# Apparent elimination of overlapping divergences

The integral,

$$\iint_{\mathbb{R}^{2+}}\frac{xy}{1+x+y} \mathrm{d}y \, \mathrm{d}x$$

possesses an overlapping divergence when $x \to \infty$ and $y \to \infty$. However, under a change of variables to polar coordinates,

$$\int_{0}^{\infty}dr\int_{0}^{2\pi}du \frac{r^{3}\sin(u)\cos(u)}{1+r(\cos(u)+\sin(u))}$$

If we integrate over the angular variables numerically we are left with a sum of one dimensional divergent integrals, $$\int_{0}^{\infty}\frac{ar^{3}}{1+br}dr$$

Therefore the overlapping UV divergence is gone. Can be this done to every UV divergent multiple-loop integral?

• I'm not sure what the question is exactly. Please note that the $u$ integral doesn't converge in general. – fqq May 24 '14 at 13:22
• the integral is still divergent but it is onle a 1-dimensional divergent integral – Jose Javier Garcia May 24 '14 at 19:49
• No. The $u$ integral is in general divergent, and your expression for the $r$ integral is wrong. – fqq May 26 '14 at 17:26
• the integral in 'u' is convergent only the integral over 'r' is here divergent – Jose Javier Garcia May 26 '14 at 19:19