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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?

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

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