First, note that $\int^b_a f(x)dx = \int_{[a,b]} f(x)\ md(x)$ where $m$ is one-dimensional Lebesgue measure.
This is an exercise from Richard Bass' $\textit{Real Analysis for Graduate Students 3.1}$. It uses Fubini and Lebesgue integration:
Prove that $\int^1_0 \int^1_0 \frac{x^2 - y^2}{(x^2 + y^2)^{3/4}} \log(4 + \sin x)) \, dy\,dx = \int^1_0 \int^1_0 \frac{x^2 - y^2}{(x^2 + y^2)^{3/4}} \log(4 + \sin x) \, dx\,dy$.
My ideas:
Here, we are trying to show that Fubini-Tonelli's Theorem can be applied to this function. Once we can show this, we may conclude that equality holds. To do this, we can show, for example, that $\int^1_0 \int^1_0 \bigg\lvert \frac{x^2 - y^2}{(x^2 + y^2)^{3/4}} \log(4 + \sin x) \bigg\rvert\, d\,dx < \infty$ (we can also show the other is finite, doesn't matter). I thought that we should bound the function in order to do this, and then solve an easier integral.
We know that $\log(4 + \sin x) \leq \log 5$ since $-1 \leq \sin x \leq 1$ for all $x$. But I'm not sure what we can bound the rest of the integrand by? And I do not see an easy way of integrating this function directly.
If you're able to provide a proper solution, thank you!
So $\left|\dfrac{x^{2}-y^{2}}{(x^{2}+y^{2})^{3/4}}\right|\cdot|\log(4+\sin x)|\leq(\log 5)\cdot\dfrac{x^{2}+y^{2}}{(x^{2}+y^{2})^{3/4}}=(\log 5)(x^{2}+y^{2})^{1/4}$ and $(x,y)\rightarrow(x^{2}+y^{2})^{1/4}$ is continuous on $[0,1]\times[0,1]$ and hence the integral is bounded.