# Inverse Laplace transform. Bromwitch integral

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 $$c+i\infty$$.

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jasonRF
Gold Member
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

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.