TITLE: Is the following integral positive?
QUESTION [7 upvotes]: For each positive integer $n$, consider the following intgral:
$$\int_0^\infty\int_0^\infty\frac{(x^4-y^2)x^{2n+1}}{(x^4+y^2)[(1+x^2)^2+y^2]^{n+1}}dxdy.$$
I want to know if there is any easy way to see that it's positive. If there is no easy proof to show that it's positive, I am satisfied if I know that it is positive. I don't need to know the exact value of the integral. I try to write the integral as
$$\int_0^\infty\int_0^\infty\frac{x^{2n+1}}{[(1+x^2)^2+y^2]^{n+1}}dxdy-
\int_0^\infty\int_0^\infty\frac{2y^2x^{2n+1}}{(x^4+y^2)[(1+x^2)^2+y^2]^{n+1}}dxdy,$$
but it seems to me that it doesn't help much.

REPLY [3 votes]: Start with the substitution suggested by @alex.jordan: $u=\sqrt{x}, v=y$. Then think of $(u,v)$ as Cartesian coordinates with the integral over the first quadrant. Change to polar coordinates $u=r\cos\theta, v=r\sin\theta$ to obtain
$$ \hbox{your integral } = I_n = \frac12\int_0^\infty r^{n+1} dr \int_0^{\pi/2} \frac{\cos^n\theta(\cos^2\theta-\sin^2\theta)}{(r^2+2r\cos\theta+1)^{n+1}}d\theta.$$
Split 
$$J_n=\int_0^{\pi/2} \frac{\cos^n\theta(\cos^2\theta-\sin^2\theta)}{(r^2+2r\cos\theta+1)^{n+1}}d\theta$$
at $\theta=\pi/4$ to obtain integrands that are obviously positive and negative. In the subinterval $\theta\in[\pi/4,\pi/2]$ change the variable to $\theta^\prime=\pi/2-\theta$ - the domain for the second integral is now $\theta^\prime\in[0,\pi/4]$ - and then change the name of $\theta^\prime$ to $\theta$. Recombine the two integrals to obtain
$$ J_n=\int_0^{\pi/4} (h_n(r,\cos\theta)-h_n(r,\sin\theta))(\cos^2\theta-\sin^2\theta)d\theta$$
where 
$$h_n(r,z)=\frac{z^n}{(r^2+2rz+1)^{n+1}}.$$
For $n\geq 2$, this is an increasing function of $z$ on $[0,1]$ which is sufficient to show that the integrand in the second representation of $J_n$ is positive. I think this deals with everything but the case $n=1$.