I have a doubt about the equivalence between Fourier Transform and Laplace Transform.
It was told me that if I have a function such that:
- $f(t)=0$ if t<0
- $f\in L^1(R) \bigcap L^2(R)$
I can define
$F[f(t)]=\int_{0}^{\infty}f(t) e^{-j\omega t}dt$
$L[f(t)]=\int_{0}^{\infty}f(t) e^{-s t}dt$
I can look at the Fourier transform as the Laplace Transform evaluated in $s=j\omega$ IF AND ONLY if the abscisse of convergence is strictly less than zero. (I.e. if the region of convergence includes the imaginary axis.
If the abscisse of congence is $\gamma=0$, then (it was told me), I can have poles on the real axes, and I have to define the Fourier transform with indentations in a proper manner.
In the Papoulis's book there is written "if $\gamma=0, $ the Laplace transform has at least one of the singular points on the imaginary axes.
So, I think that the situation should be like this:
Then, if I extend the frequency in the complex plane, I can consider that -regarding the Fourier transform- the axes are rotated with respect to the axis of the s-plane:
So I should have:
Finally,
I think that these two last steps could explain the word "real" related to the poles at the beginning of the question…
Please, tell me If the reasoning is wrong and where.. many thanks
Best Answer
If $f \ge 0$ then $$F(s) = \int_0^\infty f(t)e^{-st}dt$$ converges for $\Re(s) > \sigma$ iff $F(s)$ is bounded for $s > \sigma$. Whence if $\sigma_c$ is the abscissa of convergence then $$\lim_{ s \to \sigma_c} |F(s)| = \infty$$ And $F(s)$ is analytic for $\Re(s) > \sigma_c$ and has a singularity at $s= \sigma_c$.
Proof : $$|F(\sigma+i\omega)| \le \int_0^\infty |f(t)e^{-(\sigma+i\omega)t}|dt=\int_0^\infty f(t)e^{-\sigma t}dt = F(\sigma)$$
If $f$ is not non-negative then this is not true anymore.
Let $$f(t) = \sum_{n=1}^\infty (-1)^{n+1} e^n 1_{t \in [\ln n,e^{-n}+\ln n]}$$
Then $$F(s) = \int_0^\infty f(t) e^{-st}dt$$ converges iff $\Re(s) > 0$ but its analytic continuation is entire.
Proof, as $s \to 0^+$ : $$e^n\int_{\ln n}^{e^{-n}+\ln n} e^{-st}dt-e^{n+1}\int_{\ln (n+1)}^{e^{-(n+1)}+\ln (n+1)} e^{-st}dt\sim n^{-s}-(n+1)^{-s}$$
And $\sum_{n=1}^\infty (-1)^{n+1}n^{-s}$ diverges for $\Re(s) = 0$.