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.

  • $\begingroup$ For checking convergence, the integral obviously converges on the region between $[0,1]\times[0,1]$ and $\rho \le 1$. So for convenience, you only need to check convergence on the disk $\rho\le 1$, which has nicer limits of integration. $\endgroup$ Commented May 9, 2022 at 17:18

1 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: $$ \int\frac{x-y}{(x^2+y^2)^{3/2}}dx = -\frac{x + y}{y \sqrt{x^2 + y^2}} $$ $$ \int \frac{\sqrt{1+y^2}-1-y}{y\sqrt{1+y^2}}dy = \ln\left(\frac{1+\sqrt{1+y^2}}{y+\sqrt{1+y^2}}\right) $$

  • $\begingroup$ Got it! As you said, it is important to remark that cancellations might occur, but not in this case. As for part b, I had already tried to solve the iterated integral in cartesian coordinates, but my mistake was trying to work on the hyperbolic expression of it. Now that you give me this another expression, it is clear to me that the iterated integral converges. Thank you for your help! $\endgroup$
    – Mths
    Commented May 9, 2022 at 21:02

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