Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question

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.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.