Prove $f(x,y)=\frac{x-y}{x^2+y^2}$ is integrable on $(0,1)^2$

Question

So It is continuous on the given square and measurable so I just need to show that the integral of positive - integral of negative part is finiate. SO $\int_0^1\int_0^x|\frac{x-y}{x^2+y^2}|+\int_0^1\int_x^1|\frac{x-y}{x^2+y^2}|<\infty$ I can evaluate this but it gets ugly and I am pretty sure there is some better approach and I have a feeling that $\int_{(0,1)^2}f(x,y)=0$, because the second part of the question asks to calculate this integral

Answer 1

In polar coordinates, the integrand is $\frac{\cos(\theta)-\sin(\theta)}{r}$. When integrating with polar coordinates, there is a factor of $r$ in the $dA$. So in polar coordinates, this is $$\iint_{\text{unit square}}\left(\cos(\theta)-\sin(\theta)\right)\,dr\,d\theta.$$ This line at $x=1$ is $r=\sec(\theta)$ in polar coordinates, and the line $y=1$ is $r=\csc(\theta)$. So this can be calculated as

\begin{align*}&\int_{\theta=0}^{\pi/4}\int_{r=0}^{\sec(\theta)}\left(\cos(\theta)-\sin(\theta)\right)\,dr\,d\theta + \int_{\theta=\pi/4}^{\pi/2}\int_{r=0}^{\csc(\theta)}\left(\cos(\theta)-\sin(\theta)\right)\,dr\,d\theta \\ =& \int_{\theta=0}^{\pi/4}\left(1-\tan(\theta)\right)\,d\theta + \int_{\theta=\pi/4}^{\pi/2}\left(\cot(\theta)-1\right)\,d\theta\end{align*}

These are both finite integrals but I'll leave it to you to conclude with their explicit values, and see that they cancel out.