A question about the Weyl transformation for the vertex operator of the closed-string tachyon I met a problem of derving the Weyl transformation on the closed-string tachyon, Eq. (3.6.8) in Polchinski's string theory, vol 1, p 103.
Given the vertax operator of the closed-string tachyon

$$V_0 = 2g_c \int d^2 \sigma g^{1/2} e^{ik \cdot X}  \tag{3.6.1}$$ 

It is said the Weyl transformation of Eq. (3.6.1) is

$$\delta_W V_0 = 2g_c \int d^2 \sigma g^{1/2} \left( 2 \delta \omega( \sigma) - \frac{k^2}{2} \delta_W \Delta (\sigma, \sigma) \right) [ e^{ik \cdot X(\sigma)} ]_r \tag{3.6.8}$$

here renormalized operator $[]_r$ and $\Delta (\sigma, \sigma)$ are defined as

$$ [ \mathcal{F}]_r = \exp \left( \frac{1}{2} \int d^2 \sigma d^2 \sigma' \Delta(\sigma, \sigma') \frac{ \delta}{\delta X^{\mu}(\sigma)}  \frac{ \delta}{\delta X_{\mu}(\sigma')}  \right) \mathcal{F} \tag{3.6.5} $$ 
  $$ \Delta(\sigma,\sigma') = \frac{ \alpha'}{2} \ln d^2(\sigma,\sigma') \tag{3.6.6} $$ 
  where $d(\sigma,\sigma')$ is the geodesic distance between points $\sigma$ and $\sigma'$.

I can get the first term in the big parathesis $()$ of Eq. (3.6.8), namely consider 
$$ \delta g^{1/2} = \frac{1}{2} \frac{1}{g^{1/2}} \delta g= \frac{1}{2} \frac{1}{g^{1/2}} g g^{ab} \delta g_{ab} = \frac{1}{2} \frac{1}{g^{1/2}} g g^{ab} 2 \delta \omega g_{ab} = 2 g^{1/2} \delta \omega $$
I try to get the second term in the big parathesis of Eq. (3.6.8)
$$ \delta e^{i k \cdot X} = e^{i k \cdot X} i k^{ab} X^{ab} \delta g_{ab} = e^{i k \cdot X} i k^{ab} X^{ab} 2 \delta \omega g_{ab} = 2 i k \cdot X e^{ik \cdot X} \delta \omega,$$ 
but where do the $\Delta(\sigma,\sigma)$ and renormalized opeator $ [ e^{ik \cdot X(\sigma)} ]_r $ in (3.6.8) come from?
 A: I think it works like this : Looking at the Weyl dependence of renormalized operators:  
$$\delta_W[  \mathcal{F}]_r = [  \delta_W \mathcal{F}]_r + \frac{1}{2}   \int d^2 \sigma ~~d^2 \sigma' \delta_W \Delta(\sigma, \sigma') \frac{ \delta}{\delta X^{\mu}(\sigma)}  \frac{ \delta}{\delta X_{\mu}(\sigma')} [  \mathcal{F}]_r \tag{3.6.7}$$
We apply this formula to the operator $ \mathcal{F} = e^{i k.X(\sigma'')}$.
There is no explicit Weyl dependence of  $e^{i k.X(\sigma'')}$, so the first term $ [  \delta_W e^{i k.X(\sigma'')}]_r$ is zero.
The second term is, using the fact that we keep only the $\sigma = \sigma' =\sigma''$ terms because $e^{i k.X(\sigma'')}$ depends only of $\sigma''$:
$$\frac{1}{2} \delta_W \Delta(\sigma'', \sigma'') (-ik^\mu)(-ik_\mu) [   e^{i k.X(\sigma'')}]_r$$
So, finally :
$$\delta_W[ e^{i k.X(\sigma'')}]_r = -\frac{ k^2}{2} \delta_W \Delta(\sigma'', \sigma'')  [   e^{i k.X(\sigma'')}]_r \tag {1}$$
Now, we have : 
$$V_0 = 2g_c \int d^2 \sigma g^{1/2} [e^{ik \cdot X(\sigma)}]_r  \tag{2}$$
You already get the the Weyl dependence of $V_0$ relatively to $g^{\frac{1}{2}}$.
The Weyl dependence of $V_0$ relatively to $[e^{ik \cdot X(\sigma)}]_r$ is : 
$$\delta_W  V_0 =  2g_c \int d^2 \sigma g^{1/2} \delta_W[e^{ik \cdot X(\sigma)}]_r\tag{3}$$
So, finally, this dependence is, using $(1)$ : 
$$\delta_W  V_0 =  2g_c \int d^2 \sigma g^{1/2} (-\frac{ k^2}{2} \delta_W \Delta(\sigma, \sigma)  [   e^{i k.X(\sigma)}]_r)\tag{4}$$
Remark : The formula $(3.6.7)$ comes directly from the formula $(3.6.5)$ and could be interpreted like this: you have 2 terms, the first corresponds to an explicit Weyl dependence of the operator, and the second term to a Weyl dependence via $\Delta(\sigma, \sigma')$
