Following the derivation of the massless free-boson two-point function given in Di Francesco, Mathieu and Sènèchal, I had an apparently stupid doubt. Look at the attached picture.

Where does the contribution $\lim_{\rho \rightarrow 0} \rho K'(\rho)$to the integral in Eq. (2.100) go? Do they suppose it is zero? In this case Eq.(2.101) is not consistent with this requirement.

The only possible explanation I found is that the behaviour in $\rho\sim 0$ is not well-defined but I am not satified.

Any ideas??

enter image description here

  • $\begingroup$ What $\lim_{\rho\to0}$ are you talking about? I can see no limit there. $\endgroup$ – AccidentalFourierTransform Dec 4 '17 at 14:13
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    $\begingroup$ $\int_0^r d\rho\, \partial/\partial\rho(\rho K'(\rho)) = rK'(r) - \lim_{\rho\rightarrow 0} \rho K'(\rho)$ in Eq.(2.100) $\endgroup$ – apt45 Dec 4 '17 at 14:18
  • $\begingroup$ Please stop making trivial edits to bump the question into the front page. Thanks. $\endgroup$ – AccidentalFourierTransform Dec 4 '17 at 17:04
  • $\begingroup$ @AccidentalFourierTransform So, do you now see the limit? $\endgroup$ – apt45 Dec 4 '17 at 17:08
  • $\begingroup$ Yes. You do have a point. I don't know the answer off the top of my head. $\endgroup$ – AccidentalFourierTransform Dec 4 '17 at 17:10

This term is not there because you are integrating over a disk, and the disk has only one boundary at $r$, and no boundary at $0$. Here is how to see it more explicitly. Note that $$ \int_D -g\partial^2K(x,0)d^2x = -g\int_{\partial D}\nabla K(x,0) \centerdot dS = -2\pi g r K'(r). $$ Then we see that if $K(r)$ has a $\log$-singularity near $r=0$, then $\partial^2K$ reproduces the delta-function, while $m^2 K$ is integrable and thus does not contribute to the delta-function. Overall, we conclude $$ 1=\int_D g(-\partial^2+m^2)K(x,0)d^2x = -g\int_{\partial D}\nabla K(x,0) \centerdot dS + \int_D g m^2 K(x,0)d^2x= 2\pi g\left\{ -r K'(r)+m^2\int_0^r d\rho\,\rho K(\rho)\right\}. $$

  • $\begingroup$ You can also see that the limit you are asking about is actually related to the coefficient of the delta-function, and thus if you add it, as you suggest, you would have to put $0$ in the left hand side, instead of $1$. $\endgroup$ – Peter Kravchuk Dec 5 '17 at 7:39
  • $\begingroup$ What if instead, I take as integration domain the set $B_{r}\setminus B_{t}$ where $B_{r,t}$ are the balls centered in $x$ with radius $r,t$ respectively with $r > t$? I would like to recover the same result for $t\rightarrow 0$. I guess the RHS is now 0 even in the limit $t\rightarrow 0$. Now the contribution of the integration in the second boundary $\rho=t$ contribuites, right? $\endgroup$ – apt45 Dec 5 '17 at 9:11
  • $\begingroup$ @apt45, yes, the second boundary will contribute, but the delta function won't, even in the limit, so you'll just get a different equation. But you can easily check that the two are consistent. $\endgroup$ – Peter Kravchuk Dec 5 '17 at 15:39

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