In using dimensional regularization in QFT calculations, one comes across integrals over propagators, they might look like $(d = \text{dimension of spacetime}, n = \text{a number})$

$$\tag{1}I(d,n)=\int \frac{\mathrm{d}^d k}{(2\pi)^d}\frac{1}{\big[k^2-\Delta\big]^n}$$ where one can consider the integral to be a function of the spacetime dimension $d$ which here need not be an integer. Now there's a formula for the integral $(1)$, which is given by (see e.g. Appendix of Peskin and Schroeder) \begin{align}\tag{2}I(d,n)&=\int \frac{\mathrm{d}^d k}{(2\pi)^d}\frac{1}{\big[k^2-\Delta\big]^n} \\ &=\frac{(-1)^n\mathrm{i}}{(4\pi)^{d/2}}\frac{\Gamma(n-d/2)}{\Gamma(n)}\left(\frac{1}{\Delta}\right)^{n-d/2}. \end{align}

My question is why does the integral $$I(d,0) = 0.$$

Expanding for small $n$ we get $$\frac{i 2^{-d} \pi ^{-d/2} (-1)^n \left(\frac{1}{\Delta }\right)^{n-\frac{d}{2}} \Gamma \left(n-\frac{d}{2}\right)}{\Gamma (n)} = i 2^{-d} \pi ^{-d/2} \left(\frac{1}{\Delta }\right)^{-d/2} \Gamma \left(-\frac{d}{2}\right)n+O\left(n^2\right)$$

Which is proportional to $n$ and which according to mathematica is equation to zero in the $n\rightarrow 0$ limit. So that

$$\tag{3}\int \frac{\mathrm{d}^d k}{(2\pi)^d} = 0~\text{(in dimreg)}.$$

This reminds me of that thing in string theory $$1+2+3+4+\cdots = -\frac{1}{12}.$$

So where/what in the derivation of the dimreg result equation $(2)$ does the assumption(s) come into the picture making equation (3) possible? Because $(3)$ looks like the volume of spacetime which should be infinite, all this is really strange.

  • 2
    $\begingroup$ As such equation (2) is only true when $n > \frac{d}{2}$ since otherwise the integral is divergent (proportional to the volume of spacetime, as expected). However, once we have the finite answer, we analytically continue (2) to all $n$. Under this analytic continuation, the value of the function as $n \to 0$ is 0. $\endgroup$
    – Prahar
    Nov 21, 2014 at 17:46
  • $\begingroup$ For your information, the article of Gorishnii and Isaev "AN APPROACH TO THE CALCULATION OF MASSLESS FEYNMAN INTEGRALS OF MANY-LOOP" offers an extension of the result you are trying to understand when $d =n/2$. i.e when the integral has (energy) dimension 0. $\endgroup$
    – Frédéric
    Nov 21, 2014 at 19:54

2 Answers 2


As Prahar said in a comment, dim. reg. allows us to see that two functions are equal in a half-plane, and if one of them is analytic in the whole plane, the other function may be analytically continued as well.

So the volume of the spacetime is indeed zero in dimensional regularization. More generally, any power law divergence is set to zero in dimensional regularization. Only logarithmic divergences may be "genuine".

This result, zero, is totally compatible with the values assigned by the zeta regularization. Note that $$ \sum_{n=1}^\infty n = \zeta(-1)= -\frac{1}{12} $$ but the simpler version you want here is $$ \sum_{n=1}^\infty 1 = \zeta(0)= -\frac{1}{2} $$ The latter relationship is self-consistent because it implies $$ \sum_{n=-\infty}^\infty 1 = 1 + 2 \sum_{n=1}^\infty 1 = 1+ 2\times (-1/2) = 0 $$ I've divided the sum over all integers to the sum over positive ones (which is $-1/2$), the negative ones (also $-1/2$), and an additional $1$ from $n=0$. The sum of these three parts cancels which is no coincidence.

The integral may be written as $$ \int_{-\infty}^\infty dx\,1 = \lim_{\epsilon\to 0} \sum_{n=-\infty}^{\infty} \epsilon = 0\epsilon = 0 $$ where I used the substitution $x=n\epsilon$ and a Riemann definition of the integral. In fact, because the already vanishing result from the summation was multiplied by another $\epsilon$ from the $dx$, we got a "stronger zero" than needed. For that reason, the continuous "integral" version would be zero even if we had things like the integral over $x$ from zero, and even if the integrand were a positive power of $x$. The minus one twelfth would get multiplied by $\epsilon^2\to 0$. You may check it explicitly.

  • $\begingroup$ Some very nice examples. $\endgroup$ Nov 22, 2014 at 0:43
  • 1
    $\begingroup$ Manipulation of a non-absolute convergent series is not a valid method to show some identity, since one can manipulate it in any way to get any result. Analytic continuation is a fine methode but still one shouldn't write $\sum n = -1/12$. The analytic continuation of the zeta function is unique but may have nothing in common with the defintion for convergent values on the non-convergent domain. $\endgroup$
    – image357
    Apr 25, 2015 at 13:26
  • $\begingroup$ Dear Marcel, when doing physics and related forms of calculus, the sum isn't interpreted as a limit of partial sums but in a more general, natural, and physical way, and it is often literally written to be equal to the constant for a very good reason. Divergent sums and integrals appear everywhere in quantum field theory (physics) but that doesn't mean that one can assign them with any value he wants. They have to be regularized, renormalized, and the final result may depend on some parameters. In $\sum n$ the dependence goes away and the finite part of the sum is always equal to $-1/12$. $\endgroup$ Apr 25, 2015 at 17:09

using zeta regularization and Euler maclaurin series the integral $ \int_{0}^{\infty}x^{m} $ is not zero but is related to $ \zeta (-m) $ see my paper http://vixra.org/abs/1009.0047 adn in particular $\int_{0}^{\infty}dx=1+\zeta (0) 4$ by euler maclaurin formula


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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