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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.

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1 Answer 1

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.

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Assuming your $t$ is my $x$, I don't see why $\frac{2e^{iR}}{x^3}$ would be zero, if $x$ is non zero. Have I dont the contour integral right? Here there are no singularities so my $a$ can be anything, and my contour is the straight line parallel to the imaginary axis. Also, do you have a answer to my first query. Is there any theorem, which places some restriction on my $F(s)$ so that it has a inverse laplace transform. –  user23238 Apr 30 '13 at 0:19
You are not integrating correctly over the vertical line - you are treating it like the real line. The way to do the integral is to use Cauchy's integral theorem on a closed contour in the complex $s$ plane that includes that vertical line. Thus, $f(t) = 0 \; \forall t \gt 0$. –  Ron Gordon Apr 30 '13 at 0:25

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