπ s x} ds. \tag 1$$ As a result you get an integral expression for $f$ in terms of $F$. To justify the formal calculation above, you need at least to prove that the distributions whose Fourier transforms are taken are indeed tempered distributions, so that the formulas hold in a weak sense. With some holomorphy assumptions (e.g. on the upper half-plane), you can certainly deform the real line into a contour on which the integral (1) is absolutely converging.

You have to impose some hypotheses on the domain of homomorphy of $F$ and to precise where your asymptotic assumption holds.