Question about loop integrals

  • Thread starter earth2
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Hey,

I've been reading a bit in the QFT book by Pierre Ramond about loops and i stumbled about some derivations in appendix B that I can't follow.
First Ramond derives a formula for integration of momenta in arbitrary dimensions
[tex]\int \frac{d^Nl}{(l^2+2p\cdot l+b^2)^A}=\pi^{N/2}\frac{\Gamma(A-N/2)}{\Gamma(A)}\frac{1}{(b^2-p^2)^{A-N/2}}[/tex]

This is fine with me. What I don't get is the following. He says if we differentiate this formula with respect to [tex]p^\mu[/tex] we'd get

[tex]\int \frac{d^Nl \quad l_\mu}{(l^2+2p\cdot l+b^2)^A}=\pi^{N/2}\frac{\Gamma(A-N/2)}{\Gamma(A)}\frac{-p_\mu}{(b^2-p^2)^{A-N/2}}[/tex]

I don't see how that comes about. If I differentiate the first formula wrt p I get

[tex]\int \frac{d^Nl \quad (-A)2l_\mu}{(l^2+2p\cdot l+b^2)^{A+1}}=\pi^{N/2}\frac{\Gamma(A-N/2)}{\Gamma(A)}\frac{-p_\mu(-A+N/2)}{(b^2-p^2)^{A-N/2+1}}[/tex]

I haven't found another derivation of this formula, nor do I see where I go wrong...
Can anyone help my with this?

Thanks
earth2
 
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Answers and Replies

  • #2
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I usual, we assume an infinite amount of available space under our rug (no discussion of proper definitions of integrals etc...).

I start from your formula : there is a missing factor 2 in the numerator in the rhs, which comes from the square of p inside the parenthesis in the denominator before differentiation :
[tex]
\int \frac{d^Nl \quad (-A)2l_\mu}{(l^2+2p\cdot l+b^2)^{A+1}}=\pi^{N/2}\frac{\Gamma(A-N/2)}{\Gamma(A)}\frac{-2p_\mu(-A+N/2)}{(b^2-p^2)^{A-N/2+1}}
[/tex]
So drop the factors of 2 and re-arrange the A and A-N/2 with the Gammas :
[tex]
\int \frac{d^Nl \quad l_\mu}{(l^2+2p\cdot l+b^2)^{A+1}}=\pi^{N/2}\frac{(A-N/2)}{A}\frac{\Gamma(A-N/2)}{\Gamma(A)}\frac{-p_\mu}{(b^2-p^2)^{A-N/2+1}}
[/tex]
Then make use of
[tex]
\Gamma(z+1)=z\Gamma(z)
[/tex]

[tex]
\int \frac{d^Nl \quad l_\mu}{(l^2+2p\cdot l+b^2)^{A+1}}=\pi^{N/2}\frac{\Gamma(A+1-N/2)}{\Gamma(A+1)}\frac{-p_\mu}{(b^2-p^2)^{A+1-N/2}}
[/tex]
Finally rename A+1 -> A
[tex]
\int \frac{d^Nl \quad l_\mu}{(l^2+2p\cdot l+b^2)^{A}}=\pi^{N/2}\frac{\Gamma(A-N/2)}{\Gamma(A)}\frac{-p_\mu}{(b^2-p^2)^{A-N/2}}
[/tex]
(yes, you did the hard part :smile:)
 
  • #3
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Ah, that was enlightening! Thanks a lot! :)
 

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