Does this double integral exists?

Question

I've calculated $\iint_D\frac{y}{x^2+y^2}\,dA$ where D is bounded by $y=x$, $y=2x$, $x=2$, in this way: $$\iint_D\frac{y}{x^2+y^2}\,dA=\int_0^2\int_x^{2x}\frac{y}{x^2+y^2}\,dy\,dx=\cdots = \ln \frac{5}{2}.$$ However, I wonder if the fact that the $\frac{y}{x^2+y^2}$ is not bounded on D invalidates all the calculations and in fact the double integral does not exist. All the theorems that I consult have as a hypothesis that the integrand is bounded on the region of integration and hence my doubt regarding this double integral. Can anyone help me? Does this integral exist?

Answer 1

Changing to polar coordinates reveals why the integral is finite even though the integrand is $\mathcal{O}(r^{-1})$ as $r \to 0$.

For a quick check, note that the integrand is nonnegative and $D$ is a subset of the sector

$$S=\{(r,\theta): 0 \leqslant r \leqslant 2\sqrt{5},\, \frac{\pi}{4} \leqslant \theta \leqslant \arctan (2) \}$$

Thus,

$$\int\int_D \frac{y}{x^2+ y^2} dA \leqslant \int_{\frac{\pi}{4}}^{\arctan(2)}\int_0^{2\sqrt{5}} \frac{r \sin \theta}{r^2} r \, dr \, d\theta = 2\sqrt{5}\int_{\frac{\pi}{4}}^{\arctan(2)}\sin \theta \, d\theta,$$

where the integral on the RHS is finite since the sine function is bounded.

Answer 2

$$\int \frac{ydy}{x^2+y^2} = \frac{1}{2} \int \frac{d(y^2+x^2)}{x^2+y^2} =\frac{1}{2} \ln(x^2+y^2) + C$$ So $$\int_0^2\int_x^{2x}\frac{y}{x^2+y^2}\,dy\,dx= \frac{1}{2} \int_{0}^{2}\ln(x^2+y^2)|_{x}^{2x} dx$$