Dimensional Regularization for $\phi^4$ theory When using dimensional Regularization for $\phi^4$ theory to calculate the running of the mass, why there is no quadratic divergence as expected? 
For details, see Prof. Arttu Rajantie, Lecture notes in Advanced QFT, page 51, equation 333 and 334.
 A: The integral $\int \frac{d^{4}p}{p^{2} - m^{2}}$ in dimensional regularization is
$$
I(d, m^{2}, \mu^{2}) \sim \frac{\Gamma \left( 1-\frac{d}{2}\right)}{\Gamma (d)}\left(m^{2}\right)^{\frac{d}{2}-1}\mu^{4-d}
$$
There is the problem with integral since there is no properly defined $\Gamma$ function of negative argument. We, however, "drops" the quadratic part. Why? The reason is hidden in analytic continuation procedure. Let's use the trick. Suppose you calculate integral
$$
\tag 1 \int d^{d}p\frac{d}{dp} \cdot (pf(p)) =\int d^{d}pp \cdot \frac{d}{dp}f(p) + d\int d^{d}pf(p),
$$
where
$$
f(p) \sim \frac{1}{p^{2}-m^{2}}
$$
The left handside of this equation is surface term, which contains the quadratic divergence. However, the analytic continuation procedure ignores surface terms, and thus the left handside of the integral may be dropped. Thus you have that
$$
d\int d^{d}pf(p) = -\int d^{d}pp \cdot \frac{d}{dp}f(p),
$$
and the right handside is just
$$
\sim \int \frac{d^{d}p}{(p^{2}-m^{2})^{2}}\sim \Gamma \left(2-\frac{d}{2}\right)
$$
