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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?

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  • $\begingroup$ You didn't bother to show the computation, but probably it's valid. The function you integrate is positive and for such you can carelessly change the order of integration due to Tonelli's theorem. Also, a function doesn't have to be bounded for an integral to exist, think for simplicity about one-dimensional integral $\int_0^1 x^{-1/2}\,dx$. $\endgroup$
    – Tony419
    Jun 10 '20 at 21:29
  • $\begingroup$ Why is the fraction not bounded on the region $D$? $\endgroup$
    – Allawonder
    Jun 10 '20 at 23:27
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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.

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

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