Calculate the improper double integral: double integral

Calculate the improper double integral: $$ \iint_{\mathbb R^2} \frac{dx\,dy}{(1+x^2+y^2)^{3/2}}. $$

Step 1 · Convert to polar coordinates

$$ \iint_{\mathbb R^2}\frac{dx\,dy}{(1+x^2+y^2)^{3/2}} = \int_0^{2\pi}\int_0^\infty \frac{r}{(1+r^2)^{3/2}}\,dr\,d\theta = 2\pi\int_0^\infty\frac{r}{(1+r^2)^{3/2}}\,dr. $$

Step 2 · Substitute $w=1+r^2$

With $dw=2r\,dr$:

$$ \int_0^R \frac{r}{(1+r^2)^{3/2}}\,dr = \frac12\int_1^{1+R^2} w^{-3/2}\,dw = \Big[-w^{-1/2}\Big]_1^{1+R^2} = 1-\frac{1}{\sqrt{1+R^2}}. $$

Step 3 · Let $R\to\infty$

$$ \int_0^\infty \frac{r}{(1+r^2)^{3/2}}\,dr = \lim_{R\to\infty}\left(1-\frac{1}{\sqrt{1+R^2}}\right) = 1. $$

So:

$$ \iint_{\mathbb R^2}\frac{dx\,dy}{(1+x^2+y^2)^{3/2}} = 2\pi\cdot1 = 2\pi. $$

Additional Notes

• Verified numerically: direct integration over an increasing sequence of large disks converges rapidly to $2\pi\approx6.28319$.
• This exponent, $\frac32$, is the borderline power for which the radial integral converges at infinity in exactly the simplest way (giving a finite, closed-form limit); as noted in the related divergent example, exponent $1$ is the boundary of convergence, so $\frac32$ is safely within the convergent range.