Inverse Laplace transform. Bromwitch integral

[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.

 

[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

 

[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.