I am not quite sure that my question below is appropriate for this site, probably it should be addressed to the physical commutity. But I hope that some (mathematical) physicists do attend MO. I have two questions on dimensional regularization used in the renormalization theory (they should be very basic, I am a mathematician, even not a mathematical physicist).

1) **Is there a mathematically rigorous exposition of the dimensional regularization?**

2) Let $d$ denote the dimension of the space time.
My impression is that the method of dimensional
regularization works better for even $d$ rather than for odd. Namely
for some integrals which are obviously divergent in odd dimensions,
the method of dimensional regularization gives convergent
expressions. Below I give a simple example of such a situation for
$d=3$. **Thus my second question is how to resolve this apparent contradiction, and
whether the method can be modified to work in odd dimensions as
well, even at the physical level of rigour.**

Here is the example. Consider the theory $\phi^4$ in Euclidean space-time. Consider the
Feynmann diagram with just one vertex, one self-loop, and two
exterior lines (though, I guess, one can construct many other
examples). The corresponding integral, up to some factors containing
the interaction constant, is equal to $\int \frac{1}{p^2+1}d^dp$ (we
take the mass $m=1$ for simplicity; it is physically impossible for
dimensional considerations, but does not influence the analysis of
convergence issues). We have
\begin{eqnarray*}
\int \frac{1}{p^2+1}d^dp=\frac{2\pi^{d/2}}{\Gamma(d/2)}\int_0^\infty
\frac{r^{d-1}}{r^2+1}dr=:A.
\end{eqnarray*}
*The integral $A$ diverges for $d\geq 2$.* After some
change of variables and standard computations it becomes
\begin{eqnarray*}
2\pi^{d/2} \Gamma(-\frac{d}{2}+1).
\end{eqnarray*}

*The last expression has no poles at odd $d$, in
particular at $d=3$.* This apparently contradicts the above
mentioned divergence of the integral $A$. On the other hand,
at even $d$, $\Gamma(-\frac{d}{2}+1)$ does have a pole
as expected, and the method works well.