# Integral of the classical part of a compact boson

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I'm reading a book of conformal field theory (the one by Di Francesco, Pierre Mathieu and David Senechal), and I'm having trouble understanding the math in a section (p. 468)

The aim of the section is to compute the two-point function, on a plane, of a magnetic operator, for a compact boson.

The bosonic field is decomposed between a classical and a quantum part, the classical part having the following form :

$\varphi^{cl} = m R \ln\left( \frac{z-z_1}{z-z_2} \right)$

And the action ($S[\Phi] = \frac{-1}{8\pi} \int (\nabla \Phi)^2$, with their conventions) gives :

$S[ \tilde{\varphi} + \varphi^{\text{cl}}] = S[\tilde{\varphi}] + S[\varphi^{cl}] - \frac{1}{4\pi} \int \nabla \tilde{\varphi} \nabla \varphi^{cl}$

The third term is evaluated to zero, as $\Delta \varphi^{cl} = 0$. And the second one is given by :

$\frac{1}{|z_1 - z_2|^{\frac{mR^2}{2}}}$

The result is classic and I know it can be obtained differently, but I don't understand :

1. how that final integral is computed
2. why the third term gives zero asked Jan 21, 2016
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