$\mu \rightarrow e\gamma$ in the $R_\xi$ gauge: trouble with momenta and Dirac matrices My first ever question on stackexchange! Sorry if it is clumsy... 
I am trying to follow the computation of $\mu \rightarrow e \gamma$ in Cheng and Li and am confused about the second and third ligns in equation (13.97):
$S_1 \rightarrow \tilde{S}_1 = (p\cdot \varepsilon )\left[\bar{u}_e(1+\gamma _5) u_\mu \right] 2 m_\mu \left[2(1-\alpha_1)^2 + (2\alpha _1-1)\alpha _2\right]\\
S_2 \rightarrow \tilde{S}_2 = -k^2(p\cdot \varepsilon )\left[\bar{u}_e(1+\gamma _5) u_\mu \right] (m_\mu /M^2) \times \left[(3\alpha _2-1)+\left[2\alpha_1^2-\alpha_1+\alpha_2(2\alpha_1-1/2)\right]\right]\\
S_3 \rightarrow \tilde{S}_3 = -k^2(p\cdot \varepsilon )\left[\bar{u}_e(1+\gamma _5) u_\mu \right] (m_\mu /M^2) \left[2\alpha_1^2+\alpha_1+\alpha_2(2\alpha_1-1/2)\right]$
where 
$S_1 = \Gamma ^{\mu \nu}N_{\mu \nu}\\
S_2 = (k^\lambda \Gamma ^\mu _\lambda)(k^\nu N_{\mu \nu})/M^2\\
S_3 = \left[(k+q)^\lambda \Gamma ^\mu _\lambda \right]\left[(k+q)^\nu N_{\mu \nu}\right]/M^2$
with
$\Gamma _{\alpha \beta} = \left[(2k\cdot \varepsilon) g_{\alpha \beta} -(k+2q)_\beta \varepsilon _\alpha - (k-q)_\alpha \varepsilon _\beta\right] $
(this being equation (13.81))
and 
$N_{\mu \nu} =\bar{u}_e(p-q)\gamma _\mu ((p+k)\cdot \gamma ) \gamma _\nu (1-\gamma _5)u_\mu(p)$
(equation (13.88)).
In the calculation $S_i\rightarrow \tilde{S}_i$ we shift the integration variable $k$ to $k-\alpha_1 p - \alpha_2 q$ and discard all terms that are not $\sim (p\cdot \varepsilon)$.
Whereas I managed to obtain the expression for $\tilde{S}_1$ (with a minus sign, but apparently that's alright...), I do not understand what to do with the many momenta in the calculation of $\tilde{S}_2$ (and $\tilde{S}_3$): 
Why do Cheng and Li only get (or keep?) terms $\sim k^2$? I would have thought we would get a whole bunch of terms $\sim m_\mu ^2$ from the combinations $\sim p^2$ and $\sim p\cdot q$. (The mass of the electron is neglected here.) 
I would be very grateful for some tips... Thank you in advance!
 A: I recently went through this calculation, but only working in the $\xi = 1$ gauge. I would first note that I also got a minus sign relative to Cheng and Li, for every single diagram, so it all worked out in the end. 
When you shift the integration variable (and this shift will be different if you assigned your Feynman Parameters differently) the numerator takes on new terms which are either scalar $(\sim p \cdot \varepsilon )$ or vector $(\sim \gamma \cdot \varepsilon)$ under Lorentz transformations. Integrating over your new $k^\mu$ will not change the Lorentz transformation properties of these terms so, since they found that $$T = A \overline{u}_e(p-q)(1+\gamma_5)(2p\cdot \varepsilon-m_\mu \gamma \cdot \varepsilon) u_\mu(p)$$
(assuming $m_e \rightarrow 0$) (this is $(13.79)$) they know to keep only the scalar terms (which pleseantly happen to be proportional to $p\cdot \varepsilon$). Also all the terms which were odd under $k^\mu \rightarrow - k^\mu$ simply cancel from the integrations.
Unfortunately, working in $\xi =1$, I didn't need to calculate the $S_2$ or $S_3$ terms so I can't give you any more specific advice other than to keep staring at it. Understanding why the above scalar - vector decomposition is so clever took a lot of headscratching on my part and I can only hope I didn't convince myself of something that was completely false, but I made it to the correct answer in the end. 
