Polchinski Equation (7.2.4) On page 209 of Polchinski's string theory book he writes down the expectation value of a product of vertex operators on the torus; equation $(7.2.4)$. The derivation is analogous to an earlier calculation on the sphere, equation $(6.2.17)$, and I'm perfectly happy with the result except for the factor of $2\pi/\partial_\nu \vartheta_1(\nu\vert\tau)$.
Can anyone give me an insight into how this term appears? Thanks.
EDIT: Following Lubos's answer.
The expectation value we wish to calculate is
\begin{align}
\Bigg< \prod_{i=1}^n :e^{ik_i \cdot X(z_i,\overline z_i)}:\Bigg>_{T^2} &= iC^X_{T_2}(\tau) (2\pi)^d \delta^d(\sum_i k_i) \\& \exp \Big(-\sum_{i<j} k_i \cdot k_j \, G'(w_i,w_j) - \frac{1}{2}\sum_i k_i^2 G_r'(w_i,w_i) \Big)
\end{align}
The second line follows just as in eq. $(6.2.17)$, and the Green functions are
$$ G'(w,w') = -\frac{\alpha'}{2} \ln \Bigg\vert \vartheta_1\Big(\frac{w-w'}{2\pi}\Big\vert \tau\Big) \Bigg\vert^2 + \alpha' \frac{[Im(w-w')]^2}{4\pi\tau_
2}$$
\begin{align}
G'_r(w,w)&=G'(w,w)+\alpha'\omega(w)+\frac{\alpha'}{2}\ln \vert w-w'\vert^2 \\&= -\frac{\alpha'}{2}\ln\Bigg\vert \frac{\partial_\nu\vartheta_1(0|\tau)}{2\pi} \Bigg\vert^2 +\alpha'\omega(w)
\end{align} 
Where we have used
$$ \left. \vartheta_1 \left( \frac{w-w'}{2\pi} | \tau  \right)\right|_{w\to w'} \to \partial_\nu\vartheta_1(0|\tau)\cdot \left(\frac{w-w'}{2\pi} \right) $$
as explained by Lubos. Substituting these into the original equation and taking the curvature to infinity $\omega\to 0$, we find
\begin{align}
\Bigg< \prod_{i=1}^n :e^{ik_i \cdot X(z_i,\overline z_i)}:\Bigg>_{T^2} &= iC^X_{T_2}(\tau) (2\pi)^d \delta^d(\sum_i k_i) \\& \times\prod_{i<j} \Bigg\vert \frac{2\pi}{\partial_\nu \vartheta_1(0\vert\tau)}\vartheta\Big(\frac{w_{ij}}{2\pi}\Big\vert\tau\Big)\exp\Big[-\frac{(Im w_{ij})^2}{4\pi\tau_2}\Big] \Bigg\vert^{\alpha' k_i \cdot k_j}
\end{align}
As in equation $(7.2.4)$.
 A: This extra factor arises from the analogy of the conformal factor $\alpha'\omega$ term in (6.2.16). The required $\omega$ is 
$$\omega = \ln \left (   \frac{2\pi}{\partial_\nu\vartheta_1}\right) $$
and substituting it to the exponential we get
$$ \exp\left( -\frac{\alpha'}{2}\sum_ik_i^2 \cdot \ln \frac{2\pi}{\partial_\nu\vartheta_1}  \right) = \left(\frac{2\pi}{\partial_\nu\vartheta_1} \right)^{-\alpha'  \sum_i k_i^2/2} =  \left(\frac{2\pi}{\partial_\nu\vartheta_1} \right)^{+\alpha'  \sum_{i\lt j} k_i k_j} $$
which gives exactly the factor whose origin you wanted to trace. As Joe says, this factor you asked about comes from "normalized self-contractions", which refers to the $\sum_i$ in the exponent, but because $\sum_i k_i = 0$ (and its inner-product square is zero, too), we may convert this $\sum_i$ to $\sum_{i\lt j}$ above.
The aforementioned required $\omega$ is determined as follows. For the self-contractions, the following factor we use for the contractions should be substituted with $w-w'=0$
$$ \vartheta_1 \left( \frac{w-w'}{2\pi} | \tau  \right) $$
but it vanishes at that point, so the value has to be computed by l'Hospital rule – or the first term from the Taylor expansion as a function of $w-w'$, if you wish:
$$ \left. \vartheta_1 \left( \frac{w-w'}{2\pi} | \tau  \right)\right|_{w\to w'} \to \partial_\nu\vartheta_1(0|\tau)\cdot \left(\frac{w-w'}{2\pi} \right) $$
This expression must coincide with the $\omega$-dependent "self-contraction" exponential from (6.2.17) which fixes the value of $\omega$. 
Effectively, if you got a result omitting the factor mentioned in the question, it is analogous to saying that $f(0)\to x$ if $f(0)=0$ but the right leading approximation is $f(0)\to f'(0) x$ for such functions.
