Examine the convergence of the improper double integral
Examine the convergence of the improper double integral: $$ \iint_{\mathbb R^2} \frac{dx\,dy}{1+x^2+y^2}. $$
Explanation of Method
The integrand depends only on $r=\sqrt{x^2+y^2}$, so polar coordinates turn this into a single radial integral. The radial factor $\dfrac{r}{1+r^2}$ behaves like $\dfrac1r$ for large $r$, and $\displaystyle\int^\infty\frac{dr}r$ is the borderline logarithmically divergent case — so the double integral diverges.
- Convert to polar coordinates.
- Integrate with respect to $r$ over $[0,R]$ and let $R\to\infty$.
- Observe the logarithmic divergence.
Step 1 · Convert to polar coordinates
With $x=r\cos\theta$, $y=r\sin\theta$, $dx\,dy=r\,dr\,d\theta$:
$$ \iint_{\mathbb R^2}\frac{dx\,dy}{1+x^2+y^2} = \int_0^{2\pi}\int_0^\infty \frac{r}{1+r^2}\,dr\,d\theta = 2\pi\int_0^\infty\frac{r}{1+r^2}\,dr. $$
Step 2 · Examine the radial integral
$$ \int_0^R \frac{r}{1+r^2}\,dr = \frac12\Big[\ln(1+r^2)\Big]_0^R = \frac12\ln(1+R^2). $$
Step 3 · Let $R\to\infty$
$$ \frac12\ln(1+R^2) \longrightarrow +\infty \quad\text{as } R\to\infty. $$
So the radial integral diverges, and therefore so does the double integral.
Additional Notes
- This is the borderline case in a family: $\displaystyle\iint_{\mathbb R^2}\frac{dx\,dy}{(1+x^2+y^2)^k}$ converges for $k>1$ and diverges for $k\leqslant1$, since the radial integrand $\dfrac{r}{(1+r^2)^k}\sim r^{1-2k}$ for large $r$, and $\displaystyle\int^\infty r^{1-2k}\,dr$ converges iff $1-2k<-1$, i.e. $k>1$. Here $k=1$ sits exactly on the boundary, producing the logarithmic (rather than power-type) divergence.
Result
The integral diverges: