Why this integral diverge?

Question

For $D=[0,+\infty)×[0,+\infty)$ and $f(x,y)=(1+x+y)^{-1}$, why does the integral over $D$ of $f(x,y)$ not converge?

It's not like $\rho/(1+\rho|\sin\theta+\cos\theta|)$ which to infinite is like $1/(|\sin\theta+\cos\theta|)$ and, close to $0$ is like $0$?

Answer 1

Consider

$$D'=[0,+\infty)\times [0,1]$$

and note that

$$\int_D \frac1{1+x+y} \,dx\,dy \ge \int_{D'} \frac1{1+x+y} \,dx\,dy\ge \int_{D'} \frac1{2+x} \, dx \, dy$$

Answer 2

First, we see that $f$ is positive on all of $D$. This means that we can consider a smaller region, and if it diverges there it still diverges on ll of $D$. Consider the region $E\subseteq D$ given by $E = [0,\infty)\times [0,1]$. On $E$, we have $f(x, y)\geq \frac{1}{2+x}$, and we have $$ \iint_Df(x, y)\,dx\,dy \geq \iint_Ef(x, y)\,dx\,dy = \int_0^\infty \int_0^1f(x, y)\,dy\,dx\\ \geq \int_0^\infty\int_0^1\frac{1}{2+x}dy\,dx = \int_0^\infty\frac{1}{2+x}dx $$ which diverges.

Answer 3

Intuitively, the area of the domain grows like $xy$, while the integrand decreases like $\dfrac1{1+x+y}$, which is insufficient to compensate.