Computing $\int_Q \frac{xy}{x^2+y^2}dxdy$

Question

I'm asked to compute the following integral: $$\int_Q \frac{xy}{x^2+y^2}dxdy \qquad Q=[0,1]^2$$

Solution: First I'm going to study if the integral is convergent. To do this, we notice that $$\int_Q \frac{xy}{x^2+y^2}dxdy < \infty \iff \int_S \frac{xy}{x^2+y^2}dxdy<\infty$$ where $S=\{(x,y) \in \mathbb{R}^2 | x\geq0, y\geq 0, x^2+y^2\leq1\}$ and this is clear because the difference between $\int_Q f(x,y)dxdy$ and $\int_S f(x,y)dxdy$ is a proper integral.

Computing we have: $$\int_S \frac{xy}{x^2+y^2}dxdy=\lim_{\epsilon \to 0}\int_{\epsilon}^1\int_0^{\frac{\pi}{2}}\rho^3\sin\theta\cos\theta d\theta d\rho=\frac{1}{8}$$ and so I'm granted the convergence.

Now I have to compute the real integral, as we have assured that it's convergent. $$\int_Q \frac{xy}{x^2+y^2}dxdy=\int_0^1 \int_0^1 \frac{xy}{x^2+y^2}dxdy= \int_0^1 \frac{y}{2} \int_0^1 \frac{2x}{x^2+y^2}dxdy = \frac{1}{2} \int_0^1 y(log(1+y^2)-log(y^2))dy =$$ $$ = \frac{\log2}{2}-\frac{1}{2}\lim_{\epsilon \to 0}\int_{\epsilon}^1y\log(y^2)dy= \frac{\log2}{2}$$

I checked the result and it's correct, but I'm asking for a review of the process: did I do anything wrong?

Answer 1

As you wrote it, it looks like you claim $$ I = \frac{1}{2} \int_0^1 y \ln (y^2) dy = 0.$$

But this is not the case: \begin{align*} \frac{1}{2} \int_0^1 y \ln(y^2) dy &= \frac 1 4 \int_{0}^1 \ln(y^2)2y dy \\ &= \frac 1 4 \int_0^1 \ln(u) du. \end{align*} Using the fact the $u \ln u - u $ is a primitive of $\ln u$ you get

$$ I = \frac 1 4 \big(- 1 - (0 - 0)\big) = \frac {-1}{4}.$$

Same goes for

$$J = \frac 1 2 \int_0^1 y \ln (1 + y^2) dy \neq \frac {\ln(2)} {2}.$$

but rather

\begin{align*} \frac{1}{2} \int_0^1 y \ln (1 + y^2) &= \frac{1}{4} \int_0^1 \ln(1+y^2) 2ydy \\ &= \frac{1}{4} \int_1^2 \ln(u) du \\ &= \frac{1}{4} (2 \ln 2 - 2 - (1\cdot\ln 1 - 1)) \\ &= \frac 1 4 (2 \ln 2 -1) \\ &= \frac{\ln(2)}{2} - \frac{1}{4}. \end{align*}

So the integral is $$J - I = \left(\frac{\ln(2)}{2} -\frac{1}{4}\right) - \frac{-1}{4} =\frac{\ln(2)}{2}.$$

Concerning the first part it seems easier to me to do the following

$$ \left\vert \int_Q f(x,y)dxdy \right\vert \leq \int_Q \vert f(x,y)\vert dxdy \leq \int_{D_R^+} \vert f(x,y)\vert dxdy$$

where $D_R^+$ is the positive par of a disc of radius $R$ with $R$ such that $Q \subset D_R^+.$

Using polar coordinates you will have the following inside the integral

$$ \left \vert \frac{r^2 \cos \sin}{R^2} r \right \vert\leq r \leq R$$ and $ \int_Q f(x,y) dxdy < \infty.$

Answer 2

I'm not too impressed by the way you proved convergence. The place where the denominator of the integrand becomes $0$ is $x=0, y=0$ and $(0,0)$ is in both $Q$ and $S.$ A better way is let $$Q'=\{(x,y)|x^2+y^2<b^2\}$$ where $b<1,$ integrate over $Q\backslash Q'$ and take the limit as $b \to 0.$ Split the region $Q\backslash Q'$into two parts depending on whether $y \le x$ or $y \ge x.$ Here's how to do the integral over the 'lower' part where $y \le x$. Note that the boundary $x=1$ is $r=\sec\theta$ in polar coordinates. Then the integral over the lower part is, in polar coordinates, $$\int_0^{\pi/4}\int_b^{\sec \theta}\frac{(r\cos\theta)(r\sin \theta)}{r^2}rdrd\theta$$ which is easily integrated. The top half, where the boundary $y=1$ is $r=\csc \theta$ is equally easy to integrate. Add the top and bottom parts, take the limit as $b \to 0$ and you're done.