Electronic-vibrational-rotational Transition I'm trying to simulate the fluorescent spectrum for the first time and run into several problems. The dipole matrix element for a transition between 2 different electronic states is as follow: 
$$\begin{aligned}
\boldsymbol{M}_{i k} &=\int \chi_{i}^{*}\left[\int \Phi_{i}^{*} \boldsymbol{p}_{\mathrm{el}} \Phi_{k} \mathrm{d} \tau_{\mathrm{el}}\right] \chi_{k} \mathrm{d} \tau_{\mathrm{N}} \\
&=\int \chi_{i}^{*} \boldsymbol{M}_{i k}^{\mathrm{el}}(R) \chi_{k} \mathrm{d} \tau_{\mathrm{N}}
\end{aligned}$$
The total molecular wave function is $\psi(\boldsymbol{r}, \boldsymbol{R})=\Phi(\boldsymbol{r}, \boldsymbol{R}) \times \chi(\boldsymbol{R})$ and the nuclear wave function is $\chi_{\mathrm{N}}=\psi_{\mathrm{vib}} \cdot Y(\vartheta, \varphi)$.
The matrix element becomes
$$\begin{array}{c}
\boldsymbol{M}_{i k}=\boldsymbol{M}_{i k}^{\mathrm{el}}\left(R_{\mathrm{e}}\right) \int \psi_{\mathrm{vib}}^{*}\left(v_{i}\right) \psi_{\mathrm{vib}}\left(v_{k}\right) \mathrm{d} R \\
\cdot \int Y_{J_{i}}^{M_{i}} \hat{\boldsymbol{p}}_{\mathrm{el}}(\vec{r}) Y_{J_{k}}^{M_{k}} \sin \vartheta \mathrm{d} \vartheta \mathrm{d} \varphi
\end{array}$$
Why does the electronic contribution to the molecular dipole moment reappears in the second integral? I thought when we take $\boldsymbol{M}_{i k}^{\mathrm{el}}$ out of the integral since it's an integral over electronic coordinates, what left should be the integral of $\psi_{\mathrm{vib}} \cdot Y(\vartheta, \varphi)$ 
$$\int Y_{J_{i}}^{M_{i}} Y_{J_{k}}^{M_{k}} \sin \vartheta \mathrm{d} \vartheta \mathrm{d} \varphi \cdot \int \psi_{\mathrm{vib}}^{*}\left(v_{i}\right) \psi_{\mathrm{vib}}\left(v_{k}\right) \mathrm{d} R$$
[What i think the matrix element should be]
The formulas are from "Atoms, Molecules and Photons
An Introduction to Atomic-, Molecular- and Quantum Physics(Wolfgang Demtröder), p.346"
 A: I believe the key fact is that the electronic transition dipole moment is a vector quantity, and the unit vector in its direction $\hat{p}$ is what is included in the angular integral. So the electronic contribution does not reappear in the second integral; this integral over a unit vector $\hat p$ in the direction of $\bf p$ can only be $\leq 1$. 
See equation 9.129 in the same textbook and the description that immediately follows for what I believe is an equivalent situation. 
Edit:
You make a good point in your comments and I wasn't clear enough before. The key fact is that ${\bf M}_{ik}$ is a vector quantity in the lab frame. This means its vector components can be multiplied by electric fields in the lab frame, for example laser fields whose polarization is best expressed in the lab frame. As you mentioned, ${\bf p}_{el}$ is a function of the electronic coordinates only, but only when expressed in the molecule-fixed frame. If you want to talk about the interaction of a laser field with the electrons of a rotating molecule, you will need to rotate the components of ${\bf p}_{el}$ into the lab frame. To do this rotation, you need a Wigner rotation matrix -- something like ${\mathcal D}_{M\Omega}^{(J)}(\vartheta,\varphi,0)^*$. The rotation matrix is a function of the nuclear orientation, and so when he writes $\hat p_{\rm el}$ in the second integral, he is suppressing this fact that $\hat p_{\rm el}$ is implicitly a function of the nuclear orientation $(\vartheta,\varphi)$. Demtroder appears to have swept this complexity under the rug, but all the gory detail is in the textbook by Brown & Carrington, Chapter 5. 
In the last sentence of the paragraph following Eq. (9.135b) in Demtroder, he very briefly alludes to this issue. He says:

Different from the nuclear part of the molecular dipole moment in
  Eq. (9.124), which is directed along the internuclear axis, the electronic part of the dipole moment can have any direction in the
  molecular coordinate system.  The transformation to the space fixed
  system is therefore more complicated and has to use a transformation
  matrix, which contains the Euler angles.

Here "the Euler angles" refers to those of the nuclei. 
