Wiener Khinchin theorem : struggle in the derivation
Let me state at the beginning that the details that make this rigorous do not bring any extra understanding of the statistical behavior of signals, so the desire to use the Dirac delta distribution is completely understandable. If I could have figured out how to use it here rigorously, I would have.




I am going to repeatedly skip proof of the validity of "bringing expectation inside the integral". Note that computing an expectation is yet another integral (and over a finite measure space), so it really is just another change in the order of integration. To do so rigorously, appeal to Tonelli's Theorem and then Fubini's Theorem.




Let $x$ be a continuous-time wide-sense stationary stochastic process.
\begin{eqnarray}
r_{xx}(\tau) &amp;=&amp; \mathsf{E}\left[x(t)x(t+\tau)\right]\\
X_T(\omega) &amp;=&amp; \frac{1}{\sqrt{T}}\int_{-T/2}^{T/2}x(t)e^{-i\omega t}dt
\end{eqnarray}
We assume that the integral defining $X_T(\omega)$ converges for almost all realizations of the process $x$.
\begin{equation}
\begin{split}
\mathsf{E}\left[X_{T}(\omega)\overline{X_{T}(\omega)}\right] &amp;=~
\mathsf{E}\left[\frac{1}{\sqrt{T}}\int_{-T/2}^{T/2}x(t)e^{-i\omega t}dt\overline{\frac{1}{\sqrt{T}}\int_{-T/2}^{T/2}x(t')e^{-i\omega t'}dt'}\right]\\
&amp;=~ \mathsf{E}\left[\frac{1}{T}\int_{-T/2}^{T/2}\int_{-T/2}^{T/2}x(t)x(t')e^{-i\omega(t-t')}dtdt'\right]\\
&amp;=~ \frac{1}{T}\int_{-T/2}^{T/2}\int_{-T/2}^{T/2}\mathsf{E}\left[x(t)x(t')\right]e^{-i\omega(t-t')}dtdt'\\
&amp;=~ \frac{1}{T}\int_{-T/2}^{T/2}\int_{-T/2}^{T/2}r_{xx}(t-t')e^{-i\omega(t-t')}dtdt'\\
&amp;=~ \frac{1}{T}\int_{-T/2}^{T/2}\int_{-T/2}^{T/2}\color{red}{r_{xx}(t-t')e^{-i\omega(t-t')}}dtdt'
\end{split}
\end{equation}

Note the presence of $t-t'$ as the argument of factors in the integrand. Our final goal is to have the Fourier transform of $r_{xx}$ on the right-hand side. Perhaps we can convert this to an iterated integral in $t-t'$ and a different variable. Let
\begin{eqnarray}
p &amp;=&amp; t - t',\\
q &amp;=&amp; t + t'.
\end{eqnarray}
We want to change this to an iterated integral over $p$ and $q$. We are parametrizing the $(t,t')$ points with $p$ and $q$, and the Jacobian determinant of the map from $(p,q)$ to $(t,t')$ is $\frac{1}{2}$, so $dt~dt' = \frac{1}{2}dp~dq$.

The messy details that remain are describing the region of integration in terms of $p$ and $q$. The set over which we integrate is the diamond with corners at $(0,T)$, $(T,0)$, $(0,-T)$, and $(-T,0)$ in the $(p,q)$-plane. For $-T\leq p \leq 0$, we integrate $dq$ from $q = -T-p$ up to $q = T+p$. Hence the integral over $-T \leq p \leq 0$ is
\begin{equation}
\begin{split}
&amp;~\frac{1}{T}\int_{-T}^{0}\left[\int_{-T-p}^{T+p}dq\right]r_{xx}(p)e^{-i\omega p}\frac{1}{2}dp\\
~=&amp;~\frac{1}{T}\int_{-T}^{0}\left[T+p-(-T-p)\right]r_{xx}(p)e^{-i\omega p}\frac{1}{2}dp\\
~=&amp;~ \frac{1}{T}\int_{-T}^{0}\left(2T + 2p\right)r_{xx}(p)e^{-i\omega p}\frac{1}{2}dp\\
~=&amp;~ \int_{-T}^{0}\left(1 + \frac{p}{T}\right)r_{xx}(p)e^{-i\omega p}dp
\end{split}
\end{equation}

Now I add a restriction that a more seasoned professional might not need: I assume that
\begin{equation}
\int_{-\infty}^{\infty}\left|p~r_{xx}(p)\right|dp &lt; \infty.
\end{equation}
Then
\begin{equation}
\begin{split}
&amp;~\lim_{T\to\infty}\int_{-T}^{0}\left(1 + \frac{p}{T}\right)r_{xx}(p)e^{-i\omega p}dp\\
&amp;=~\lim_{T\to\infty}\int_{-T}^{0}r_{xx}(p)e^{-i\omega p}dp
+ \lim_{T\to\infty}\frac{1}{T}\underbrace{\color{red}{\int_{-T}^{0}p~r_{xx}(p)e^{-i\omega p}dp}}_{\color{red}{\textrm{bounded as $T\to\infty$}}}\\
&amp;=~ \int_{-\infty}^{0}r_{xx}(p)e^{-i\omega p}dp
\end{split}
\end{equation}

A similar calculation for $0\leq p \leq T$ (for which $-T+p\leq q\leq T-p$) yields
\begin{equation}
\int_{0}^{\infty}r_{xx}(p)e^{-i\omega p}dp.
\end{equation}
Adding these shows that
\begin{equation}
\lim_{T\to\infty}\mathsf{E}\left[\left|X_T(\omega)\right|^2\right]
~=~ \int_{-\infty}^{\infty}r_{xx}(p)e^{-i\omega p}dp.
\end{equation}