Condition for the inverse laplace transform of a function to exist and bromwich integral Given any function, is there any way of determining from the nature of the function, if it is the laplace transform of a piecewise continuous function of exponential order? For e.g. say the function $\hat{F}(s)=s^2$? One method is too find the inverse using the bromwich integral. I am not sure how to do this. For e.g. for $s^2$
$$\mathcal{L^{-1}}=\lim_{R \to \infty} \frac{1}{2 \pi i}\int_{a-iR}^{a+iR} s^2 e^{sx} dx=[\frac{x^2 s^2 e^{sx}-2xse^{sx}+2e^{sx}}{x^3}]_{a-iR}^{a+iR}=\frac{e^a}{x^3}e^{iR}[4airx^2-4xaiR+2]$$
Now, clearly $a$ can be anything, and if I set it to 0, I get the limit of $\frac{2 e^{iR}}{x^3}$. This limit doesn't exist. Is it right? and so $s^2$ doesn't have an inverse laplace transform?
Am I going wrong somewhere, because I am skeptical of whether by limit is independent of my $a$ in the last expression. 

The idea behind the value of $a$ is that all the poles of thr LT should be to the left of the line $\Re{z}=a$.  In this way, the ILT will be zero for $t < 0$, which is what we need from the ILT.
As far as your specific $F(s)=s^2$ is concerned, the ILT will be zero unless $t=0$.  In that case, the integral over the Bromwich contour diverges.  It turns out that the ILT is a distribution $f(t) = \delta''(t)$, where $\delta(t)$ is the Dirac delta function.
