Estimating the PSD (Periodogram): How do you find the Expected Value
TLDR;

The mistake is here:


  If I calculate the expected value, which is according to
  definition an average, this way 
  
  $E\left\{\widehat{P_x}(e^{jω})\right\} = \frac{1}{2π}\int_{&lt;2π&gt;}
&gt; \widehat{P_x}(e^{jω})dω$


The mistake is in your interpretation of expectation. It is not an average over all frequencies.

For each fixed $\omega$, the expectation is taken over many realizations of the random signal $x$.

Gory details

Suppose you are given a complex valued random signal $x(n)$ for $0\leq n\leq N-1$. I would emphasize that this is a random signal. If you repeat your experiment you'll obtain a different set of $N$ values each time (called "realizations" of $x$).

Recall how the periodogram $\hat P_x(e^{j\omega})$ is computed. You first compute the sample autocorrelation at each lag $m$:
$$
\hat r_x(m) = \frac{1}{N-m}\sum_{n=0}^{N-m-1}x(n)x^*(n+m) \;\;\;\;\;\;\;\;\;\;\;\;(1)
$$
and then get the periodogram estimate by taking the DTFT:
$$
\hat P_x(e^{j\omega}) = \sum_{m=-(N-1)}^{N-1} \hat r_x(m) e^{j \omega m}.
$$
Since we started with a random $x$ we would like to know how this estimate performs on average i.e. average over the randomness in $x$:. So we take the expectation:
\begin{eqnarray}
\hat P_x(e^{j\omega}) &amp;=&amp; \mathbf E\left[\sum_{m=-(N-1)}^{N-1} \hat r_x(m) e^{j \omega m}\right]\\
&amp;=&amp; \sum_{m=-(N-1)}^{N-1} \mathbf E[ \hat r_x(m)] e^{j \omega m}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;(2)\\
&amp;=&amp; \sum_{m=-(N-1)}^{N-1} \mathbf E\left[ \frac{1}{N-m}\sum_{n=0}^{N-m-1}x(n)x^*(n+m) \right ] e^{j \omega m} \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;(3)\\
&amp;=&amp; \sum_{m=-(N-1)}^{N-1} \frac{1}{N-m}\sum_{n=0}^{N-m-1}\underbrace{\mathbf E[x(n)x^*(n+m)]}_{\mbox{autocorrelation function of }x} \; e^{j \omega m}\\
&amp;=&amp; \sum_{m=-(N-1)}^{N-1} r_x(m) e^{j\omega m}  \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;(4) \\
&amp;=&amp; P_x(e^{j\omega})    \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;(5)
\end{eqnarray}

where (3) follows by substituting (1) in (2), (4) is by definition of autocorrelation $r_x$ of the random sequence $x$ and (5) is by definition of $P_x$.

Once again, note that the expectation was taken to "average out" the randomness introduced by $x$. It is not an average over frequencies because the frequencies are not random here.