Help setting up integral: $\iint_{\mathbb R^2} \frac{1}{(1+4x^2+9y^2)^2} dx dy$

Question

Evaluate $\iint_{\mathbb R^2} \frac{1}{(1+4x^2+9y^2)^2} dx dy$

If I use polar coordinates I get $$\int_0^{2\pi}\int_0^\infty \frac{r}{(1+4r^2\cos^2\theta+9r^2\sin^2\theta)^2} drd\theta.$$

However, I am not sure where to go from here.

Answer 1

We substitute first $2x=s$, $3y=t$, so $6dx\wedge dy = ds\wedge dt$. The integral is after this easily computed. $$ \begin{aligned} \iint_{\mathbb R^2} \frac{1}{(1+4x^2+9y^2)^2} \; dx\; dy &= \frac 16\iint_{\mathbb R^2} \frac{1}{(1+s^2+t^2)^2} \;ds\; dt \\ &= \frac 16\int_0^\infty\int_0^{2\pi} \frac{1}{(1+r^2)^2} \; r\;dr\; d\theta \\ &= \frac 16\cdot2\pi\cdot \frac 12 \int_0^\infty \frac{1}{(1+r^2)^2} \;d(r^2) \\ &= \frac 16\cdot2\pi\cdot \frac 12 \left[\ -\frac{1}{1+r^2}\ \right]_0^\infty \\ &=\frac \pi6\ . \end{aligned} $$