Evaluation of a path integral I am stuck weeks in the following calculation from a paper:
Path integral $I$ is written as
$$ I = \int_{\mathbf{r}_0 = \mathbf{0}}^{\mathbf{r}_L=\mathbf{R}}\exp\left[-A\int_0^L\dot{\mathbf{r}}_t^2\,dt+\frac{i\eta}{2L^2}\int_0^L\int_0^L (\mathbf{r}_t-\mathbf{r}_s)^2\,ds\,dt\right]{\cal D}\mathbf{r}.$$
Since the contribution of the path $\mathbf{r}_t$ to $I$ becomes most dominant when $\mathbf{r}_t$ makes the exponent stationary, we have $\mathbf{r}_t$ as the solution of the following Euler-Lagrange equation subject to the boundary conditions $\mathbf{r}_0 = \mathbf{0}$ and $\mathbf{r}_L = \mathbf{R}$:
$$\ddot{\mathbf{r}}_t + \alpha^2\mathbf{r}_t - \beta\int_0^L\mathbf{r}_s\,ds=0,$$
where $\alpha^2 = \dfrac{i\eta}{AL}$ and $\beta = \dfrac{i\eta}{AL^2}$. With this solution, $I$ is expressed as 
$$I = \exp(-A\mathbf{r}_L\dot{\mathbf{r}}_L).$$
I don't have any problem with the Euler-Lagrange equation but cannot obtain the last expression. Anyone help me out of this?  

I assume the bounds of the integral with respect to $s$ are also $0$ and $L$.
First, we integrate the first term inside the exponential by parts. We get
$$-A\int_0^L \dot{\mathbf{r}}_t^2 dt = \left[ -A\mathbf{r}_t \dot{\mathbf{r}}_t \right]_0^L +A\int_0^L \mathbf{r}_t \ddot{\mathbf{r}}_t dt = -A\mathbf{r}_L \dot{\mathbf{r}}_L +A\int_0^L \mathbf{r}_t \ddot{\mathbf{r}}_t dt,$$
since $\mathbf{r}_0 = 0$.
Next, we expand the square inside the second term. We get
$$\frac{i\eta}{2L^2} \int_0^L \int_0^L \left( \mathbf{r}_t^2 + \mathbf{r}_s^2 - 2\mathbf{r}_t\mathbf{r}_s \right) ds\: dt.$$
The $\mathbf{r}_t^2$ term doesn't depend on $s$, so we can integrate it with respect to $s$ getting a factor of $L$. As for the $\mathbf{r}_s^2$ term, we can switch the dummy variables $s$ and $t$ so it becomes identical to the $\mathbf{r}_t^2$ term. With the definitions of $\alpha$ and $\beta$, we arrive at
$$\frac{i\eta}{2L^2} \int_0^L \left( L \mathbf{r}_t^2 + L \mathbf{r}_t^2 - 2\mathbf{r}_t \int_0^L \mathbf{r}_s ds \right) dt = \int_0^L \mathbf{r}_t \left(A\alpha^2 \mathbf{r}_t -A\beta \int_0^L \mathbf{r}_s ds \right) dt.$$
Finally, putting all together we find that the whole expression inside the exponent is
$$-A\mathbf{r}_L \dot{\mathbf{r}}_L + A \int_0^L \mathbf{r}_t\left(\ddot{\mathbf{r}}_t + \alpha^2 \mathbf{r}_t -\beta \int_0^L \mathbf{r}_s ds \right) dt.$$
The expression inside the parentheses clearly vanishes for the "classical" path satisfying the Euler-Lagrange equation. Since we are only considering this path's contribution to the path integral, we arrive at the desired solution
$$I=\exp(-A\mathbf{r}_L \dot{\mathbf{r}}_L).$$
