Help With Deriving Caldeira-Leggett's Influence Functional I am attempting to retrace steps performed numerous times before and derive Caldeira-Leggett's influence functional found in their paper "Path Integral Approach To Quantum Brownian Motion".
However, I'm getting caught up in what I am sure is a fundamental misunderstanding on my part. I'll pose my question after laying out the mathematical "path" I have followed to get to the current point:
[One of the multiple resources online I've found, and have been following, see pages 16-18 of :
http://web.science.uu.nl/itf/Teaching/2006/MxWakker.pdf]
Following the path integral formulism, and obtaining the action for a  forced harmonic oscillator from R.P. Feynman and A.R. Hibbs, Quantum Mechanics and Path Integrals,
$ \begin{equation} S^{(k)}_{cl}[x(t),R_{k},R^{\prime}_{k}]=\frac{m\omega_{k}}{2\sin(\omega_{k} T)} \left[ (R_{k}^{2}+R_{k}^{\prime2})\cos(\omega_{k} T)-2R_{k}R^{\prime}_{k} +  \\ R_{k}\frac{2}{m \omega_{k}}\int_{t_{a}}^{t_{b}}x(t)\sin(\omega_{k}(t-t_{a}))dt + R^{\prime}_{k}\frac{2}{m \omega_{k}}\int_{t_{a}}^{t_{b}}x(t)\sin(\omega_{k}(t_{b}-t))dt \\ - \frac{2}{m^{2} \omega_{k}^{2}}\int_{t_{a}}^{t_{b}}\int_{t_{a}}^{t}x(t)x(s)\sin(\omega_{k}(t_{b}-t))\sin(\omega_{k}(s-t_{a}))dsdt \right] \end{equation} $ $ \tag{1} $
where $T=t_{b}-t_{a}$, and $x(t)$ is an external force, then the resulting kernel is
$K^{(k)}[x(t),R_{k},R^{\prime}_{k}]= \left(\frac{m\omega_{k}}{2\pi i \sin(\omega_{k} T)}\right)^{\frac{1}{2}} \exp \left[\frac{i}{\hbar} S^{(k)}_{cl}[x(t),R_{k},R^{\prime}_{k}] \right] $ $ \tag{2}$
With this information, the object I am trying to obtain is the influence functional,
$F[x(t),y(t)] \equiv \int \int \int dR^{\prime}dQ^{\prime}dR \rho_{B}(R^{\prime},Q^{\prime},0) \int_{R^{\prime}}^{R} \int_{Q^{\prime}}^{Q} [dQ(t)][dR(t)] \\  
\exp \left[\frac{i}{\hbar} (( S^{(k)}_{I}[x(t),R(t)]-S^{(k)}_{I}[y(t),Q(t)]) \\ + ( S^{(k)}_{B}[x(t),R(t)]-S^{(k)}_{B}[y(t),Q(t)])) \right]  $ $ \tag{3}$
which can be further simplified using
$ \rho_{B}(R^{\prime},Q^{\prime},0) \equiv \prod_{k} \rho^{(k)}_{B}(R_{}^{\prime},Q_{k}^{\prime},0) \\ \equiv  \prod_{k} \left(\frac{m\omega_{k}}{2\pi \hbar \sinh\left(\frac{m\omega_{k}}{k_{b}T}\right)}\right) \exp \left[  \left(\frac{-m\omega_{k}}{2 \hbar \sinh\left(\frac{m\omega_{k}}{k_{b}T}\right)}\right) \left( (R_{k}^{\prime2}+Q_{k}^{\prime2})\cosh\left(\frac{m\omega_{k}}{k_{b}T}\right)    - 2 R_{k}^{\prime}Q_{k}^{\prime}\right)\right] $ $ \tag{4}$
which can be substituted into equation (3). A further simplification of (3) is the substitution
$ \int_{R^{\prime}}^{R} \int_{Q^{\prime}}^{Q} [dQ(t)][dR(t)] \\
\exp \left[\frac{i}{\hbar} (( S^{(k)}_{I}[x(t),R(t)]-S^{(k)}_{I}[y(t),Q(t)]) + ( S^{(k)}_{B}[x(t),R(t)]-S^{(k)}_{B}[y(t),Q(t)])) \right] \\ = \prod_{k} K^{(k)}[x(t),R_{k},R^{\prime}_{k}] K^{*(k)}[y(t),R_{k},Q^{\prime}_{k}]   $ $\tag{5}$
$K^{(k)}$ is the same as that found in equation (2). 
All this substitution will leave a simple, yet tedious, series of Gaussian integrals through $R_{k},R^{\prime}_{k}$ and $Q^{\prime}_{k}$.
My question is this: It appears to me that when you multiply the memory kernel in equation (2) by its conjugate, as is done in equation (5), then the common factor of $R_{k}^{2}$ should cancel, leaving terms of $R_{k}$ in the first order. This would end up being a dirac delta function when being integrated through $R_{k}$. When I did this, first by hand and then with Mathematica, I could not reproduce the answer found in the Caldeira paper. It also goes against the many comments I have read on this derivation, particularly regarding the "multiple tedious gaussian integrals". Is my understanding of calculating the influence functional completely off? Particularly I am wondering if I have misinterpreted equation (5), was simply conjugating the memory kernel (alongside substituting $x(t)$ to $y(t)$), and substituting all the elements I have listed here, naive of me?
Thank you for reading all of that, and any help you can provide!
 A: The integrals you obtained after doing the Gaussians are correct. All you need is an integration trick and a bit of algebra to rearrange them in the final form. 
The integration trick: 
A double integral over a 2-dimensional square domain decomposes as
$$
\int_0^t dt' \int_0^t dt" f(t', t") = \int_0^t dt' \int_0^{t'} dt" f(t', t") + \int_0^t dt" \int_0^{t"} dt' f(t', t") 
$$ 
But if we interchange the labeling of the integration variables in the 2nd term we get an even simpler form:
$$
\int_0^t dt' \int_0^t dt" f(t', t") = \int_0^t dt' \int_0^{t'} dt" f(t', t") + \int_0^t dt' \int_0^{t'} dt" f(t", t') = \\
= \int_0^t dt' \int_0^{t'} dt" \Big( f(t', t") +  f(t", t') \Big)
$$
For the rest of the answer I will only refer to the integrals in your handwritten notes, leaving aside any irrelevant factors. 
Let us apply the above decomposition to your first set of integrals, which reads
$$
I_1 =\frac{1}{\sin \omega_k t} \left[ - 2 \int_0^t{dt' \int_0^{t'}{ dt" \Big( x(t')x(t") - y(t')y(t") \Big) \sin\omega_k(t-t') \sin\omega_k t" }} + \\
+ \int_0^t{dt'\int_0^t{dt" \Big( x(t') + y(t')\Big) \Big( x(t") - y(t")\Big) \sin\omega_k(t-t') \sin\omega_k t" }} \right]
$$
We only want to write the 2nd term, since the 1st is already confined to half the square $[0,t]x[0,t]$: 
$$
I_1 = \frac{1}{\sin \omega_k t} \left[ - 2 \int_0^t{dt' \int_0^{t'}{ dt" \Big( x(t')x(t") - y(t')y(t") \Big) \sin\omega_k(t-t') \sin\omega_k t" }} +  \right.\\
+ \int_0^t{dt'\int_0^{t'}{dt" \left[ \Big( x(t') + y(t')\Big) \Big( x(t") - y(t")\Big) \sin\omega_k(t-t') \sin\omega t"  + \right.}} \\
+ \left.\left.\Big( x(t') - y(t')\Big) \Big( x(t") + y(t")\Big)  \sin\omega_k t' \sin\omega_k(t-t") \right]\right]
$$
Now we can rewrite everything under the same integrals, and proceed to regroup and simplify the first 2 terms. We obtain eventually
$$
I_1 = \frac{1}{\sin \omega_k t}   \int_0^t{dt' \int_0^{t'}{ dt" \left[ - \Big( x(t') - y(t') \Big) \Big( x(t") + y(t")\Big) \sin\omega_k(t-t') \sin\omega_k t" +  \right.}}
$$
$$
+ \left. \Big( x(t') - y(t')\Big) \Big( x(t") + y(t")\Big)  \sin\omega_k t' \sin\omega_k(t-t") \right]
$$
$$
= \frac{1}{\sin \omega_k t}   \int_0^t{dt' \int_0^{t'}{ dt" \Big( x(t') - y(t') \Big) \Big( x(t") + y(t")\Big)\cdot \\
\cdot \left[  \sin\omega_k t' \sin\omega_k(t-t") - \sin\omega_k(t-t') \sin\omega_k t"  \right]}}
$$
But doing the trigonometry in the last square brackets gives
$$
\sin\omega_k t' \sin\omega_k(t-t") - \sin\omega_k(t-t') \sin\omega_k t" = 
$$
$$
= \sin\omega_k t' \sin\omega_k t \cos\omega_k t" - \sin\omega_k t' \cos\omega_k t \sin\omega_k t" -  
$$
$$
- \sin\omega_k t \cos\omega_k t' \sin\omega_k t" +  \cos\omega_k t \sin\omega_k t' \sin\omega_k t" = \sin\omega_k t \sin\omega_k(t'-t")
$$
and we finally have
$$
I_1 =  \int_0^t{dt' \int_0^{t'}{ dt" \Big( x(t') - y(t') \Big) \Big( x(t") + y(t")\Big)\sin\omega_k(t'-t")}}
$$
Now for the 2nd set of integrals, which reads
$$
I_2 = \frac{1}{\sin^2 \omega_k t} \left[ \int_0^t{dt' \int_0^{t}{ dt" \Big( x(t') - y(t') \Big) \Big( x(t") + y(t")\Big)\sin\omega_k(t-t')\sin\omega_k(t-t')  }} + \right.
$$
$$
+  \int_0^t{dt' \int_0^{t}{ dt" \Big( x(t') - y(t') \Big) \Big( x(t") + y(t")\Big)\sin\omega_kt' \sin\omega_kt"  }} +
$$
$$
\left.+ 2 \cos\omega_kt \int_0^t{dt' \int_0^{t}{ dt" \Big( x(t') - y(t') \Big) \Big( x(t") + y(t")\Big)\sin\omega_kt'\sin\omega_k(t-t')  }} \right]
$$
First bring everything under one integral and simplify the trigonometry by expanding the sines as before. I leave the details as an exercise (lots of typing), but the result is 
$$
I_2 = \frac{1}{\sin^2 \omega_k t} \int_0^t{dt' \int_0^{t}{ dt" \Big( x(t') - y(t') \Big) \Big( x(t") + y(t")\Big) \left[ \sin\omega_k(t-t')\sin\omega_k(t-t') + \right. }}\\
\left. + \sin\omega_kt' \sin\omega_kt" + 2\cos\omega_kt \sin\omega_kt'\sin\omega_k(t-t') \right]  = \\
= \frac{1}{\sin^2 \omega_k t} \int_0^t{dt' \int_0^{t}{ dt" \Big( x(t') - y(t') \Big) \Big( x(t") + y(t")\Big) \cdot }}\\
\cdot \left[ \sin\omega_k t \sin\omega_k t' \sin\omega_k t \sin\omega_k t" + \sin\omega_k t \cos\omega_k t' \sin\omega_k t \cos\omega_k t' \right] = 
$$
So
$$
I_2 = \int_0^t{dt' \int_0^{t}{ dt" \Big( x(t') - y(t') \Big) \Big( x(t") + y(t")\Big) \cos\omega_k(t'-t")}}
$$
The rest is just a matter of summing up the different contributions so as to write the influence integral neatly.
