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**[SOLVED] One-particle irreducibles in P&S**

I'm going through the derivation of the [tex]-i\Sigma_2 (p)[/tex] correction in Peskin & Shcroeder. On the top of page 218, eq. 7.17 they say

[tex]

-i\Sigma_{2}\left(p\right) = -e^{2}\int_{0}^{1}dx\int\frac{d^{4}\ell}{\left(2\pi\right)^{4}}\frac{-2x pslash+4m_{0}}{\left[\ell^{2}-\Delta+i\epsilon\right]^{2}}

[/tex]

where the denominator is O.K. Have that [tex] \ell \equiv k-xp [/tex].

Before the Feynman parameter was introduced, the correction had the appearance (p. 217, eq. 7.16)

[tex]

-i\Sigma_{2}\left(p\right)=\left(-ie\right)^{2}\int\frac{d^{4}k}{\left(2\pi\right)^{4}}\gamma^{\mu}\frac{i\left(kslash+m_{0}\right)}{k^{2}-m_{0}^{2}+i\epsilon}\gamma_{\mu}\frac{-i}{\left(p-k\right)^{2}-\mu^{2}+i\epsilon}

[/tex]

**To make a long story short:**I get the numerator to be [tex]-2\gamma^{\mu}\ell_{\mu}-2x pslash+4m_{0}[/tex] as I suspect the authors did to. But they're dropping the linear terms in [tex]\ell[/tex], and I think the reason is to be found on p. 191, eq. 6.45.

"... This task is simplified by noting that since [tex]D[/tex] depends only on the magnitude of [tex]\ell[/tex],

[tex]

\int\frac{d^{4}\ell}{\left(2\pi\right)^{4}}\frac{\ell^{\mu}}{D^{3}}=0

[/tex]

The (...) identity follows from symmetry."

(They have defined [tex]D \equiv \ell^2 - \Delta +i\epsilon[/tex], so in my case the numerator will be [tex]D^2[/tex])

I don't really see how this follows from symmetry. Any suggestions?

Btw: sorry about the kslash/pslash notation.

Thanks!