If I have a function and I want to represent it as being the Laplace transform of another, that is, I want to be sure that there is $\hat{f}(s)$ such that my function $f(x)$ can be written as:

$f(x) = \int ds \hat{f}(s) \exp(-sx)$

what conditions should I impose over $f(x)$?

In other words, what are the conditions for the Fourier-Mellin-Bromwich integral

$\hat{f}(s) = \frac{1}{2\pi i} \int_{\gamma - i\infty}^{\gamma + i\infty} f(x) \exp(sx) dx$

to exist?

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    $\begingroup$ Another formula for the inverse Laplace transform... en.wikipedia.org/wiki/Post%27s_inversion_formula $\endgroup$ Nov 3 '10 at 20:40
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    $\begingroup$ "In other words" isn't correct, as these are two different questions. The existence of the Bromwich integral does not imply that a function is a Laplace transform. A simplest example is when the values differ for different $\gamma$'s. I doubt that it is a Laplace transform even if the values agree for all $s$ and $\gamma$. $\endgroup$
    – zhoraster
    Nov 3 '10 at 23:17
  • $\begingroup$ I've added the Complex Variables and Fourier Analysis tags, since (I think) they're definitely relevant here - I hope no-one objects! $\endgroup$
    – Zen Harper
    Dec 3 '10 at 7:56
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    $\begingroup$ -zhoraster: you are correct; there are analytic functions $f$ where the Bromwich integral converges for all $\gamma$ and gives you a nice, consistent function $g(s)$, whose Laplace transform is NOT $f$. For example, take $f(z) = \exp(z^2)$ which cannot be represented as the bilateral Laplace transform of any function (or even any distribution), even though it has very rapid vertical decay (but non-uniformly in $x$). However, if we assume that the analytic function $f$ can be represented by some Laplace transform, then the Bromwich inversion integral does work under reasonable conditions. $\endgroup$
    – Zen Harper
    Dec 3 '10 at 10:22

The answer depends on the class of functions $\phi(t):(0,\infty)\to\mathbb R$ where you want to define the Laplace transform. A standard assumption is that $$e^{-ct}\phi(t)\in L^2(0,\infty)\qquad\qquad\qquad(1) $$ for some $c\in \mathbb R$. In this case the Laplace transform $$f(s)=\int_{0}^{\infty}e^{-st}\phi(t)dt\qquad\qquad\qquad(2)$$ can be extended analytically to the right half-plane $\{s=\sigma+i\tau:\ \sigma>c\}$. Moreover, it is easy to check that $$\sup\limits_{\sigma>c}\int_{-\infty}^{\infty}|f(\sigma+i\tau)|^2d\tau<\infty.\qquad(3)$$ Now rewriting $(2)$ as $$f(\sigma+i\tau)=\int_{0}^{\infty}e^{-it\tau}e^{-\sigma\tau}\phi(t)dt,$$ we observe that $f$ is just the Fourier transform of the function $e^{^{-\sigma t}}\phi(t)$ (trivially extended by $0$ to $t\leq 0$) belonging to $L^2(\mathbb R)$ for $\sigma=c$ and to $L^1(\mathbb R)\cap L^2(\mathbb R)$ for $\sigma>c$. Taking the inverse Fourier transform, we get that $$ e^{-\sigma t}\phi(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty}e^{it\tau}f(\sigma+i\tau)d\tau,\qquad t>0,$$ and $$0=\frac{1}{2\pi}\int_{-\infty}^{\infty}e^{it\tau}f(\sigma+i\tau)d\tau,\qquad t<0,$$ or, equivalently, $$\lim\limits_{T\to\infty}\frac{1}{2\pi i}\int_{\sigma-iT}^{\sigma+iT}e^{st}f(s)ds=\begin{cases} \phi(t), & t>0 \\\ \\\ 0, & t\leq 0 \end{cases} $$

One can show also that the Parseval identity holds

$$\frac{1}{2\pi}\int_{-\infty}^{\infty}|f(\sigma+i\tau)|^2d\tau=\int_{0}^{\infty}e^{-2\sigma t}|\phi(t)|^2dt,$$ so there is a complete analogy with the standard Fourier transform.

Executive summary. A function $f$ is the Laplace transform of some function $\phi$ satisfying condition (1), if and only if it can be extended analytically to the right half-plane $\{s=\sigma+i\tau:\ \sigma>c\}$ and condition (3) holds. This class of functions is known as the Hardy space on a (right) half-plane.

  • $\begingroup$ Yes; although this answer sketches the proof of only the easier part of the equivalence (needing only Parseval's Theorem). Given a function $F$ in the Hardy space $H^2$ of some right half-plane, it is the Laplace transform of some $\phi$; but this result, although not too hard, is non-trivial and requires a little bit more than just Parseval's Theorem. (It's sometimes called the Paley-Wiener Theorem, but it seems not everyone uses this name, and there is another related but different result also called the Paley-Wiener Theorem). $\endgroup$
    – Zen Harper
    Dec 3 '10 at 10:11
  • $\begingroup$ ...Sorry to keep self-referencing, but I'll also point out that a natural weighted generalisation of the Paley-Wiener theorem, by applying Parseval's Theorem to each vertical line, does NOT always hold - which was a big surprise to me, anyway. See my comment to the original question. This Laplace transform business can be quite subtle at times; I discuss this a little bit in my Documenta Math. paper referenced in my answer (and also another paper of mine appearing in the journal Complex Analysis and Operator Theory in 2009, which is also referenced by that paper). $\endgroup$
    – Zen Harper
    Dec 3 '10 at 10:29

This kind of question is very interesting, and I too would like to know answers.

Sorry to self-publicise; I hope it's not regarded as impolite, but since I have also considered this exact kind of question, it's quickest just to refer to my own paper (and the references I give in there):

Laplace Transform Representations and Paley-Wiener Theorems for Functions on Vertical Strips by Zen Harper. Documenta Math. 15 (2010) 235-254.

This is freely available online from the journal:


Given an analytic function on a vertical strip, I try to find conditions which guarantee it can be represented by a bilateral Laplace transform (in various senses). I definitely don't claim to have any kind of complete answer, but it's the best I know of (by definition! If I knew any better answers, I would have written them in my paper!)

It seems like there are still many open questions about this.

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    $\begingroup$ The link appears to be broken. $\endgroup$ Jul 6 '19 at 16:15

In Audrey Terras' "Harmonic Analysis on Symmetric Spaces and Applications, I" she has on p. 21 the following: Suppose $\exp(-cx)f(x)$ lies in $L^1(\mathbb R)$ and $f(x)$ vanishes for negative $x$. Assume also that $f(x)$ is piecewise differentiable. Then

$(f(x+)+f(x-))/2=\lim_{T\to\infty}\frac{1}{2\pi i}\int_{c-i T}^{c+i T}\exp(sx)\mathcal Lf(s)ds$

(NB traditional Laplace transform notation seems to be the reverse of yours; $x>0$ and $s\in\mathbb C$)


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