Calculate the improper double integral: double integral
Calculate the improper double integral: $$ \iint_{x\geqslant0,\,y\geqslant0} \frac{dx\,dy}{(a^2+x^2+y^2)^2}, \qquad a>0. $$
Explanation of Method
The integrand depends only on $r=\sqrt{x^2+y^2}$, and the region is the first quadrant — a quarter of the full plane in angular measure. Polar coordinates reduce the problem to a single radial integral, elementary via $w=a^2+r^2$.
- Convert to polar coordinates, with $\theta$ ranging over the first quadrant only.
- Substitute $w=a^2+r^2$ in the radial integral.
- Combine the angular and radial results.
Step 1 · Convert to polar coordinates
The first quadrant corresponds to $\theta\in\left[0,\frac\pi2\right]$, $r\in[0,\infty)$:
$$ \iint_{x,y\geqslant0}\frac{dx\,dy}{(a^2+x^2+y^2)^2} = \int_0^{\pi/2}\int_0^\infty \frac{r}{(a^2+r^2)^2}\,dr\,d\theta = \frac\pi2\int_0^\infty\frac{r}{(a^2+r^2)^2}\,dr. $$
Step 2 · Evaluate the radial integral
With $w=a^2+r^2$, $dw=2r\,dr$:
$$ \int_0^R \frac{r}{(a^2+r^2)^2}\,dr = \frac12\int_{a^2}^{a^2+R^2} w^{-2}\,dw = \frac12\left[\frac1{a^2}-\frac1{a^2+R^2}\right] \xrightarrow[R\to\infty]{} \frac1{2a^2}. $$
Step 3 · Combine
$$ \iint_{x,y\geqslant0}\frac{dx\,dy}{(a^2+x^2+y^2)^2} = \frac\pi2\cdot\frac{1}{2a^2} = \frac{\pi}{4a^2}. $$
Additional Notes
- As a check, the full-plane version of this integral is exactly four times as large (all four quadrants contribute equally by symmetry): $\displaystyle\iint_{\mathbb R^2}\frac{dx\,dy}{(a^2+x^2+y^2)^2}=4\cdot\frac{\pi}{4a^2}=\frac{\pi}{a^2}$, which matches direct computation using the full angular range $2\pi$ in Step 1.
Result
The solution to the integral is: