How exactly do I calculate this correlation function? I found a research paper (from 1977) that has a particular equation I need to reproduce. The paper essential calculates dynamic light scattering correlation functions. 
The full equations I need to work out is this:
$$
f(q,\tau) = \frac{\int_{-1}^{+1}(1-2w^2\nu^2) \exp(-w^2\nu^2)\ d\nu \times \int_{0}^{2\pi} A[q,\mu(t)]A^*[q,\mu(0)]\ d\psi}{\int_{0}^{2\pi} |A[q,\mu(0)]|^2\ d\psi}
$$
Essentially I need to calculate this for values of the delay time $t$. The bit I am not sure how to do is the  $\int_{0}^{2\pi} A[q,\mu(t)]A^*[q,\mu(0)]\ d\psi$. The actual functional form of A isn't that important, but it consists of spherical Bessel functions and the term $\psi$ which is to be numerically integrated over. 
Now, where I am stuck is what do I actually do to calculate that function for the delay times. I have tried a lot of things but I just cant get it looking like they have in the paper. Here is what I have tried:


*

*Calculate $\int_{0}^{2\pi} A[q,\mu(t)]$ for all delay times, and then use a built in autocorrelation function to obtain the results.

*Take my first $t$ value (say 0.01 sec), and my second $t$ value (say 0.02 sec) and then calculate $\int_{0}^{2\pi} A[q,\mu(0.02)]A^*[q,\mu(0.01)]\ d\psi$... etc for all t values (so $t0$ lags 1 behind t.

*A few other things.
As you can see I am a bit confused, if anyone has any insight as to how I would calculate this by hand that would be extremely helpful.

Edit:
The full functional form of $A(q,\mu(t))$ is 
$$
A(q,\mu(t))\propto \frac{3j_1(u)}{u} + F^3 \frac{m_2-m_1}{m_1-1}\frac{3j_1(f u)}{fu}
%Old version: \frac{F^3 (m_2-m_1) (3 j_1(F \mu(t)))}{(m_1-1) (F \mu(t))}+\frac{3 j_1(\mu(t))}{\mu(t)}
$$
where $j_1$ is a spherical Bessel function of the first kind, $u=q\sqrt{a^2\mu^2(t)+b^2(1-\mu^2(t))}$, $F=1-\frac ta$, and $\mu(t)=\cos(\psi)\sin(\theta)\sin(\alpha(t))+\cos(\theta)\cos(\alpha(t))$, for $\nu=\cos(\theta)$ and $\alpha(t)=\omega t+\phi$.
Yes, I am after $f$ vs $t$ for a $q$ of my choice. I am just not sure I am computing the integral properly for the values of $t$.
For example, lets say my $t$ of interest was $0.1, 0.2,0.4, 0.8, 1.6$ (just an example).
For the very first $t$, would I simply do, 
$$
\int_{0}^{2\pi}A[q,\mu(0.2)]\times A[q,\mu(0.1)]
$$
Then for the next $t$
$$
\int_{0}^{2\pi}A[q,\mu(0.4)]\times A[q,\mu(0.1)]
$$
or 
$$
\int_{0}^{2\pi}A[q,\mu(0.4)]\times A[q,\mu(0.2)]
$$
And dont I need to average over the same delay time $\tau$? So for $\tau 1$ its
$$
\int_{0}^{2\pi}A[q,\mu(0.2)]\times A[q,\mu(0.1)]\\
\int_{0}^{2\pi}A[q,\mu(0.4)]\times A[q,\mu(0.2)]\\
\int_{0}^{2\pi}A[q,\mu(0.8)]\times A[q,\mu(0.4)]\\
\int_{0}^{2\pi}A[q,\mu(1.6)]\times A[q,\mu(0.8)]
$$.
I am using Mathematica to evaluate these functions.
As you can see I am rather confused. Thanks so much for your help. The paper I am referencing is: 

Rotational—translational models for interpretation of quasi-elastic light scattering spectra of motile bacteria. Michael Holz and Sow-Hsin Chen. Appl. Optics 17 no. 20, pp. 3197-3204 (1978).

 A: If the integration over $\psi$ must be done numerically, then there is most likely very little you can do to simplify this expression. The prefactor
$$
\frac{\int_{-1}^{+1}(1-2w^2\nu^2) \exp(-w^2\nu^2)\ d\nu }{\int_{0}^{2\pi} |A[q,\mu(0)]|^2\ d\psi}
$$
you can compute once for each $q$, and then you can forget about it as you vary $\tau$ (which I presume is identical to $t$).
The main bit of your integral, though,
$$\int_{0}^{2\pi} A[q,\mu(\tau)]A^*[q,\mu(0)]\ d\psi,$$
will probably need to be computed numerically for each time $\tau$ of interest. If you post more details on the functional dependence of $A$ on $q$, $\mu$ and $\psi$ (and of $\mu$ on $t$) then it may be possible to find a simpler method which simplifies the integration and even allows you to integrate numerically only once, but this is not possible in general, and quite unlikely as a rule.
Other than that, it depends on what you want to know about $f(q,t)$. If all you want is graphs of $f$ as a function of $t$ for a few values of $q$, then you're essentially asking for the values of $f$ on a grid of times $\tau_0,\tau_0+\delta \tau,\ldots,\tau_1-\delta \tau,\tau_1$, and each of these must be computed separately by numerically integrating (1). If you want more sophisticated information on $f$ then... well, it depends.

Regarding the averaging over initial times:
Typically, autocorrelation functions of the form $$C(\tau)=C(t,t')=\left\langle c^\ast(t)c(t')\right\rangle$$
are formally and explicitly independent of time translation, so $C(t+t'',t'+t'')=C(t,t')\,\forall t'$ and they can be reduced to a function of $\tau=t-t'$. In this case the formal time translation invariance has been lost, probably by enforcing initial conditions at $t'=0$, which means that here you should not change the $t'=0$ inside $A^\ast$ to other times (generally you don't need to, but it doesn't hurt). Take the products $A(q,\mu(t))A^\ast(q,\mu(0))$ as you've been given and don't tamper with them. 
