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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 $f(x)$.

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

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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.

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