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?



When I can safely assume that a function is a Laplace transform of other function?
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.




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**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](http://en.wikipedia.org/wiki/Hardy_space) on a (right) half-plane.


