The task is to calculate the orientational average of a phase factor $<\mathrm{e}^{i\mathbf{Q}\cdot\mathbf{r}}>$, with $\mathbf{r}$ being a randomly oriented vector of constant length r.
To calculate the rotational average, I assume you do the integral \begin{equation} <\mathrm{r}^{i\mathbf{Q}\mathbf{r}}> = \frac{\int{\mathrm{e}^{i\mathbf{Q}\mathbf{r}}}\mathrm{sin}(\theta)d\theta d\phi}{4\pi}, \end{equation}
where according to my source the integral simplifies to
\begin{equation} \int{\mathrm{e}^{iQr\ \mathrm{cos}(\theta)}\mathrm{sin}(\theta)d\theta d\phi}, \end{equation}
which, of course, is easy to solve.
But my question corresponds to the dot product in the exponent, which, i guess, generally looks like \begin{equation} \mathbf{Q\cdot r} = Qr\left( \sin{\phi_q}\sin{\phi_r}\cos(\theta_q - \theta_r) + \cos{\phi_q}\cos{\phi_r}\right), \end{equation}
which, even if i choose $\phi_q = \theta_q = \pi/2$ leaves me with an integral of the form \begin{equation} \iint\mathrm{e}^{iQr\left( \sin{\phi\cos{(\phi+\pi/2)}}\right)}\sin{\theta}d\phi d\theta, \end{equation}
which looks different to the above case and in which case I am not able to solve the integral myself :(
My question is are the two integrals the same after all and I am missing some important simplification or am I doing something wrong from the beginning in assuming the particular shape of the dot product? Why does $\mathbf{Q \cdot r} = \mathrm{Q}\mathrm{r}\cos{\theta}$ take this form without any $\phi$-dependence??
Hope I made my case clear, thanks in advance.
Just orient $\vec Q$ along the z axis and then the integration over $\varphi$ becomes multiplication by $2 \pi$. The rest is trivial.
Using this particular orientation does not result in any loss of generality as the dot product is invariant to 3D rotations.