Possible typo in Jaynes probability Theory 7.6 
Probability of errors in orthogonal directions should be independent. 
  $$p(x,y)\,dx\,dy = f(x)\,dx\,\times f(y)\,dy \tag{7.3}$$ We can write the distribution equally well in polar coordinates $r, \theta$ defined by $x = r\,\cos\theta,\ \ y = r\,\sin\theta$: $p(x,y)\,dx\,dy = g(r,\theta)\,r\,dr\,d\theta \tag{7.4}$
$(P2)$ This probability should be independent of the angle: $g(r,θ)= g(r)$
Then (7.3) and (7.4) yield the functional eqaution
  $f(x)f(y) = g\left(\sqrt{x^2+y^2}\,\right) \tag{7.5}$
  and, setting $y = 0$, this reduces to $g(x) = f(x)\,f(0)$, so (7.5) becomes the functional equation
  $\log\left[\frac{f(x)}{f(0)}\right] + \log\left[\frac{f(y)}{f(0)}\right] = \log\left[\frac{f(\sqrt{x^2+y^2})}{f(0)})\right] \tag{7.6}$

By looking at (7.6), when we set $x = 0$, it becomes $\log\left[\frac{f(x)}{f(0)}\right] + \log\left[\frac{f(0)}{f(0)}\right] = \log\left[\frac{f(\sqrt{x^2+0^2})}{f(0)}\right]$.The same for setting $y = 0$.
But if neither $x$ nor $y$ is $0$, whether (7.6) is still right? 
or (7.6) should be $\log\left[\frac{f(x)}{f(0)}\right] + \log\left[\frac{f(y)}{f(0)}\right] = \log\left[\frac{g(\sqrt{x^2+y^2})}{f^2(0)})\right]$
Any help will be appreciated. Thanks.
 A: Let variables $x,y$ take values in $\mathbb{R}$. If we have, for all $x,y$,
$$f(x)\,f(y)=g\left(\sqrt{x^2+y^2}\,\right)\tag{1}$$
then we have, for all $x$,
$$f(x)\,f(0)=g\left(|x|\right),\tag{2}$$
assuming the principal square root function.
From $(2)$, therefore, for all $x,y$,
$$f\left(\sqrt{x^2+y^2}\,\right)\,f(0)=g\left(\sqrt{x^2+y^2}\,\right).\tag{3}$$
Substituting $(3)$ into $(1)$ gives, for all $x,y$,
$$f(x)\,f(y)=f\left(\sqrt{x^2+y^2}\,\right)\,f(0)\tag{4}. $$
Assuming $f(0)\neq 0$, dividing both sides of $(4)$ by $f(0)^2$ gives, for all $x,y$,
$$\frac{f(x)}{f(0)} \, \frac{f(y)}{f(0)} =\frac{f\left(\sqrt{x^2+y^2}\,\right)}{f(0)}\tag{5},$$
from which the desired result (Jaynes' equation 7.6) follows.

From this result, we conclude that for some constant $a$,
$$f(x) = f(0) e^{a\,x^2}.$$
Now in order for $f$ to be a probability density on $\mathbb{R}$, it must satisfy
$$ \int_\mathbb{R}f(x)\,dx = 1,$$
which requires that $a<0$, say $a=-\alpha$ for some positive $\alpha$ (otherwise the integral diverges). Thus we must have
$$\int_\mathbb{R}f(0)\,e^{-\alpha\,x^2}\,dx = 1,$$
hence
$$f(0)= \frac{1}{\int_\mathbb{R}\,e^{-\alpha\,x^2}\,dx}=\sqrt{\frac{\alpha}{\pi}}.$$
