# Where do $X_0^{-d}$ and $G'_r$ come from in Polchinski Eq. (6.2.13)?

I have a question about deriving Eq. (6.2.13) in Polchinski's string theory book volume I. It is claimed that

Now consider the path integral with a product of tachyon vertex operators, $$A_{S_{2}}^{n}(k,\sigma)=\left\langle [e^{ik_{1}\cdot X(\sigma_{1})}]_{r}[e^{ik_{2}\cdot X(\sigma_{2})}]_{r}\cdots[e^{ik_{n}\cdot X(\sigma_{n})}]_{r}\right\rangle _{S_{2}}\tag{6.2.11}$$ This corresponds to $$J(\sigma)=\sum_{i=1}^{n}k_{i}\delta^{2}(\sigma-\sigma_{i})\tag{6.2.12}$$ The amplitude (6.2.6) then becomes $$\begin{multline}A_{S_{2}}^{n}(k,\sigma)=iC_{S_{2}}^{X}(2\pi)^{d}\delta^{d}\biggl(\sum_{i}k_{i}\biggr)\\\times\exp\biggl(-\sum_{i,j=1;i<j}^{n}k_{i}\cdot k_{j}G'(\sigma_{i},\sigma_{j})-\frac{1}{2}\sum_{i=1}^{n}k_{i}^{2}G_{r}'(\sigma_{i},\sigma_{i})\biggr)\qquad\end{multline}\tag{6.2.13}$$ where $C_{S_{2}}^{X}=X_{0}^{-d}(\det'\frac{-\nabla^{2}}{4\pi^{2}\alpha'})_{S_{2}}^{-d/2}$ and $G_{r}'(\sigma,\sigma')=G'(\sigma,\sigma')+\frac{\alpha'}{2}\ln d^{2}(\sigma,\sigma')$

Eq. (6.2.6) is

$$\begin{multline}Z[J]=i(2\pi)^{d}\delta^{d}(J_{0})\biggl(\det'\frac{-\nabla^{2}}{4\pi^{2}\alpha'}\biggr)^{-d/2}\\\times\exp\biggl(-\frac{1}{2}\int d^{2}\sigma d^{2}\sigma'J(\sigma)\cdot J(\sigma')G'(\sigma,\sigma')\biggr)\end{multline}\tag{6.2.6}$$

My question is: where do $X_0^{-d}$ and $G_r'$ come from in Eq. (6.2.13)? I could try to plug (6.2.12) into (6.2.6) to see all other term appears, but not $X_0^{-d}$ nor $G_r'$.

The renormalization $[...]_r$ of the exponentials is defined in equation $(3.6.5)$

$$[\mathcal{F}]_r = \exp \left( \frac{1}{2} \int d^2\sigma d^2\sigma' \Delta(\sigma,\sigma')\frac{\delta}{\delta X^\mu(\sigma)}\frac{\delta}{\delta X_\mu(\sigma')}\right) \mathcal{F}$$

Here

$$\Delta(\sigma,\sigma') =\frac{\alpha'}{2} \ln d^2(\sigma,\sigma')$$

where $d(\sigma,\sigma')$ is the geodesic distance between points $\sigma$ and $\sigma'$. So,

$$[e^{ik \cdot X}]_r = \exp \left(- \frac{1}{2} \int d^2\sigma d^2\sigma' \frac{\alpha'}{2}k^2 \ln d^2(\sigma,\sigma') \right) e^{ik \cdot X}$$

This is the origin of the last term in $G'_r$. The $X_0^{-d}$ term comes from the delta function $\delta^d(J_0)$, where

$$J_0^\mu = \int d^2\sigma J^\mu(\sigma) X_0$$

($X_0$ is a constant $(6.2.5)$). Plugging in $J(\sigma)$ this becomes

\begin{align} J_0^\mu &= \int d^2\sigma \sum_{i=1}^n k_i^\mu\delta^2(\sigma-\sigma_i) X_0 \\&= X_0\sum_{i=1}^n k_i^\mu \end{align}

Using the formula

$$\delta[g(x)]=\sum_{i}\frac{\delta(x-x_i)}{g'(x_i)}$$

where $g'(x)$ is the x-derivative and the sum is over the roots of the function $g(x)$, the delta function becomes $$\delta^d\left(X_0 \sum_i k_i\right) = X_0^{-d} \delta^d \left(\sum_i k_i\right)$$