QCD Feynman Rules: Gluon-Gluon OPI GF

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Basically it's the diagram attached below. If one uses the Feynman rules for QCD, he gets the expression

[tex]\begin{array}{c} \left(\tilde{\Gamma}^{(2)}_{2,\mu\nu}\right)\left(p,-p\right) <br /> <br /> = \frac{1}{2}\hat{g}^{2}M^{\varepsilon }f^{d}{}_{ac}f^{c}{}_{bd}\int \frac{d^{2\omega }q}{(2\pi)^{2\omega }}\left( \frac{-g^{\lambda \rho }q^{2}+\eta q^{\lambda }q^{\rho }}{\left( q^{2}+i\epsilon \right) ^{2}}\right) \left( \frac{-g^{\sigma \tau }\left( p+q\right) ^{2}+\eta \left( p+q\right) ^{\sigma }\left( p+q\right) ^{\tau }}{\left[ \left( p+q\right) ^{2}+i\epsilon \right] ^{2}}\right) \\<br /> <br /> \times \left[ \left( q-p\right) _{\sigma }g_{\rho \mu }-\left( p+2q\right) _{\mu }g_{\sigma \rho }+\left( 2p+q\right) _{\rho }g_{\mu \sigma }\right] \left[ \left( 2p+q\right) _{\lambda }g_{\tau \nu }-\left( p+2q\right) _{\nu }g_{\lambda \tau }+\left( q-p\right) _{\tau }g_{\nu \lambda }\right] \end{array} <br /> [/tex].
 
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My question involves solving that integral.

[itex]\eta=1-\xi[/itex]

and usually QFT books deal only with the Feynman gauge in which [itex]\eta =0[/itex].

I'm interested only in the [itex]\eta[/itex] dependent part (funny, right ?) in which, after doing all the possible contractions and multiplications (using that D=dim.space-time= 2 \omega), i got 18 terms leading me to 18 Feynman integrals.

An example of an integral that is over my head is

[tex]\frac{1}{2}\hat{g}^{2}M^{\varepsilon }f^{d}{}_{ac}f^{c}{}_{bd}\eta $\int \frac{d^{2\omega }q}{\left( 2\pi \right) ^{2\omega }} \left(\frac{\left( p+2q\right) _{\mu }\left( p+q\right) _{\nu }\left( q^{2}-p^{2}\right) q^{2}}{\left( q^{2}+i\epsilon \right) ^{2}\left[ \left( p+q\right) ^{2}+i\epsilon \right] ^{2}}\right)[/tex]

So can anyone help...?

Daniel.
 
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dextercioby said:
Well, the good news is that i finally bumped into Bailin & Love's book which have an interesting table of Feynman integrals, which can be used especially for QCD.

Daniel.

Ok, so you're good then? I'm happy to say something about the integral if you still need help.
 
Thx for the help offer, but i could manage on my own. As a bonus, i could actually rigorously prove the first formula of that appendix:

[tex]\int \frac{d^{2\omega}k}{(2\pi)^{2\omega}} \left(k^2 \right)^{-n} \left[(k+p)^2 \right]^{-m} =\frac{i (-)^{n+m}}{(4\pi)^{\omega}} \frac{\Gamma (\omega-n-m) B(\omega-n, \omega-m)}{\Gamma (n)\Gamma (m)} \left(-p^2\right)^{\omega-n-m}[/tex]

Daniel.