I'm trying to prove the cross correlation theorem for the single-sided Laplace transform. Define the cross correlation for two functions $f, g: [0, \infty) \to \mathbb{C}^n$ to be
$$
(f\star g)(t) = \int_0^\infty f(\tau)^* g(t + \tau)d\tau.
$$
The cross correlation theorem (from Wikipedia) then states that the single-sided Laplace transform of $f\star g$ is given by
$$
F^*(-s^*)\cdot G(s).
$$
Proof attempt
$$
\mathcal{L}(f\star g)(s) = \int_0^\infty e^{-st} \int_0^\infty f(\tau)^*g(t + \tau)d\tau dt\\
= \int_0^\infty f(\tau)^* \int_0^\infty e^{-st} g(t + \tau) dt d\tau.
$$
Set $u = t + \tau$ and evaluate the inner integral:
$$
\int_0^\infty e^{-st} g(t + \tau) dt = \int_0^\infty e^{-s(u – \tau)} g(u)du\\
= e^{s\tau} \int_0^\infty e^{-su} g(u) du\\
= e^{s\tau} G(s).
$$
We are then left with
$$
\int_0^\infty f(\tau)^* e^{s\tau} d\tau G(s).
$$
If we were using the two-sided Laplace transform, the result would then follow by substituting $\omega = – \tau$. In the single-sided setting, however, this doesn't work, as $f$ only takes positive arguments.
I suppose one way around this issue is to extend $f$ and $g$ to $(-\infty, \infty)$, setting them to $0$ on $(-\infty, 0]$. The result can then be derived using the two-sided Laplace transform, which is equal to the single-sided Laplace transform in this case. It is a bit unsatisfying though, a direct proof would be nice.
Thanks in advance for pointing out where I'm being stupid!
Best Answer
You can't make $\tau$ negative but what if we play around with $s$? $$\begin{align}\int_{0}^{\infty}f^*(\tau)e^{s\tau}d\tau&=\int_{0}^{\infty}f^*(\tau)e^{-(-s)\tau}d\tau\\\\&=\mathcal{ULT}\{f^*(t)\}_{s=-s}\\\\&=\mathcal{F}^*(s^*)_{s=-s}\\\\&=\mathcal{F^*}(-s^*)\end{align}$$
(As, $(-s)^*=-s^*$)