Simplify quantum electrodynamics expression I have a quantum electrodynamics exercise on the one-loop electron self-energy correction in which I need to show that
$$
\tag{1}
ie^{2}\Sigma(p)=\frac{\left(-ie\right)^{2}}{\left(2\pi\right)^{4}}\int d^{4}qD_{\lambda\sigma}\left(q\right)\gamma^{\lambda}S_{F}\left(p-q\right)\gamma^{\sigma}
$$
where
$$
\tag{2}
D_{\lambda\sigma}\left(q\right)=\frac{-i\eta_{\lambda\sigma}}{q^2-i\epsilon}  \qquad\qquad S_{F}\left(p\right)=\frac{i\left(\gamma^{\alpha}p_{\alpha}-m\right)}{p^{2}+m^{2}-i\epsilon}
$$
can be written as
$$
\tag{3}
ie^{2}\Sigma(p)=\frac{e^{2}}{\left(2\pi\right)^{4}}\int d^{4}q\frac{1}{q^2-i\epsilon}\frac{2\left(\gamma^{\alpha}\left(p-q\right)_{\alpha}-2m\right)}{\left(p-q\right)^{2}+m^{2}-i\epsilon}
$$
I can get close to show this using $\gamma_{\sigma}\gamma^{\sigma}=4$ and $\gamma_{\mu}\gamma^{\sigma}\gamma^{\mu}=-2\gamma^{\sigma}$, but I wasn't able to do it completely yet. So, the first thing I would like to know is if $D_{\lambda\sigma}\left(q\right)$ and $S_{F}\left(p\right)$ are well defined with this formulas, because the formulas I've studied are different,
$$
D_{\lambda\sigma}\left(q\right)=\frac{-i\eta_{\lambda\sigma}}{q^2+i\epsilon}  \qquad\qquad S_{F}\left(p\right)=\frac{i\left(\gamma^{\alpha}p_{\alpha}+m\right)}{p^{2}-m^{2}+i\epsilon}
$$
so I want to know if both forms are equivalent. If the first ones are correct, then I cannot show what the exercise asks for. Could you advise?
 A: To answer my own question, the closest I've come to a solution using the given propagators in $(2)$ was showing $(3)$ up to a minus sign that I'm assuming is missing on the term $\gamma^{\alpha}\left(p-q\right)_{\alpha}$.
In my first derivation I was using the gamma matrix identities for signature $(+---)$. However, for metric choice $(-+++)$ the identities one should use are
$$\gamma_{\sigma}\gamma^{\sigma}=-4$$
$$\gamma_{\mu}\gamma^{\sigma}\gamma^{\mu}=2\gamma^{\sigma}$$
A: In the 'mostly plus' signature, you find a modification to the propagators as shown in the OP with a corresponding sign flip to the rhs of the Clifford algebra (contains the metric tensor). This is all summarised in e.g ref Srednicki. I'll address the second part of the OP, now v2.
Let $V^{(1)}$ be the one loop contribution to the one particle irreducible three point QED vertex function. The UV divergence ( = pole in dimensional regularisation) manifest in the one loop integral is subtracted via the one loop charge renormalisation constant. If you have never done such a computation before, it might be useful to do e.g the four point vertex function in scalar phi^4 theory so that intricacies in Dirac algebra manipulations do not detract from the conceptual idea.
I would write the full QED vertex function $V^{\mu}$ as $$V^{\mu} = -iZ_1 e_0 \gamma^{\mu} + V^{\mu,(1)}(p,p’) + \dots, $$ with $e_0$ the bare parameter appearing in the lagrangian.  Then, at tree level, the charge does not renormalise so there is an expansion in the $Z_1$ of the form $$Z_1-1 =  Z_1^{(1)} e_0^2 + \dots $$ Inserting this into $V^{\mu}$ gives $$ V^{\mu} = -ie_0\gamma^{\mu}  - iZ_1^{(1)} e_0^3 \gamma^{\mu} + V^{\mu,(1)}(p,p’)$$ which is the tree level bare interaction vertex, the counter term vertex and the one loop correction respectively.
Writing down the corresponding integral representation for the one loop diagram as you have done for the self energy contribution to the electron, you should find in a minimal subtraction, $$Z_1^{(1)} \propto -\frac{1}{4 \pi} \frac{1}{\epsilon},$$ that is minus the UV divergence of the one loop integral.
