How does $\iint_{\mathbf{ℝ^2}}\frac{x}{1+x^2+y^2}dxdy$ diverge? I have a question regarding a improper double integral which will diverge but I cannot seem to understand how to reach that conclusion. The integral is the following: $$\iint_{\mathbf{ℝ^2}}\frac{x}{1+x^2+y^2}dxdy$$ 
I can see that $f(x,y) \geq 0 \space \forall x\geq0$ and $f(x,y)\leq 0 \space \forall x\leq 0$ and therefore I split $\mathbf{ℝ}$ into to parts such that: $ℝ=\Omega_1\cap \Omega_2 = \{(x,y)\in ℝ^2 \mid x\geq0\} \cap \{(x,y)\inℝ^2 \mid x\leq0\}$ and now we integrate over both $\Omega_1$ and $\Omega_2$. However, when I attempt this using polar coordinates I find that both the integrals have the final form of $ \space"0 \cdot\infty "$. How can I conclude that the integral diverges from this? Or am I doing something completely wrong?
 A: Hint. Since the integrand is non negative in $\Omega_1$, we have that, for $R>0$,
$$\iint_{\Omega_1}\frac{x}{1+x^2+y^2}dxdy\geq \iint_{\Omega_1\cap B(0,R)}\frac{x}{1+x^2+y^2}dxdy=
\int_{-\pi/2}^{\pi/2}\int_{\rho=0}^R\frac{\rho^2\cos(\theta)}{1+\rho^2}d\theta d\rho\\=2\int_{\rho=0}^R\frac{\rho^2}{1+\rho^2} d\rho=2(R-\arctan(R)).$$
So, what may we conclude about the integral over $\Omega_1$?
A: Let's ask a related question: what's $\int_{-1}^1\frac{dx}{x}$? Naively, it's $[\ln x]^1_{-1}=0$. But you're adding a $+\infty$ and a $-\infty$, and we say the integral diverges. In the case at hand, the same thing happens. The $x\ge 0$ region can be written as $|\theta|\le \pi/2$, if you'll permit me to use the integration range $[-\pi,\,\pi]$ for the full integration range. But $\int_{-\pi/2}^{\pi/2}\cos\theta d\theta\ne 0$.
A: Note that
$$\iint_{\mathbf{\Omega_1}}\frac{x}{1+x^2+y^2}dxdy\ge \int_{0}^{\pi/4}d\theta \int_0^{\infty}\frac{r\cos\theta}{1+r^2}dr\ge\frac{\pi}4\frac{\sqrt 2}2\int_0^{\infty}\frac{r}{1+r^2}dr>\infty$$
indeed since $\frac{r}{1+r^2}\sim \frac1r$
$$\int_0^{\infty}\frac{r}{1+r^2}dr>\infty$$
