Understanding Loop Integrals and Differentiation in Quantum Field Theory

[earth2]

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
$$\int \frac{d^Nl}{(l^2+2p\cdot l+b^2)^A}=\pi^{N/2}\frac{\Gamma(A-N/2)}{\Gamma(A)}\frac{1}{(b^2-p^2)^{A-N/2}}$$

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

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

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

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

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

 

[humanino]

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 :
$$\int \frac{d^Nl \quad (-A)2l_\mu}{(l^2+2p\cdot l+b^2)^{A+1}}=\pi^{N/2}\frac{\Gamma(A-N/2)}{\Gamma(A)}\frac{-2p_\mu(-A+N/2)}{(b^2-p^2)^{A-N/2+1}}$$
So drop the factors of 2 and re-arrange the A and A-N/2 with the Gammas :
$$\int \frac{d^Nl \quad l_\mu}{(l^2+2p\cdot l+b^2)^{A+1}}=\pi^{N/2}\frac{(A-N/2)}{A}\frac{\Gamma(A-N/2)}{\Gamma(A)}\frac{-p_\mu}{(b^2-p^2)^{A-N/2+1}}$$
Then make use of
$$\Gamma(z+1)=z\Gamma(z)$$

$$\int \frac{d^Nl \quad l_\mu}{(l^2+2p\cdot l+b^2)^{A+1}}=\pi^{N/2}\frac{\Gamma(A+1-N/2)}{\Gamma(A+1)}\frac{-p_\mu}{(b^2-p^2)^{A+1-N/2}}$$
Finally rename A+1 -> A
$$\int \frac{d^Nl \quad l_\mu}{(l^2+2p\cdot l+b^2)^{A}}=\pi^{N/2}\frac{\Gamma(A-N/2)}{\Gamma(A)}\frac{-p_\mu}{(b^2-p^2)^{A-N/2}}$$
(yes, you did the hard part )

 

[earth2]

Ah, that was enlightening! Thanks a lot! :)