How does $\iint_{\mathbf{ℝ^2}}\frac{x}{1+x^2+y^2}dxdy$ diverge?

Question

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?

Answer 1

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 $\Omega_1$?

Answer 2

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 $\frac{r}{1+r^2}\sim \frac1r$

$$\int_0^{\infty}\frac{r}{1+r^2}dr=\infty$$