Use a <Feynman parameter> with $A=k^2+m^2$ and $B=(p-k)^2$. Then
$$
xA+(1-x)B=[k-(1-x)p]^2+\Delta,
\qquad
\Delta=xm^2+x(1-x)p^2.
$$
The identities for <gamma matrices> give
$$
\gamma^\mu(-i\not k+m)\gamma_\mu
=i(d-2)\not k+dm.
$$
After the shift $\ell=k-(1-x)p$, the term odd in $\ell$ integrates to zero. The rotationally symmetric loop integral is
$$
\int\frac{d^d\ell}{(2\pi)^d}\frac1{(\ell^2+\Delta)^2}
=\frac1{(4\pi)^{d/2}}\Gamma\left(\frac\epsilon2\right)
\Delta^{-\epsilon/2}.
$$
Consequently
$$
\boxed{\Sigma(\not p)=-\frac{e^2}{(4\pi)^{d/2}}
\Gamma\left(\frac\epsilon2\right)
\int_0^1dx\,
\frac{i(2-\epsilon)(1-x)\not p+(4-\epsilon)m}
{[xm^2+x(1-x)p^2]^{\epsilon/2}}.}
$$
Thus
$$
\boxed{C=i(2-\epsilon)(1-x),\qquad F=4-\epsilon,
\qquad\Delta=xm^2+x(1-x)p^2.}
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 304 / 3 / d / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/iii/paper-304/3/d/solution