# Inverse Laplace transform. Bromwitch integral

1. Nov 19, 2014

### LagrangeEuler

Inverse Laplace transform
$$\mathcal{L}^{-1}[F(p)]=\frac{1}{2\pi i}\int^{c+i\infty}_{c-i\infty}F(s)e^{st}dp=f(t)$$

Question if we integrate along a straight line in complex plane where axis are $$Re(p)$$, $$Im(p)$$, why we integrate from $$c-i \ínfty$$ to $$c+\infty$$? So my question is, because $$Im(p)$$ are also real numbers why we integrate from $$c-i\infty$$ to $$c+i\infty$$.

Last edited: Nov 19, 2014
2. Nov 24, 2014

### Greg Bernhardt

Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?

3. Nov 24, 2014

### jasonRF

I'm guessing that you meant that the integral is with respect to $s$, not $p$. Anyway, for the traditional Laplace transform,
F(s) = \int^{\infty}_{0} dt e^{-s t} f(t)

$F(s)$ is only analytic for some right half plane (that is $\Re(s)>s_0$). So for the inverse transform you pick a $c>s_0$ for your path.

Does that help?

jason

4. Nov 25, 2014

### LagrangeEuler

Yes. Mistake. I understand that. I am not very good in complex analysis. My question is if [tex[Im(s)[/tex] are real numbers why I integrate from $$c-i\infty$$ to $$c+i\infty$$? Why I have this $$i$$? Thanks for the answer.