Convergence and value of the integral $\iint_{\mathbb{R}^2} \frac{x^2}{(1+x^2)(x^2 + y^2)^{3/2}}\,dx\,dy$

Question

I am asked to determine if the double integral: $$I=\iint_{\mathbb{R}^2} \frac{x^2}{(1+x^2)(x^2 + y^2)^{3/2}}\,dx\,dy$$ converges or not.

My attempt: I let $$x = r\cos\phi.$$ $$y = r\sin\phi.$$ $$E_k : (\frac{1}{k} \leq r \leq k, 0 \leq \phi \leq 2\pi).$$

I notice that the integrand is positive which means the value of the integral is independent of the suite. I calculate the integral:
$$\lim_{k \to \infty}\iint_{E_k} \frac{r^2\cos^2\phi\cdot r}{(1+r^2\cos^2\phi)(r^3 )}\,dr\,d\phi $$

$$= \lim_{k \to \infty}\iint_{E_k} \frac{\cos^2\phi}{1+r^2\cos^2\phi}\,dr\,d\phi$$

However, I get stuck here and can't compute the integral. The answer is that the integral converges.

What can I change about my setup to make the integral easier to compute?

Is my setup flat out wrong?

Is there an easier way of determining if the integral converges?

Answer 1

To compute the integral $I$ on $(x,y)$ in $\mathbb R^2$, first note that, by symmetry of the integrand, $I$ is four times the integral on $x>0$, $y>0$, then define $(x,y)=(u,uv)$ and note that $dx\,dy=u\,du\,dv$ hence $$I=4\iint_{u>0,v>0}\frac{u^2}{(1+u^2)u^3(1+v^2)^{3/2}}\,u\,du\,dv=4KL$$ with $$K=\int_0^\infty\frac{du}{1+u^2}\qquad L=\int_0^\infty\frac{dv}{(1+v^2)^{3/2}}$$ One sees that $K$ and $L$ both converge hence, by Tonelli(-Fubini), $I$ converges as well, and it suffices to compute the values of $K$ and $L$ to deduce the value of $I$ by the formula above. Can you do that?

Answer 2

You have $$ \int_0^{2\pi} \left( \int_{1/k}^k \frac{\cos^2\varphi}{1+r^2\cos^2\varphi} \, dr \right) \, d\varphi. $$ The inner integral is $$ (\cos\varphi) \int_{1/k}^k \frac{(\cos\varphi)\, dr}{1 + ((\cos\varphi) r)^2} = (\cos\varphi) \int_{(\cos\varphi)/k}^{k\cos\varphi} \frac{du}{1+u^2} \to \frac \pi 2 \left|\cos\varphi\right| \text{ as } k\to\infty $$ where the absolute value in $\left|\cos\varphi\right|$ comes from thinking about the two cases $\cos\varphi>0$ and $\cos\varphi<0.$

Therefore \begin{align} & \lim_k \int_0^{2\pi} \left( \int_{(\cos\varphi)/k}^{k\cos\varphi} \cdots\cdots \,dr \right) \, d\varphi \\[10pt] = {} & \int_0^{2\pi} \left( \lim_k \int_{(\cos\varphi)/k}^{k\cos\varphi} \cdots\cdots \, dr \right) \,d\varphi \text{ by the monotone convergence theorem} \\[10pt] = {} & \int_0^{2\pi} \frac \pi 2 \left|\cos\varphi\right|\,d\varphi. \end{align}