Given a general random variable $X$ with density function $f(x)$ and characteristic function $\phi_X(u)$ we can go back and forth between the density and the characteristic by using the Fourier transform relations : \begin{equation} \phi_X(u)=\int_{\infty}^{\infty}e^{-iux}f(x)dx \end{equation}and \begin{equation} f(x)=\int_{-\infty}^{\infty}e^{-iux}\phi_X(u)du \end{equation}
However, if X is a non-negative random variable the Laplace is said to be more useful but I don't see exactly why.
First of all, the Laplace transform is defined in terms of a complex variable $s$, meaning that the characteristic function \begin{equation} \phi_X(s)=\int_{0}^{\infty}e^{-sx}f(x)dx \end{equation}
is a function of a complex variable $s$.
Moreover, the inversion Laplace transform is ugly: \begin{equation} f(x)=\frac{1}{2\pi i}\lim_{T\to \infty}\int_{\gamma-iT}^{\gamma+iT}e^{sx}\phi_X(s)ds \end{equation}
To me, using the Fourier versions \begin{equation} \phi_X(u)=\int_{0}^{\infty}e^{-iux}f(x)dx \end{equation}and \begin{equation} f(x)=\int_{0}^{\infty}e^{-iux}\phi_X(u)du \end{equation} is simpler (but is it correct?).
I know that $\phi_X(s)$ is the extended characteristic function, does this mean that we just set the real part of $s$ equal to zero in order to obtain the characteristic function? Could someone shade some light on this please.