Double integral. Iterated integration and polar coordinates

Question

everyone! I was proposed in class the following exercise. I hope you can help me.

a) Solve the following integral: $$\iint_D \frac{\lvert x-y \rvert} {(x^{2}+y^{2})^{\frac{3}{2}}}$$ $D= {[0,1] \times [0,1]} $

b) Study the iterated integrals of the double integral and its integrability: $$\iint_D \frac{x-y} {(x^{2}+y^{2})^{\frac{3}{2}}}$$ $D= {[0,1] \times [0,1]} $

My attempt:

a) At first, I tried to solve the integral directly, but, seeing that the calculus was getting harder, I considered changing to polar coordinates. Using the Fubini theorem for positive functions, one iterated integral resulted in:

$$\int_{0}^\frac{\pi}{4} \int_{0}^{\sec\varphi} \frac{\lvert \cos \varphi -\sin \varphi \rvert} {\rho} d\rho d\varphi + \int_ {\frac{\pi} {4} } ^\frac{\pi}{2} \int_{0}^{\csc\varphi} \frac{\lvert \cos \varphi -\sin \varphi \rvert} {\rho} d\rho d\varphi$$

When calculating both, I had to evaluate the neperian logarithm in zero, getting $\infty$. Can I conclude that the integral diverges?

b) As for the one without absolute value, I proceeded in the same way when solving the first iterated integral. In fact, I got the same result, as I had to evaluate the neperian logarithm in zero as well.

$$\int_{0}^\frac{\pi}{4} \int_{0}^{\sec\varphi} \frac{\cos \varphi -\sin \varphi} {\rho} d\rho d\varphi $$

Regarding the other iterated integral, I am not sure how to rewrite it. I know that rho goes from zero to square root of 2 and phi, from zero to arccos(1/rho), but I believe there is more to it than that.

Anyway, I think there is no need to calculate one of the iterated integrals since it is a continuous function on a compact set, hence it is integrable and the result of the iterated integrals is the same, right?

In any case, I think I might be missing something, so any help would be greatly appreciated. Thank you.

Answer 1

Part $a$ diverges for the reason you found: the integral blows up logarithmically as $\rho\rightarrow 0$. Since the integrand is strictly nonnegative, there's no need to worry about cancellations between positive and negative.

Part $b$ is more interesting. You can set it up as an iterated integral in cartesian or polar, integrating $x$ first, $y$ first, $\rho$ first, or $\varphi$ first. Try it and see what you get.

If you're having trouble with the cartesian integral, here's a handy integral table: $$ \int\frac{x-y}{(x^2+y^2)^{3/2}}dx = -\frac{x + y}{y \sqrt{x^2 + y^2}} $$ $$ \int \frac{\sqrt{1+y^2}-1-y}{y\sqrt{1+y^2}}dy = \ln\left(\frac{1+\sqrt{1+y^2}}{y+\sqrt{1+y^2}}\right) $$