If $D$ is critical dimension of Bosonic strings, a particular derivation goes like the following, where we arrive finally at $$ \frac{D-2}{2}\sum_{n=1}^\infty n + 1 = 0. $$ Now mathematically this is clearly a divergent series, but using zeta function regularization here we are taking $$ \sum_{n=1}^\infty n = \zeta(-1) = -\frac{1}{12}. $$ And obtain $ D = 26 $ where $\zeta $ is the analytic continuation of the zeta function we know. But it makes no sense in putting $ s = -1 $ in the formulae $$ \zeta(s) = \sum_{n=1}^\infty \frac{1}{n^s}. $$ As the above is only valid for $ Re(s) > 1 $. So what is going on in here? Can anyone give me a reasonable explanation about obtaining $ -1/12 $?

  • 3
    $\begingroup$ Related: physics.stackexchange.com/q/4994 $\endgroup$
    – twistor59
    Commented May 29, 2013 at 14:46
  • $\begingroup$ @twistor59 I have seen that, but it doesn't seem to answer my question $\endgroup$
    – smiley06
    Commented May 29, 2013 at 14:49
  • 4
    $\begingroup$ This particular sum is also discussed here and here, and on Math.SE here. See also this Phys.SE post. Also related Phys.SE post here. $\endgroup$
    – Qmechanic
    Commented May 29, 2013 at 15:09
  • 2
    $\begingroup$ There is a mention of this sum and its regularization in page 39 here - damtp.cam.ac.uk/user/tong/string/two.pdf. The statement is: Quantization of the string action requires going to $d=2+\epsilon$, to get rid of the ultraviolet divergences. This introduces a cosmological constant term in the action. However, this breaks Weyl invariance. In order to restore this Weyl invariance, one must introduce a counterterm. It is this counterterm that precisely cancels the divergence in the sum above. I have never worked this out myself, so if my understanding is wrong, please correct me. $\endgroup$
    – Prahar
    Commented May 29, 2013 at 16:12
  • $\begingroup$ I understand the mathematical part saying $\zeta(-1) = -1/12 $ but still I am pretty much in the dark about getting rid of such infinite terms in context of physics, can anyone provide some details perhaps in an answer so that a student of mathematical background with a knowledge of basic QM and QFT can understand (if it is possible of course to give such an explanation) ? $\endgroup$
    – smiley06
    Commented May 30, 2013 at 7:48

2 Answers 2


A way to do this is using regularization by substracting a continuous integral, ,with the help of the Euler-MacLaurin formula:

You can write :

$$ \sum_{Regularized} =(\sum_{n=0}^{+\infty}f(n) - \int_0^{+\infty} f(t) \,dt) = \frac{1}{2}(f(\infty) + f(0)) + \sum_{k=1}^{+\infty} \frac{B_k}{k!} (f^{(k - 1)} (\infty) - f^{(k - 1)} (0))$$ where $B_k$ are the Bernoulli numbers.

With the function $f(t) = te^{-\epsilon t}$, with $\epsilon > 0$, you have $f^{(k)}(\infty) = 0$ and $f(0) = 0$, so with the limit $\epsilon \rightarrow 0$, you will find : $$\sum_{Regularized} = - \frac{B_1}{1!} f (0) - \frac{B_2}{2!} f' (0) = - \frac{1}{12}$$

because $f(0) = 0$ and $B_2 = \frac{1}{6}$

  • 2
    $\begingroup$ But we have the Hamiltonian as $$ H = \frac{1}{2}\sum_{n=1}^\infty \alpha_{-n}\alpha_n $$ And then replacing $\alpha_n$ with terms having $\sqrt{n} $ we arrive at $$ \frac{D-2}{2}\sum_{n=1}^\infty n $$ Now it seems to be the normal $\sum_n $ and not $\sum_{Regularised} $ in the expression of Hamiltonian because that is further obtained by using normal Fourier series $$ X^\mu(\tau,\sigma) = X^\mu_{COM} + \sum_{m\neq 0} \frac{1}{m}\alpha^\mu_m e^{-im\tau}\cos(m\sigma) $$ in the expression $$ H = \frac{T}{2}\int_0^\pi (\dot{X}^2+ X'^2)d\sigma $$ So how do we have $ \sum_{Regularised} $ above ? $\endgroup$
    – smiley06
    Commented May 30, 2013 at 19:17
  • $\begingroup$ Am I mistaken somewhere above in the derivations ? $\endgroup$
    – smiley06
    Commented May 30, 2013 at 19:25
  • $\begingroup$ First, I made an error on the last line, it is $f(0) =0$ and not $B_1= 0$ !!! See the edit. $\endgroup$
    – Trimok
    Commented May 31, 2013 at 9:20
  • 2
    $\begingroup$ You cannot escape some kind of regularization. The best you can do, is to find a acceptable physical explanation, for this regularization. The same kind of calculus is used, for instance, in the 1-dimensional Casimir effect. The modes between the plates are quantized, and they are not quantized outside the plates, so, to have the net result, you must make the difference of the continous modes and the discrete modes. So you can have a kind of analogy here, sometimes people speak of Casimir Energy even in the string domain. $\endgroup$
    – Trimok
    Commented May 31, 2013 at 9:29
  • $\begingroup$ Maybe a way to see that, is to consider a kind of infinite string, where only a finite part (a segment) is concerned by discrete modes. If you accept this, it will be an analogy with the 1-dimensional Casimir effect. Of course, we have to be cautious about the limit of these analogies, but I have nothing more clever. Just an additional remark, fermionic and bosonic calculus are different, in original Casimir effect, as in string theory. $\endgroup$
    – Trimok
    Commented May 31, 2013 at 9:33

I know some derivations in which one can track the emergence of the concrete value, without having to buy that the second order contribution in the Euler-MacLaurin formula (see other answer) is $-\frac{1}{2!}$ times the second Bernoulli number $B_2$.

The limit $\lim_{z\to 1}$ of the sum


diverges, because of the pole in

$\sum_{k=0}^\infty k\,z^k=z\frac{{\mathrm d}}{{\mathrm d}z}\sum_{k=0}^\infty z^k=z\frac{{\mathrm d}}{{\mathrm d}z}\frac{1}{1-z}=\frac{z}{(z-1)^2}, \hspace{1cm} z\in(0,1)$

We are instead going to consider the sum of smooth deviations of the above, using the local mean

$\langle f(k)\rangle:=\int_{k}^{k+1}f(k')\,{\mathrm d}k'$.

for which $\langle k\,z^k\rangle=z\frac{{\mathrm d}}{{\mathrm d}z}\langle z^k\rangle=z\frac{{\mathrm d}}{{\mathrm d}z}\langle {\mathrm e}^{k \log(z)}\rangle=z\frac{{\mathrm d}}{{\mathrm d}z}\frac{z^{k'}}{\log(z)}\left|_{k}^{k+1}\right.$.

Because of canceling upper and lower bounds, the sum $\sum_{k=0}^n\langle k\,z^k\rangle$ is $\frac{z^0}{\log(z)^2}$ plus terms suppressed by $z^n$. Finally, using the expansion


we find

$\sum_{k=0}^\infty \left(k\,z^k-\langle k\,z^k\rangle\right)=\dfrac{z}{(z-1)^2}-\dfrac{1}{\log(z)^2}=-\dfrac{1}{12}+{\mathcal O}\left((z-1)^1\right).$

The picture shows the two functions $\dfrac{z}{(z-1)^2}$ and $\dfrac{1}{\log(z)^2}$, as well as their difference (blue, red, yellow). While the functions themselves clearly have a pole at $z=1$, their difference converges against

$$-\frac{1}{1!\,2!\,3!}=-\frac{1}{12}=-0.08{\dot 3}.$$

enter image description here

  • $\begingroup$ This is better @NikolajK Thanx :) $\endgroup$
    – smiley06
    Commented Jun 18, 2015 at 16:17
  • $\begingroup$ @smiley06: I have a bunch of other perspectives too. It's interesting to point out that the $\frac{1}{12}$ here is "the same" as the one in Baker–Campbell–Hausdorff formula or the Todd class. $\endgroup$
    – Nikolaj-K
    Commented Jun 18, 2015 at 17:02
  • $\begingroup$ @smiley06: Or say you want to compare the difference $\Delta_h f(x)=f(x+h)-f(x)$ with it's first order approximation $f'(x)\,h$. You'll find $$\dfrac{f'(x)\,h}{\Delta_h f(x)}=1-\dfrac{f''(x)}{2!}\left(\dfrac{h}{f'(x)}\right)+\left(\dfrac{f''(x)\,f''(x)}{2!\,2!}-\dfrac{f'(x)\,f'''(x)}{1!\,3!}\right)\left(\dfrac{h}{f'(x)}\right)^2+{\mathcal O}(h^3).$$ Note that $\frac{1}{2!\,2!}-\frac{1}{1!\,3!}=\frac{1}{1!\,2!\,3!}(3-2)=\frac{1}{12}$! :) $\endgroup$
    – Nikolaj-K
    Commented Jun 18, 2015 at 17:05
  • $\begingroup$ @smiley06: In that light, it's not so surprising that it pops in connection to the Riemann zeta function, which has the integral representation $\zeta(s) =\frac{1}{\Gamma(s)}\int_0^\infty\frac{x^{s}}{e^x-1}\frac{{\mathrm d}x}{x}$. And now take another look at Planck's law $B_\nu(\nu, T) = \frac{ 2 h \nu^3}{c^2} \frac{1}{e^\frac{h\nu}{k_\mathrm{B}T} - 1}$. I can go on.. $\endgroup$
    – Nikolaj-K
    Commented Jun 18, 2015 at 17:13
  • $\begingroup$ @smiley06: Ah, here's another factoid: The sum was exploding as $\frac{1}{(1-z)^2}$, and the finite limit emerges after subtracting $\frac{1}{\log(z)^2}$. More broadly, $$\dfrac{1}{\log(z)^n}=\dfrac{1}{(z-1)^n} \left(1+ \frac{n}{2} (z-1)+ \frac{n}{2} \frac{3n-5}{12} (z-1)^2+\frac{n}{2}\frac{(n-2)(n-3)}{24}(z-1)^3+\dots\right)$$ and use may use this to produce limits for high powers too. Above, for $n=2$, you got $-\frac{2}{2}\frac{3\cdot 2-5}{12}=-\frac{1}{12}$. Or plug in $n=1$ and you find $-\frac{1}{2}$. $\endgroup$
    – Nikolaj-K
    Commented Jun 19, 2015 at 9:02

Your Answer

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

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