I'm reading this article and having a question.
Consider a function $f$ and its Laplace transform
$\hspace{3.0cm} F(s) = \int_0^\infty f(t) e^{-st} dt$, with $\{s|\text{Re}(s) = 0\} \in \text{ROC}[F(s)]$
The inverse Laplace transform would be $f(t) = \lim_{\omega \to \infty} \frac{1}{2\pi i}\int_{\sigma - i \omega}^{\sigma + i \omega} F(s) e^{st} ds$.
Now, consider the case where $\text{Re}(s) = 0$, we can see that
$\hspace{3.0cm} f(t) = \lim_{\omega \to \infty} \frac{1}{2\pi i}\int_{- i \omega}^{i \omega} F(s) e^{st} ds = \frac{1}{2\pi}\int_{-\infty}^\infty F(i \omega) e^{i \omega t} d\omega$
is an inverse Fourier transform.
According to Fourier transform, we know that
$\hspace{3.0cm} F(i \omega) = \cal{F}\{f(t)\} = \int_{-\infty}^\infty f(t) e^{-i\omega t} dt$
But the Laplace transform of $f$, when $\text{Re}(s) = 0$, suggests that
$\hspace{3.0cm} F(i \omega) = \int_0^\infty f(t) e^{-i \omega t} dt$
It turns out that $\int_{-\infty}^\infty f(t) e^{-i \omega t} dt = \int_0^\infty f(t) e^{-i \omega t} dt$ ? This seems wrong to me ? Am I getting wrong somewhere ?