Question about improper integral, Munkres Problem 15.5 I have a question about improper integral from Munkres' Analysis on Manifold chapter 15. This is problem 5:

What I have done was set up the double iterated integral over A and B and showed that the first one goes to infinity and the second one has a defined value and exist. My question is whether that is sufficient to prove that the first (improper) integral does not exist and the second one exist? The reason is because the integral may or may not exist, while the iterated integral may exist but be different than the integral. Both A and B are simple regions so Fubini's theorem can apply, but I think it needs to be a bounded region. So what I should do is to build a increasing sequence of compact rectifiable sets $U_n$  so that $U_1$ is in $U_2$ and so on and the union of all $U_n$ is the region A or B. How should I construct these sequence of sets?
 A: Here is a rather generous hint for your first integral. You can proceed in a similar manner for the second integral:
$$\begin{align}
\int_A\frac{1}{(y+1)^2}\,dx\,dy&=\int^\infty_0\Big(\int^{2x}_x\frac{1}{(y+1)^2}\,dy\Big)\,dx=-\int^\infty_0\frac{1}{y+1}\Big|^{2x}_x\,dx\\
&=\int^\infty_0\frac{1}{x+1}-\frac{1}{2x+1}\,dx
\end{align}$$
Truncating the integral to say $\int^a_0$ and then passing to the limit $a\rightarrow\infty$ gives
$$\int^a_0\frac{1}{x+1}-\frac{1}{2x+1}\,dx=\log(a+1)-\frac{1}{2}\log(2a+1)=\log\big(\tfrac{a+1}{\sqrt{2a+1}}\big)\xrightarrow{a\rightarrow\infty}?$$
As for the second integral
$$\begin{align}
\int_B\frac{1}{(y+1)^2}\,dxdy&=\int^\infty_0\Big(\int^{2x^2}_{x^2}\frac{1}{(y+1)^2}\,dy\Big)\,dx=-\int^\infty_0\frac{1}{y+1}\Big|^{2x^2}_{x^2}\,dx\\
&=\int^\infty_0\frac{1}{1+x^2}-\frac{1}{1+2x^2}\,dx\\
&=\lim_{a\rightarrow\infty}\int^a_0\frac{1}{1+x^2}-\frac{1}{1+2x^2}\,dx\\
&=\lim_{a\rightarrow\infty}\Big(\arctan(a)-\frac{1}{\sqrt{2}}\arctan(\sqrt{2}a)\Big)=?
\end{align}$$
