Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

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 http://www.maths.tcd.ie/~cblair/notes/list.pdf, 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.

share|improve this question
add comment

1 Answer 1

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.

share|improve this answer
    
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.