# Redundancy in the Laplace transform and Mellin's inverse formula

As I understand it, Mellin's inverse formula relates a sufficiently 'nice' function $f$ and its Laplace transform $F$ as follows: $$f(t)=\frac1{2\pi i}\lim_{T\to\infty}\int_{-T}^{T}e^{i\omega t}e^{\sigma_0 t}F(\sigma_0+i\omega)\ d\omega$$ for sufficiently large $\sigma_0$. In other words, the behavior of the function $f$ is entirely represented by $F$ along a single 'line' of its domain, $(\sigma_0-i\infty,\sigma_0+i\infty)$. This follows from viewing the Laplace transform here as a sort-of Fourier transform of $f$ modulated/damped/tamed by a decaying exponential $e^{-\sigma t}$ where $s=\sigma+i\omega$ since $e^{-st}=e^{-\sigma t}e^{-i\omega t}$.

Since Mellin's inversion transform suggests the Laplace transform encodes all the necessary behavior of $f$ along only a single line in its domain, is it not highly redundant? Each line $(\sigma_0-i\infty,\sigma_0+i\infty)$ for all sufficiently large $\sigma_0$ seems to encode more-or-less the same information. On the other hand, for example, the Fourier transform of $f$ seems significantly less redundant since the behavior of $f$ is encoded along only the unit circle.

Is there a comparative benefit to this 'redundancy'? Is it that it allows the same Laplace transform to be applicable to functions with significantly different asymptotic behavior all at once?

## 1 Answer

You're dealing with a holomorphic function. If you want to evaluate a path integral of a holomorphic function between points, then you can deform the contour between the two points in any way without changing its value, provided you don't cross any singularities.

Holomorphic functions themselves are redundant because a holomorphic function $F$ on a simply connected domain is fully determined by its values $\{ F(z_{n})\}_{n=1}^{\infty}$ on any infinite sequence $\{ z_{n}\}_{n=1}^{\infty}$ with a limit point in the domain.

What's going on is that the Fourier transform of an $L^{2}(\mathbb{R})$ function which vanishes for $t < 0$ is a holomorphic function on the lower half plane: $$F(s) = \int_{0}^{\infty}f(t)e^{-ist}dt,\;\; \Im s < 0.$$ So the inversion integral--which is the inverse Fourier transform--may be viewed as a complex contour integral over the boundary of the lower half plane, and the contour you use can be shifted into the lower half plane using complex function theory.

The Paley-Wiener theorem very precise the connection between $f \in L^{2}[0,\infty)$ and $$\mathscr{L}\{ f\} = \int_{0}^{\infty}f(t)e^{-st},\;\;\;\Re s > 0.$$ A Holomorphic function $F(s)$ is the Fourier transform of such a function $f$ iff (a) $F$ is square integrable on every vertical line $\Re s = \alpha > 0$, and (b) the $L^{2}$ norms are uniformly bounded for all $\alpha > 0$. This theorem can be used in a very precise way to deal with the Laplace inversion for Laplace transforms of functions $f$ of exponential order on $[0,\infty)$.

• Hmm... that seems to make sense. – oldrinb Jul 2 '15 at 17:25