# Dimensional regularization of Electron self-energy from Ryder's book

I am Studying Electron self-energy using Ryder's textbook, In page 334 we can see

Defining $$k'=k-pz$$ and avoiding the term linear in $$k'$$(because it integrates to zero) gives $$$$\Sigma(p)=-ie^2\mu^{4-d}\int_0^1dz\gamma_\mu({\not} p-{\not}p z+m)\gamma^\mu\int\frac{d^dk'}{(2\pi)^d}\frac{1}{[k'^2-m^2z+p^2z(1-z)]^2}.\label{r2.7}$$$$ [...] This integral is performed with the help of equation (9A.5), giving $$$$\Sigma(p)=\mu^{4-d}e^2\frac{\Gamma(2-\frac{d}{2})}{(4\pi)^{d/2}}\int_0^1dz\gamma_\mu[{\not}p(1-z)+m]\gamma^\nu[-m^2z+p^2z(1-z)]^{d/2-2}.$$$$

The equation 9A.5 is $$$$\int\frac{d^dp}{(p^2+2pq-m^2)^{\alpha}}=(-1)^{d/2}\imath\pi^{d/2}\frac{\Gamma\left(\alpha-\frac{d}{2}\right)}{\Gamma(\alpha)}\frac{1}{[-q^2-m^2]^{\alpha-d/2}} .\tag{9A.5}$$$$ I don't understand how he applied this integral (9A.5) to obtain the result $$$$\Sigma(p)=\mu^{4-d}e^2\frac{\Gamma(2-\frac{d}{2})}{(4\pi)^{d/2}}\int_0^1dz\gamma_\mu[{\not}p(1-z)+m]\gamma^\nu[-m^2z+p^2z(1-z)]^{d/2-2}.$$$$ please help me to get an idea.

It's just a matter of applying the result (9A.5) to the integral in $$d^d k^\prime$$. In fact call $$M^2 = m^2z-p^2z(1-z)$$ and put $$q=0$$ in the integral (9A.5) $$\int\frac{d^dk'}{(2\pi)^d}\frac{1}{[k'^2-m^2z+p^2z(1-z)]^2} = \int\frac{d^dp}{(2\pi)^d}\frac{1}{[p^2-M^2]^2}=\frac{1}{(2\pi)^d}(-1)^{d/2}i\pi^{d/2}\frac{\Gamma\left(2-\dfrac{d}{2}\right)}{\Gamma(2)}\frac{1}{[-M^2]^{2-d/2}}$$
where we just changed the integration variable from $$k^\prime$$ to $$p$$ to make it clearer from the result 9A.5. Using the fact that $$\Gamma(2) = 1$$, using the above definition of $$M^2$$ and simplifying a bit you get $$\frac{(-1)^{d/2}}{2^d}i\pi^{-d/2}\Gamma\left(2-\frac{d}{2}\right)[-m^2z+p^2z(1-z)]^{d/2-2} = \frac{i(-1)^{d/2}}{(4\pi)^{d/2}}\Gamma\left(2-\dfrac{d}{2}\right)[-m^2z+p^2z(1-z)]^{d/2-2}$$ where we used the fact that $$2^d = 4^{d/2}$$
Compare the second integrand in the first equation with ty he in the grand in 9A5. You see that $$\alpha \rightarrow 2$$, $$q \rightarrow 0$$, $$-m^2 \rightarrow etc.$$ will transform one integrand into the other. Making the same substitutions in the rhs of 9A5 should give you the desired result.