Why inverse laplace is line integral?

1. May 1, 2014

Jhenrique

Watching this video http://youtu.be/1JnayXHhjlg?t=5m30s, I understood the ideia the fourier transform, that is a continuous summation of sinusoids. But now If I have amplitude and phase as function of σ and ω, the summation wouldn't be $\sum_\sigma \sum_\omega A_{\sigma \omega} \exp(i \varphi_{\sigma \omega}) \exp((\sigma + i \omega)t)$? And in its continous form why the inverse laplace transform isn't a double integral wrt sigma and omega? $$\int_{-\infty }^{+\infty} \int_{-\infty }^{+\infty} F(\sigma, \omega) \exp((\sigma + i \omega)t)d\sigma d\omega$$

2. May 1, 2014

lurflurf

I take it you are talking about the Bromwich integral? It is just one way of calculating the inverse laplace transform. It is a line integral and an area integral, they are the same thing because of Stokes theorem.
$$\int \! \! \int \, \nabla\times\mathbf{F}\cdot\mathrm{d}A=\oint \mathrm{F}\cdot\mathrm{d}l$$

3. May 2, 2014

Jhenrique

Yeah! But the Bromwich integral isn't a closed line in the complex plane, therefore, this formula haven't connection with a double integral over a bidimensional region (like is stated in the green theorem).

4. May 2, 2014

lurflurf

Yes that is true the closed path integral has two parts, the Bromwich part and another part. Two common cases are when the other part vanishes and the branch point case where we can calculate the other part.

5. May 4, 2014

Jhenrique

I did not understand.

6. May 4, 2014

lurflurf

$$\int \! \! \int \, \nabla\times\mathbf{F}\cdot\mathrm{d}A=\oint \mathrm{F}\cdot\mathrm{d}l=\int_\mathrm{Bromwich} \mathrm{F}\cdot\mathrm{d}l+\int_{other} \mathrm{F}\cdot\mathrm{d}l$$

So there is always an area interpretation if you want it. Just close the contour like when you use the residue theorem. Often the other part is zero or empty, sometimes it is not due to a branch cut or something.

By empty I mean we can often ignore the other contour by giving arguments like the Bromwhich contour is closed because the ends meet at infinity or something. All that stuff is not as important as knowing if the term is zero or not.

7. May 8, 2014

Jhenrique

Btw, is because this formula that the Bromwich integral is scaled by $\frac{1}{2 \pi i}$ ?

8. May 10, 2014

^yes