Computing $\int_Q \frac{xy}{x^2+y^2}dxdy$ 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:
$$\int_S \frac{xy}{x^2+y^2}dxdy=\lim_{\epsilon \to 0}\int_{\epsilon}^1\int_0^{\frac{\pi}{2}}\rho^3\sin\theta\cos\theta d\rho d\theta=\frac{1}{8}$$
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?
 A: 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.$
A: 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. 
