Vertex operators in string theory It is well known that operators in string theory are inserted at the origin. So, for example, the vertex operator for tachyon looks like
$$:e^{ik.X(0, 0)}:$$ However, while computing the string amplitude we use an integrated version of the vertex operator i.e. $$\int d^{2}\sigma \sqrt{det(g)} e^{ik.X}$$ This seems like a contradiction to me. Please explain why are these two equivalent ? 
 A: Under the state-operator mapping we have the following identification  $$   |k;0,0    \rangle \longleftrightarrow   :e^{ik.X(0, 0)}:$$ where $:e^{ik.X(0, 0)}:$ is the normally ordered closed string tachyon vertex operator and $   |k;0,0    \rangle $ is the closed string tachyon ground state. 
Now, why is $:e^{ik.X(0, 0)}:$ not used in computing string S-matrix elements? 
The answer is that any S-matrix element must be invariant under Weyl and diffeormphism transformations, something like $$S(k)= \sum_{compact \\ topologies} \int \frac{[dXdg]}{V_{diff \times Weyl}} exp(-S_{x}-\lambda\chi):e^{ik.X(0, 0)}: $$
 can't work for obvious reasons, namely that $S(k)$ is not a scalar under diff and Weyl symmetries because $:e^{ik.X(0, 0)}:$ is an operator of conformal dimensions $(\Delta=\frac{l_{S}^{2}k^{2}}{4} , \Delta=\frac{l_{S}^{2}k^{2}}{4})$. 
The solution is to integrate $:e^{ik.X(0, 0)}:$ over the relevant worldsheet as
$$\int d^{2}{\sigma} \sqrt{g} :e^{ikX}:.$$ Now is easy to see why this is the correct guess.
In locally flat coordinates we have the replacement $$\int d^{2}{\sigma} \sqrt{g} e^{ikX} \rightarrow \int d^{2}{z}  e^{ikX}$$ where we can turn $d^{2}{z}:e^{ikX}:$ into a tensor of type $(0,0)$ by noticing that $d^{2}{z}$ is of type $(1,1)$ and choosing $k^{2}=-m^{2}=\frac{4}{l_{S}}$. 
Sleeping beauty: Please notice how beautiful is the fact that the precise value of the mass of the tachyon is enforced by consistency.
We conclude that the correct expression for a stringy S-matrix element should be $$S(k)= \sum_{compact \\ topologies} \int \frac{[dXdg]}{V_{diff \times Weyl}} exp(-S_{x}-\lambda\chi)\prod_{a=1}^{n} \int d^{2}{\sigma} \sqrt{g} :e^{ik_{a}X_{a}}: $$ As desired.
