When I can safely assume that a function is a Laplace transform of other function? 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?
 A: 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.
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.
A: 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 $s\in\mathbb C$.)
A: 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
$$e^{-ct}\phi(t)\in L^2(0,\infty)\tag{1}\label{1} $$
for some $c\in \mathbb R$. In this case the Laplace transform
$$f(s)=\int_{0}^{\infty}e^{-st}\phi(t)dt\tag{2}\label{2}$$
can be extended analytically to the right half-plane $\{s=\sigma+i\tau:\ \sigma>c\}$. Moreover, it is easy to check that
$$\sup\limits_{\sigma>c}\int_{-\infty}^{\infty}|f(\sigma+i\tau)|^2d\tau<\infty.\tag{3}\label{3}$$
Now rewriting \eqref{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
$$\frac{1}{2\pi}\int_{-\infty}^{\infty}|f(\sigma+i\tau)|^2d\tau=\int_{0}^{\infty}e^{-2\sigma t}|\phi(t)|^2dt$$
holds, 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 \eqref{1}, if and only if it can be extended analytically to the right half-plane $\{s=\sigma+i\tau:\ \sigma>c\}$ and condition \eqref{3} holds. This class of functions is known as the  Hardy space  on a (right) half-plane.
