Let $x, y$ be i.i.d. standard normal random variables. Is the following expectation bounded?
$$\mathbb{E}\Big[ \frac{xy}{x^2 + (x+y)^2}\Big]$$
I used Wolfram alpha and simulation to compute the above expectation and in both cases, I got the value -0.2. I'm not sure if I can trust this result because ratio distributions are often heavy-tailed.
Another thing that I've observed is that although the ratio $x/y$ has Cauchy distribution whose expectation is undefined, Wolfram Alpha computes the expectation $\mathbb{E}[x/y]$ as zero.
Not just the expectation is bounded; there isn’t even a singularity in the function itself. Numerator and denominator are both homogeneous functions of degree $2$, so in polar coordinates the value depends only on the angular coordinate and not on the radial coordinate. With $x=r\cos\phi$ and $y=r\sin\phi$, we have
$$ \frac{xy}{x^2+(x+y)^2}=\frac{\cos\phi\sin\phi}{\cos^2\phi+(\cos\phi+\sin\phi)^2}=\frac{\sin2\phi}{3+\cos2\phi+2\sin2\phi}\;. $$
The denominator is strictly positive for all angles. In fact, we can integrate over $\phi$ to compute the expectation; the result is $-\frac15$, in agreement with what you found. This applies to any rotationally symmetric joint distribution of $x,y$, not just to the standard normal one, since the function is independent of the radial coordinate.