You can think this of the following as a 3d QFT where we try to calculate the self-energy of two fields. $I$ is a this external self-energy and let us assume it does not depend on the loop momenta which I denote a $\vec{x}=(x,y)$. Consider the following propagator $$ P(\vec{r}) = \frac{x-ia}{2i^2[\,(x-ia)^2+(y-ib)^2+(z-ic)^2]\,} \,\, \gamma_i $$ Then we want to calculate $$ I(x) = \int \frac{\,dx \,dy\, dz}{(2\pi)^3} \text{Tr} \left( \frac{x-ia}{2i^2[\,(x-ia)^2+(y-ib)^2+(z-ic)^2]\,}\gamma_x \gamma_x \right) $$ Where Tr$(\gamma_i \gamma_j)=2\delta_{ij}$. Therefore I am trying to resolve the following integral where $x,y,z \in \mathbb{R}$ and $a,b,c$ are constants (internal self-energies if you wish) in $\mathbb{R}$. Then, after simplifying the propagator $P(x,y)$ we end up with the following expression for the $x$ component: $$ I = \frac{1}{2i^2} \int \frac{dxdy}{(2\pi)^2} (x-ia) \int \frac{dz}{2\pi} \frac{1}{[\,(x-ia)^2+(y-ib)^2+(z-ic)^2]\,} $$ but it can be shown that $$ \int \frac{dq}{(q+iv)^2+w^2} = \frac{\pi}{|w|}\theta(|w|-|v|). $$ [e.g. see here Does contour integral over $\mathbb{R}$ give a step function? ] Thus we get that for $|(y-ib)^2+(x-ia)^2|>|a|$, $\theta=1$ (for this is what we are interested about) and then we are left with $$ I = \frac{1}{4i^2} \int \frac{dxdy}{(2\pi)^2} \frac{(x-ia)}{|(x-ia)^2+(y-ib)^2|} $$
The question is wether I can make the transformation $$ (x-ia) = q \cos \theta $$ $$ (y-ib) = q \sin \theta $$ in polar coordinates. Then $dx dy = qdqd\theta$ and the previous integral becomes, assuming $|q|>0$, $$ I = \int_0^{\infty} dq \int_0^{2\pi} d\theta \frac{q \cos \theta}{|q|} =0 $$ Is this correct? I am confused since one can find complex poles in the $I$. If not, what would the correct way to resolve this be? I have a suspicion that this integral should indeed give zero but how can we find this?