What is the momentum squared in QFT self interaction integrals? $$
\int \dfrac{d^dp}{p^2+m^2}
$$
Source: Regularization of Feynman Integrals
I have seen this integral mentioned several times in Feynman integral regularization papers, but I do not understand what $p^2$ is. Is it Euclidean or hyperbolic? Is it $p^2 = p^2_1+p^2_2+p^2_3$ or $p^2 = p^2_1+p^2_2+p^2_3-p^2_0$. Can I naively evaluate this integral as euclidean over all variables, or must I evaluate its real poles?
 A: $p^2$ is $p_1^2+p_2^2+p_3^2 - p_0^2$. This integrand does have poles, the intuition behind the poles is that you expect a small amplitude when $p^2$ is very different than $m^2$, because then the particle has more (or less) momentum than you would conclude by looking at the energy. The intuition is that the particle has "borrowed'' this extra kinetic energy using the uncertainty principle, but the more energy it borrows, the less likely the process. In this case where $p^2 \ne m^2$, the process is called off shell. 
In the other case where you do have the equality $p^2=-m^2$, there is a divergence, because the particle does not need to "borrow" from the uncertainty principle, so it has a very high chance of occurring. You can read more in the wikipedia article linked here and above: On shell and off shell
In spite of this function's poles, its integration can be done using a contour in the complex plane. There are multiple inequivalent choices of contours which correspond to advanced, retareded, and so-called Feynman propagators. This is explained in the wikipedia page on propagators
