Study the convergence of the following integral $$\iint_A\frac{dxdy}{x+y}$$
Where $A=[0,1]^2$. If we consider $C=\{(x,y)\in \mathbb R : x^2+y^2\le1,\ x>0, y>0 \}$, we can say that $$\iint_A\frac{dxdy}{x+y}=\iint_{A-C}\frac{dxdy}{x+y}+\iint_C\frac{dxdy}{x+y}$$
Now the integral on $A-C$ is a proper integral, while the integral on $C$ is improper in $(x,y)=(0,0)$. Applying polar coordinates: $$\iint_C \frac{dxdy}{x+y}=\lim_{\varepsilon \to 0^+}\int_\varepsilon^1d\rho\int_0^{\frac \pi2}\frac{d\theta}{\cos \theta + \sin \theta}=\lim_{\varepsilon\to0^+}(1-\varepsilon)\int_0^{\frac \pi2}\frac{d\theta}{\cos \theta + \sin \theta}$$
Now we can say that $\cos \theta + \sin \theta$ is a continuous function and $C$ is compact, so for the Weierstrass theorem $\exists m>0 : \cos \theta + \sin \theta \ge m, \ \forall \theta \in[0,\frac \pi2]$, and he concludes that this last integral is convergent. I've doubts about:
1) Why do we need to say all the stuff about $m$ in the end? It's because we're doing a limit in two dimensions so we have to prove that it is uniform in $\theta$ so it exists? Or is for another reason?
2) After this affirmation about $m$ he just applies squeeze theorem like this $$\lim_{\varepsilon\to0^+}(1-\varepsilon)\int_0^{\frac \pi2}\frac{d\theta}{\cos \theta + \sin \theta}\le\lim_{\varepsilon\to0^+}(1-\varepsilon)\int_0^{\frac \pi2}\frac{d\theta}{m}$$
And since the last integral is convergent he concludes that the other one is convergent too, or is he using something else?
Thanks for your time.
(2) is the reason for (1).
We need to affirm that (1) holds so that we can say (2) does.
$$\exists m{>}0\ ~\forall \theta\in[0;\pi/2] : (m\leq \cos \theta+\sin\theta)\tag 1$$
Which is true since the trig. sum is strictly positive over that interval.
$$0~\leq ~\lim_{\epsilon\to 0^+}(1-\epsilon)\int_0^{\pi/2}\dfrac{\mathrm d\theta }{\cos\theta+\sin\theta}~\leq~ \lim_{\epsilon\to 0^+}(1-\epsilon)\int_0^{\pi/2}\dfrac{\mathrm d\theta }{m}\tag 2$$
Which means the integral we seek must be less than a finite value, whatever it is.