# partition function on annulus

I'm considering a two dimensional Eucilidian massless free boson theory defined on an annulus. i.e.

$$S=\int d^2r\sqrt{g}(\nabla{\phi})^2$$

This is a CFT with c=1. Denote the inner radius of the annulus as $a$ and the outer radius as $b$. The two boundary conditions on the two circles are all Direclet: $$\phi(r=a)=\phi(r=b)=0$$ I want to know how does the following partition function (path-integral) scale with $a$ and $b$? $$Z(a,b)=\int_{\phi(a)=0}^{\phi(b)=0} D[\phi]e^{-S[\phi]}$$

First, make a scale transformation to recale the two radius $(a,b)$ to $(1,\frac{b}{a})$. As this is a CFT, the partion function should be invariant under this transformation up to the anomaly contribution. However, the anomaly is proportional to the integral of geodesic curvature $k$ of the two boundaries. $$\int_{\partial} ds \sqrt{g} k$$ and this integral is proportional to the Euler number of the annulus (since the bulk integral $\int d^2r \sqrt{g} R=0$), which is zero. So we have $$Z(a,b)=Z(1,\frac{b}{a})$$ Then I replace the inner boundary by an operator $O(0)$ inserted at origin. $$Z(1,\frac{b}{a})=\int ^{\phi(r=\frac{b}{a})=0} D[\phi]e^{-S[\phi]}O(0)$$
Here I have used the correspondence between conformal boundary condition and boundary state as well as the state and operator correspondence. Now the LHS is a path integral on an annulus with two boundaries, and the RHS is a path integral on a disk with an outer boundary as well as an operator inserted at origin. Finally, I apply a scale transformation to the RHS to recale $\frac{b}{a}$ to $b$. Now the path integral is no longer invariant. Write $O(0)$ as combination of operators with well defined scaling dimension: $$O(0)=\sum_if_iO_i(0)$$ Then after the rescaling, $O(0)$ becomes $O(0)'=\sum_if_ia^{\Delta_i}O_i(0)$. The anomaly in this time is proportional to the Ruler number of the disk times central charge. Combining this two effect, we should have
$$Z(a,b)=\int ^{\phi(r=\frac{b}{a})=0} D[\phi]e^{-S[\phi]}O(0)=a^{-\frac{1}{3}c}\int^{\phi(r=b)=0}D[\phi]e^{-S[\phi]}\sum_if_ia^{h_i}O_i(0)$$
From this expression, we can see that as $a\rightarrow 0$, $\sum_if_ia^{h_i}O_i(0)$ is dominated by the operator with $h_i=0$ and can be replaced by $f_II$. The problem is that $a^{-\frac{1}{3}c}$ seems to induce an infinity in this limit. Does anyone know what is the reason of this divergence or where I was wrong in my deduction?