# Reproducing Ramond's sunset diagram calculation for $\phi^4$ theory

I am unable to reproduce the calculation of the sunset diagram for $\phi^4$ theory in Pierre Ramond's Fied Theory a Modern Primer. This is the second edition chapter 4.4. He starts with eq. (4.4.19) $$\Sigma(p) = \frac{\lambda^2 (\mu^2)^{4-2\omega}}{6} \int \frac{d^{2\omega}\ell}{(2\pi)^{2\omega}} \frac{d^{2\omega}q}{(2\pi)^{2\omega}} \frac{1}{\ell^2+m^2} \frac{1}{q^2+m^2} \frac{1}{(q+p-\ell)^2+m^2}$$ He introduces 1 in the form $$1=\frac{1}{4\omega}\left[ \frac{\partial \ell_\mu}{ \partial \ell_\mu}+ \frac{\partial q_\mu}{ \partial q_\mu}\right]$$ to get $$\Sigma(p) = \frac{\lambda^2 (\mu^2)^{4-2\omega}}{6} \int \frac{d^{2\omega}\ell}{(2\pi)^{2\omega}} \frac{d^{2\omega}q}{(2\pi)^{2\omega}}\frac{1}{4\omega}\left[ \frac{\partial \ell_\mu}{ \partial \ell_\mu}+ \frac{\partial q_\mu}{ \partial q_\mu}\right] \frac{1}{\ell^2+m^2} \frac{1}{q^2+m^2} \frac{1}{(q+p-\ell)^2+m^2}$$ then uses partial integration and discards the boundary terms to get $$\Sigma(p) = -\frac{\lambda^2 (\mu^2)^{4-2\omega}}{6} \times\\ \int \frac{d^{2\omega}\ell}{(2\pi)^{2\omega}} \frac{d^{2\omega}}{(2\pi)^{2\omega}}\frac{1}{4\omega}\left[ \ell_\mu\frac{\partial}{ \partial \ell_\mu}+ q_\mu \frac{\partial}{ \partial q_\mu}\right] \frac{1}{\ell^2+m^2} \frac{1}{q^2+m^2} \frac{1}{(q+p-\ell)^2+m^2}\qquad (1)$$ All of that is fine, but then he says that explicit differentiation gives the result $$\Sigma(p) = \frac{1}{2\omega-3}\frac{\lambda^2 (\mu^2)^{4-2\omega}}{6}\int \frac{d^{2\omega}\ell}{(2\pi)^{2\omega}} \frac{d^{2\omega}q}{(2\pi)^{2\omega}}\frac{3 m^2 + p\cdot(p+q-\ell)} {(\ell^2+m^2) (q^2+m^2) [(q+p-\ell)^2+m^2]^2}$$ I cannot find how to reproduce this formula. In fact I do not understand how the coefficient $1/\omega$ disappears and the coefficient $1/(2\omega-3)$ can appear. Indeed, for general momenta $\ell$ and $k$ $$\ell_\mu\frac{\partial}{ \partial \ell_\mu} \frac{1}{(k-\ell)^2+ m^2} = \frac{2\ell\cdot (k-\ell)}{[(k-\ell)^2+ m^2]^2}$$ When you use this into (1) you only get inner products of momenta. Without having to do the calculation in detail, how can the dimension (dis)appear in the computation?

$$I(p)=\int\frac{d^D p_1}{(2\pi)^D}\frac{d^Dp_2}{(2\pi)^D}\frac{1}{(p_1^2+m^2)(p_2+m^2)[(p+p_1+p_2)^2+m^2]}.$$
$$I(p)=-\int\frac{d^D p_1}{(2\pi)^D}\frac{d^Dp_2}{(2\pi)^D}\frac{1}{2D}(p_1^\mu\frac{\partial }{p_1^\mu}+p_2^\mu\frac{\partial }{p_2^\mu})\frac{1}{(p_1^2+m^2)(p_2+m^2)[(p+p_1+p_2)^2+m^2]}.$$
$$\begin{eqnarray*} I(p)&=& \frac{1}{D}\int\frac{d^D p_1}{(2\pi)^D}\frac{d^Dp_2}{(2\pi)^D} \frac{p_1^2}{(p_1^2+m^2)^2(p_2+m^2)[(p+p_1+p_2)^2+m^2]}\\ && \frac{1}{D}\int\frac{d^D p_1}{(2\pi)^D}\frac{d^Dp_2}{(2\pi)^D}\frac{p_1(p+p_1+p_2)}{(p_1^2+m^2)(p_2+m^2)[(p+p_1+p_2)^2+m^2]^2}\\ && \frac{1}{D}\int\frac{d^D p_1}{(2\pi)^D}\frac{d^Dp_2}{(2\pi)^D} \frac{p_1^2}{(p_1^2+m^2)(p_2+m^2)^2[(p+p_1+p_2)^2+m^2]}\\ && \frac{1}{D}\int\frac{d^D p_1}{(2\pi)^D}\frac{d^Dp_2}{(2\pi)^D} \frac{p_2(p+p_1+p_2)}{(p_1^2+m^2)(p_2+m^2)[(p+p_1+p_2)^2+m^2]^2}\\ &=& \frac{3}{D}\int\frac{d^D p_1}{(2\pi)^D}\frac{d^Dp_2}{(2\pi)^D} \frac{1}{(p_1^2+m^2) (p_2+m^2)[(p+p_1+p_2)^2+m^2]}\\ &&-\frac{1}{D}\int\frac{d^D p_1}{(2\pi)^D}\frac{d^Dp_2}{(2\pi)^D} \frac{m^2}{(p_1^2+m^2)^2 (p_2+m^2)[(p+p_1+p_2)^2+m^2]} \\ &&-\frac{1}{D}\int\frac{d^D p_1}{(2\pi)^D}\frac{d^Dp_2}{(2\pi)^D} \frac{m^2}{(p_1^2+m^2) (p_2+m^2)^2[(p+p_1+p_2)^2+m^2]} \\ &&-\frac{1}{D}\int\frac{d^D p_1}{(2\pi)^D}\frac{d^Dp_2}{(2\pi)^D} \frac{m^2}{(p_1^2+m^2)(p_2+m^2)[(p+p_1+p_2)^2+m^2]^2 } \\ &&-\frac{1}{D}\int\frac{d^D p_1}{(2\pi)^D}\frac{d^Dp_2}{(2\pi)^D} \frac{p (p+p_1+p_2)}{(p_1^2+m^2)(p_2+m^2)[(p+p_1+p_2)^2+m^2]^2}\\ &=& \frac{3}{D} I(p) - \frac{1}{D}\int\frac{d^D p_1}{(2\pi)^D}\frac{d^Dp_2}{(2\pi)^D} \frac{3m^2+p (p+p_1+p_2)}{(p_1^2+m^2)(p_2+m^2)[(p+p_1+p_2)^2+m^2]^2} \end{eqnarray*}$$
$$I(p)= -\frac{1}{D-3}\int\frac{d^D p_1}{(2\pi)^D}\frac{d^Dp_2}{(2\pi)^D} \frac{3m^2+p (p+p_1+p_2)}{(p_1^2+m^2)(p_2+m^2)[(p+p_1+p_2)^2+m^2]^2}$$