For Witsenhausens counterexample, I want to compute the first term of:
$$ J = k^2E[u_1(x_0)^2] + \text{something positive} \tag{1} $$
where $x_0 \sim N(0,\sigma^2)$ is a random variable drawn from a normal distribution with zero mean and variance $\sigma^2$. Take the standard values $k=0.2$ and $\sigma = 5$.
It is known (due to Witsenhausen), that choosing $u_1(x_0) = \sigma\, \text{sgn}(x_0)$, together with $u_2(x) = \sigma \tanh(\sigma x)$ gives a value of $J \approx 0.4$ (since $u_2$ enters only into the second term of $(1)$, it is not of interest here).
However, now I want to compute the first term of $(1)$, which should be given by:
$$ k^2E[u_1(x_0)^2] = k^2\int_{-\infty}^{\infty} \big(\sigma\, \text{sgn}(x_0) \big)^2\frac{1}{\sqrt{2 \pi\sigma^2}}\exp \Big(-\frac{x_0^2}{2\sigma^2}\Big) d x_0 $$
However, for the given values of $k$ and $\sigma$, this integral evaluates to $1$.
Therefore, something has to be wrong here, since $J$ can never reach a value of around $0.4$ as the first term is already too large and the second term is also positive, thus would make $J$ even larger.
Question: Where is the mistake here?
After checking again, I found the mistake: $u_1$ is not the "final" function which is instead given by $u_1 = \gamma_1(x_0) = f(x_0) - x_0 = \sigma \, \text{sgn}(x_0) - x_0$. So for convinience, only $f$ is ususally refered to, while the actual function $\gamma_1$ requires an extra $-x_0$. With this addition, the integral yields the correct value.