Extremes of integral of the autocorrelation theorem I am trying to apply the correlation theorem (here attached) to this definition of correlation of a finite continuous signal $g(t)$:
$\rho(\tau)=\frac{1}{T} \int_{t_{0}}^{ t_{0}+T}g(t)^{.}g(t+\tau)dt  \tag{1}$
where $t$ is the time, T is the time duration of the signal, $t_{0}$ is the starting time, $\tau$ is the time shift of the correlation.
As you can see, in the attached demonstration the correlation is defined from $[-\infty,+\infty]$. Obviously, the signals we work with are always finite. So I defined the interval as $[t_{0},t_{0}+T]$, as I also see in a book of Turbulence.
If I do the fast Fourier transform of both the member of the Eq (1) I obtain
$E(f)= \frac{1}{T} \int_{\color{red}{t_{0}}}^{\color{red}{t_{0}+T}}\int_{\color{red}{t_{0}}}^{ \color{red}{t_{0}+T}}g(t)^{.}g(t+\tau) e^{-i2\pi f \tau}d\tau dt \tag{2}$
where $E(f)$ is the Fourier Transform of $\rho(\tau)$. Usually the Fourier transforms, that I have seen in books, are defined in the interval $[-\infty,+\infty]$, but, as I said before, the signal we work with are defined in a time window such as $[t_{0},t_{0}+T]$, so I tried to apply them to finite signal
Introducing the variable
$y=t+\tau$
The Eq. (2) can be rewritten:
$E(f)= \frac{1}{T} \int_{\color{red}{t_{0}}}^{\color{red}{ t_{0}+T}}\int_{\color{red}{t_{0}}}^{ \color{red}{t_{0}+T}}g(t)^{.}g(y) e^{-i2\pi f (y-t)}dy dt \tag{3}$
Furthermore the Eq. (3) can be rewritten as:
$E(f)= \frac{1}{T} \int_{\color{red}{t_{0}}}^{\color{red}{ t_{0}+T}}g(t) e^{+i2\pi f t}dt\int_{\color{red}{t_{0}}}^{ \color{red}{t_{0}+T}}g(y) e^{-i2\pi f y}dy \tag{4}$
So E(f) can be rewritten as:
$E(f)= \frac{1}{T} G^{*}(f)^{.}G(f)\tag{5}$
I am not sure about the (red) extremes of the integral, I think they are wrong can you help me? Is it possible to apply the correlation and the Fourier transform to finite signals with interval of existence $[t_{0},t_{0}+T]$?
Maybe your signal is finite, but the function you are transforming is infinite. So you can do your demonstration using the interval $[0,T]$ or $[-\infty,+\infty]$, the result should be the same, because if your signal is finite the FT threats it as an infinite signal with period T.