Why are there no linearly divergent terms in the sunset diagram?
Consider the following 2-loop 1PI diagram for the $\lambda \phi^4$ theory.
This is given by the following integral:
$$-i\ \Sigma(p^2) = \frac{(i\lambda_0)^2}{#} \int\frac{\mathrm d^4s}{(2\pi)^4}\frac{\mathrm d^4t}{(2\pi)^4} \frac{i}{s^2 - \mu_0^²+i\varepsilon} \frac{i}{t^2 - \mu_0^²+i\varepsilon}\frac{i}{(p-s-t)^2 - \mu_0^²+i\varepsilon} \,.$$
We can see that it is quadratically divergent. Note that a derivative with respect to $p^2\,,$
$$ \frac{\partial}{\partial p^2} = \frac{1}{2p^2} p_\mu \frac{\partial}{\partial p_\mu}\,,$$
decreases the degree of divergence by $1$.
That is to say, $$\frac{\partial \Sigma}{\partial p^2}\Big|{p^2=0} \text{ is linearly divergent,} \ \quad \text{ and } \frac{\partial^2 \Sigma}{\partial^2 p^2}\Big|{p^2=0} \text{ is logarithmically divergent.} $$
Therefore, a <PERSON> series expansion of this function about $p^2 = 0$ should contain three divergent terms.
$$ \Sigma(p^2) = \color{red}{\Sigma(0)} + \color{red}{\frac{\partial \Sigma}{\partial p^2}\Big|{p^2=0}} p^2 + \color{red}{\frac{\partial^2 \Sigma}{\partial^2 p^2}\Big|{p^2=0}} (p^2)^2 + \tilde{\Sigma}(p^2) \,,$$
where $\tilde{\Sigma}(p^2)$ is UV-finite and $\tilde{\Sigma}(0) = 0\,, \tilde{\Sigma}'(0) = 0 \,.$
That is my understanding. However, the general consensus is that there is no linearly divergent term in the above series. 818
In fact, the coefficient of $p^2$ should be logarithmically divergent. I do not understand why.
Cheng and <PERSON> (p.34) say that it is because the coefficient should in fact be $\frac1{8} (\partial/\partial p_\mu) (\partial/\partial p^\mu) \Sigma(p^2)|_{p^2 =0}\,.$ Furthermore, the linearly divergent term is absent due to Lorentz invariance.
First of all, I do not understand why <PERSON> and <PERSON> have a different coefficient than mine. Secondly, what has Lorentz invariance got anything to do with <PERSON> expansion?
To summarise, why are there no linearly divergent terms in the sunset diagram? 346