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I had a question on a dimensional regularization identity. A reference or a quick sort of derivation will be greatly appreciated. I looked at some textbooks of QFT, but couldn't find the one I was looking for.

I found in, a result for $\int\frac{d^dp(p^2)^a}{(p^2+D)^b}$ (see eq. 3.2 of the above link). I wanted something which is $\int\frac{d^dp(p^2)^a}{(p^2+2pq+D)^b}$ i.e the integrand has a linear power of $p$ too. May be a derivation of the previous equation will help. But anyway, some light on $\int\frac{d^dp(p^2)^a}{(p^2+2pq+D)^b}$ or $\int\frac{d^dp(p_\mu p_\nu..p_\lambda)}{(p^2+2pq+D)^b}$ is what I need. Thanks in advance.

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up vote 8 down vote accepted

The more complicated integrals can be easily reduced to the basic integral from equation 3.2. You start with modifications that simplify the denominator. First of all, $2pq$ in the denominator may be eliminated by completing the square: $$ p^2+2pq + D = (p+q)^2 + (D-q^2) $$ which is of the same form as the original integral with $1/(p^{\prime 2}+D')^b$, as long as $p'=p+q$ and $D'=D-q^2$.

Second, the polynomials $p^\alpha p^\beta\dots$ in the numerator – which have already been rewritten in terms of the new variable $p$ so that the denominator is $1/(p^2+D)^b$ – can be easily calculated because the integral is a tensor so the integral with $2n$ copies of $p^\alpha$ in the numerator must be proportional to $$g^{\alpha\beta} g^{\gamma\delta} \dots g^{\alpha_n\beta_n}+{\rm permutations} $$ times the integral with $(p^2)^n$ replacing the product of the $p^\alpha$ factors where the overall coefficient may be calculated in a straightforward way by checking the same identity with $n$ contractions.

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I can complete the square, but the application of 3.2 needs a factor of $(p')^{2a}$ in the numerator then, and not $p^{2a}$. Isn't it? – user1349 Jan 19 '12 at 0:15
Dear user, you should make the steps in the order I indicated. You first bring the denominator to the standard form $1/(p^2)^a$ by completing the squares. Then, with this definition of the variable $p$, you get something in the numerator whatever it is (it's a new polynomial, different from the orig. one, written in terms of $p$), and this is then treated by the second step I described which is relevant for the numerator. You seem eager to randomly permute the steps or otherwise damage the procedure I carefully sketched and then you seem to be surprised that yours doesn't work. But mine does. – Luboš Motl Jan 20 '12 at 14:54

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