# Inverse Laplace Transform

1. Feb 27, 2007

### Swapnil

My book on signal processing says that:

$$f(t) = \frac{1}{2\pi j} \int_{c-j\infty}^{c+j\infty} F(s) e^{st} ds = \lim_{\Delta s \to 0} \sum_{n = -\infty}^{\infty} \Big[ \frac{F(n\Delta s)\Delta s}{2\pi j} \Big] e^{n\Delta s t}$$

I don't get this. How/Why can you write a integration over a complex variable as the above sum?

edit: I forgot a coefficient $$\frac{1}{2\pi j}$$ on the LHS. Sorry about that. Its fixed now.

Last edited: Feb 28, 2007
2. Feb 27, 2007

Isn't that definition of a Reimann integral?

3. Feb 28, 2007

### gammamcc

I understand how you could be confused! It would have to be explained what \Delta s is in the sum since the integral seems to have s lie on a vertical line c + jy in the complex plane (real y). So do we perhaps have \Delta s =
j\Delta y???

Supposing the above and not knowing anything else about F, using a Riemann sum, you should have limit as \Delta s -> 0 of sum over n of the following terms:
F(c+n\Delta s)\exp((c+n\Delta s)t)\Delta s
which is the limit as \Delta y-> 0 of the sum of terms

F(c+n j \Delta y)\exp((c+ n j \Delta y)t)j\Delta y

The 2\pi j in the denominator seems something like Cauchy's integral thm., but we need to know quite a bit more about F to get that. (Is F an entire fuction? Does it vanish very rapidly away from the y-axis in the complex plane?)

Generally speaking, the independence of c on the RHS makes the formula dubious.

Can you still get the money back for the book?

4. Mar 4, 2007

### Swapnil

Edited gammamcc's post to look nice...