 Solution

Apply a Feynman parameter and shift the loop momentum to $\ell =p+xk$. The common denominator becomes $[{\ell }^2+M_x^2{]}^2$, whereFor positive $M_x^2$, the Gamma-integral representation $a^{-2}={\int }_0^{\infty }dss{e}^{-sa}$ and a Gaussian integral giveThis formula initially converges for $\mathrm{Re}d<4$ and defines the Euclidean massive loop integral at other dimensions by analytic continuation. Consequently,With $d=6-ϵ$, the Gamma function factor is $\Gamma (-1+ϵ/2)=-2/ϵ+O(1)$. Only its pole matters: the other factors can be evaluated at $ϵ=0$ when extracting that pole. Since ${\int }_0^{1}x(1-x)dx=1/6$,This is the one-loop two-point divergence in six-dimensional cubic scalar theory. Its polynomial momentum dependence is precisely what permits subtraction by local counterterms.