Wednesday, 28 January 2015

Critical Dimension of Bosonic Strings and Regularization of sumin=1nftyn


If D is critical dimension of Bosonic strings, a particular derivation goes like the following, where we arrive finally at D22n=1n+1=0.

Now mathematically this is clearly a divergent series, but using zeta function regularization here we are taking n=1n=ζ(1)=112.
And obtain D=26 where ζ is the analytic continuation of the zeta function we know. But it makes no sense in putting s=1 in the formulae ζ(s)=n=11ns.
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?




Answer



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 12! times the second Bernoulli number B2.




The limit limz1 of the sum


0+1z1+2z2+3z3+


diverges, because of the pole in


k=0kzk=zddzk=0zk=zddz11z=z(z1)2,z(0,1)


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


f(k):=k+1kf(k)dk.


for which kzk=zddzzk=zddzeklog(z)=zddzzklog(z)|k+1k.



Because of canceling upper and lower bounds, the sum nk=0kzk is z0log(z)2 plus terms suppressed by zn. Finally, using the expansion


1(log(1+r)/r)2=11r+(111!2!3!)r2+O(r3)=1+r+11!2!3!r2+O(r3),


we find


k=0(kzkkzk)=z(z1)21log(z)2=112+O((z1)1).




The picture shows the two functions z(z1)2 and 1log(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


11!2!3!=112=0.08˙3.


enter image description here


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