Scattering amplitude of open + closed strings and D-Branes When calculating the scattering amplitude of $n$ open string tachyons and $m$ closed string tachyons on the disk, I'd like to understand why choosing the open string tachyons to be attached to $D_{25}$ - branes simplifies the formula of the correlator of the $X$ - sector to the following:
\begin{align}
\langle \prod_{i=1}^{m}:e^{ik_iX(z_i,\bar{z}_i)} : \prod_{j=1}^{n}:e^{ip_jX(x_j)}: \rangle_{D^2} &= (2\pi )^{26} \delta^{(26)} \left(\sum_{i} k_i
+ \sum_{j} p_j\right)\\\\
&\times \prod_{i<j} |z_i-z_j|^{\alpha 'k_i\cdot k_j} |z_i-\bar{z}_j|^{\alpha 'k_i\cdot k_j}\\\\&\times \prod_u |z_u-\bar{z}_u|^{\alpha 'k_{u,\parallel}^2} \prod_{v<w} |x_v-x_w|^{2\alpha 'p_v\cdot p_w}\\\\
&\times \prod_{r,s} |x_r-z_s|^{\alpha ' p_r\cdot k_s} |x_r-\bar{z}_s|^{\alpha ' p_r\cdot k_s}
\end{align}
where the $k_{u,\parallel}$ denotes the momenta of the closed strings parallel to the $D_{25}$ - brane.
 A: The general philosophy to simplify scattering computations involving both open and closed string vertex operators is to insert the first ones at the worldsheet boundary.
Notice that this philosophy is physically reasonable. Open strings are the D-brane low energy degrees of freedom, performing computations  as if the open string operators were attached to the worldsheet boundary (locii where D-branes are localized) simplify things because, the latter statement is classically true. At the end, this logic makes manifest the fact that the interior CFT operators are independent of the boundary ones.
How does that works in the present case? Consider the scalar Green function for the Poisson's equation with Neumann boundary conditions on a disk, $$\langle X^{\mu}(z_{1},\bar{z}_{1})X^{\nu}(z_{2},\bar{z}_{2}) \rangle_{D_{2}}= -\frac{{\alpha}^{´}}{2}ln|{z_{1}-z_{2}}|^{2} -\frac{{\alpha}^{´}}{2}ln|{\bar{z}_{1}-\bar{z}_{2}}|^{2}.$$
Recall the ordinary normal ordering prescription for scalars $X^{\mu}(z_{1},\bar{z}_{1})$ and $X^{\nu}(z_{2},\bar{z}_{2})$ $$:X^{\mu}(z_{1},\bar{z}_{1})X^{\nu}(z_{2},\bar{z}_{2}):=X^{\mu}(z_{1},\bar{z}_{1})X^{\nu}(z_{2},\bar{z}_{2}) \ + \ \frac{{\alpha}^{´}}{2}ln|{{z}_{1}-z_{2}}|^{2}.  $$
The problem is that any correlator of open string vertex operators computed in the disk has divengences when the instertion points approach the boundary (even if the inserted operators were normal ordered).
Example one: Recall that the disk is biholomorphic to the complex upper half-plane $H_{2}$ and consider $\langle X^{\mu}(z_{1},\bar{z}_{1})X^{\nu}(z_{2},\bar{z}_{2}) \rangle_{H_{2}}$ in the limit at which $z_{1}$ is a real number $y$, $$\langle :X^{\mu}(y)X^{\nu}(z_{2},\bar{z}_{2}): \rangle_{H_{2}}= -\frac{{\alpha}^{´}}{2}ln|{y-z_{2}}|^{2},$$ a singular expression if $y \rightarrow z_{2}$ and despite the fact that $:X^{\mu}(y)X^{\nu}(z_{2},\bar{z}_{2}):$ was normally ordered.
Example two: The correlator of $n$ open string tachyon operators \begin{align}
\langle \prod_{i=1}^{m}:e^{ik_iX(z_i,\bar{z}_i)} : \rangle_{H_{2}} \ \propto \ \delta^{26}\left(\sum_{i} k_i\right)\
&\times \prod_{i=1}^{m} |z_i-z_j|^{\alpha 'k_i\cdot k_i / 2}
\times \prod_{i,j=1 \\ i<j}^{m} |z_i-z_j|^{\alpha 'k_i\cdot k_i}\ |z_i-\bar{z}_j|^{\alpha 'k_i\cdot k_i},
\end{align}
Here the dangerous term is $\prod_{i=1}^{m} |z_i-z_j|^{\alpha 'k_i\cdot k_i / 2}$ in the same limit as in example one.
How to the above problem can be fixed?
Step one: Read the solution to problem 2.10 of Polchinski, Vol. 1 were boundary normal ordering is introduced. This prescription is nothing else that an instruction to subtract the divergences that appear when the points on what vertex operators depend on approach the boundary. But it would be beneful if you learn how to do this systematically by using boundary normal ordering for open string operators every time you insert them into correlators.
Step two: Compute \begin{align}
\langle \prod_{i=1}^{m}:e^{ik_iX(z_i,\bar{z}_i)} : \prod_{j=1}^{n}:e^{ip_jX(x_j)}: \rangle_{H^2}
\end{align} but now with the operator $: \prod_{j=1}^{n}:e^{ip_jX(x_j)}:$ normally ordered according to the boundary normal ordering prescription.
Good luck and if you have some trouble with the computation feel free to ask for details.
