MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I wanted some insights about the integral in equation A.5 (page 19) in this paper,

  • What is the derivation of this?

  • Is there something more general from where this comes?

  • If the $\vec{k}-\vec{q}$ of the denominator were replaced by $\vec{k}+\vec{q}$ would the only change be an overall factor of $(-1)^d$?

  • One sees the Laurent series kind expansion of the Gamma function as stated in equation 3.8 (page 14). How can one use that to find the possibly similar pole structure in the RHS of the the integral in A.5?

share|cite|improve this question

To clarify, for those who have not looked at the reference, the integral identity in question is $$ \int_{\mathbb{R}^d} \frac{d^dq}{(q^2)^{\nu_1} ((k-q)^2)^{\nu_2}} = \frac{\Gamma(d/2-\nu_1)\Gamma(d/2-\nu_2)\Gamma(\nu_1+\nu_2-d/2)} {\Gamma(\nu_1)\Gamma(\nu_2)\Gamma(d-\nu_1-\nu_2)} \pi^{d/2} (k^2)^{d/2-\nu_1-\nu_2} , $$ where $q^2$ is standard Euclidean norm squared. I think this is not the standard way of deriving the answer, but the quickest way I see to get it is notice that this integral is a convolution, $$(2\pi)^d \frac{1}{(k^2)^{\nu_1}}*\frac{1}{(k^2)^{\nu_2}},$$ of two distributions and to use Fourier transforms to simplify it.

Basically, you need to take the Fourier transforms of $1/(q^2)^{\nu_1}$ and $1/(q^2)^{\nu_2}$ (that would be from momentum space to position space, in physics terminology), multiply them, and Fourier transform back. Luckily, the Fourier transforms of such distributions are well known. An explicit formula is given, for instance in formula 2.5 of the Fourier Transform Table Appendix of Gelfand & Shilov's Generalized Functions. Volume 1:

\begin{align} \mathcal{F}[|x|^\lambda] &= \frac{\Gamma(\lambda/2+d/2)}{\Gamma(-\lambda/2)} 2^{\lambda+d} \pi^{d/2} |q|^{-\lambda-d} , \\ \mathcal{F}^{-1}[|q|^\nu] &= \frac{\Gamma(\nu/2+d/2)}{\Gamma(-\nu/2)} 2^\nu \pi^{-d/2} |x|^{-\nu-d} , \end{align} where $|q| = \sqrt{q^2}$, and with the conventions $\mathcal{F}[\delta(x)] = 1$, $\mathcal{F}[1] = (2\pi)^d\delta(q)$. If $q^2$ is computed using the Lorentzian inner product instead of the Euclidean one, similar formulas can be found, but they are more subtle: see formulas 2.13--16 of the same table in Gelfand & Shilov.

Putting this together with the convolution formula, and the fact that the product of two powers of $|x|$ is a power of $|x|$, the above integral is then $$ (2\pi)^d \frac{1}{(k^2)^{\nu_1}} * \frac{1}{(k^2)^{\nu_2}} = \frac{\Gamma(d/2-\nu_1)\Gamma(d/2-\nu_2)\Gamma(\nu_1+\nu_2-d/2)} {\Gamma(\nu_1)\Gamma(\nu_2)\Gamma(d-\nu_1-\nu_2)} \pi^{d/2} \frac{1}{(k^2)^{\nu_1+\nu_2 - d/2}} , $$ which is the desired form. The arithmetic can easily be checked by hand or with a computer algebra system (as I did it).

As for your last question, which asks for the pole structure of the above expression as a function of $d$, you need only know that $\Gamma(\nu)$ is never zero for real $\nu$ and that it has simple poles at $\nu=0$ and all negative integers, while the functional identity $\Gamma(\nu)\Gamma(1-\nu) = \pi/\sin(\pi \nu)$ tells you what the residues are.

share|cite|improve this answer
"$d^d q$"? Ouch! – Noam D. Elkies Mar 22 '13 at 3:50
Alas, not my notation. – Igor Khavkine Mar 22 '13 at 4:25
Hmm, any comments to go along with the downvote? – Igor Khavkine Mar 22 '13 at 9:31
@Igor Khavkine Thanks for the reply. The issue about the poles that I am confused about is this - for generic values of $\nu_1$ and $\nu_2$ all the $4$ Gamma functions in the value of the integral which have a $d$ in the argument are going to get poles. So individually each of the $4$ Gamma functions has a pole. Now how does one multiply and divide such things? – curious Mar 22 '13 at 21:00
@Igor I don't have this book you refer to with me right now. Can you kindly write down the Fourier transform and the convolution steps? At least the steps and the results. (..I didn't get why you needed to Fourier transform to do the convolution..) – curious Mar 22 '13 at 21:03

Your Answer


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.