This is my first post in math.stackexchange.
This is my Problem: I'm trying to solve the following Gaussian path integral:
\begin{equation} I=\lim\limits_{N \to \infty}\int \left(\prod_{n=1}^{N-1} \mathrm{d} u_n\right)\left(\prod_{n=1}^{N-1} \mathrm{d} \bar{u}_n\right) \cdot \\\exp \left(\sum_{n=0}^{N-1}\left[-\frac{a}{2}\left[\left(u_{n+1}-u_n\right)^2-(\bar{u}_{n+1}-\bar{u}_{n})^2 \right] -\frac{b}{2}(\bar{u}_n-u_n+r)^2\right]\right), \end{equation}
where $a,b$ and $r$ are constants, $u_0 = u_N=0$ and $\lim_{N \rightarrow \infty} a+b=a$. Then I solve the $u_n$ integral using the identity
$$ \int \, \mathrm{d}^n \textbf{u} \exp\Big\{ -\frac{1}{2}\textbf{u}^T A \textbf{u} + \textbf{J}^T \textbf{u} \Big\}=\frac{(2 \pi)^{n/2}}{\det(A)^{1/2}}e^{\textbf{J}^T A^{-1} \textbf{J}}, $$
where in my case $\textbf{J}_ i=b (\bar{u}_i+r)$ and
$$ A= a \begin{bmatrix} 2 & -1 & 0 & ... & & 0\\ -1 & 2 & -1 & ... & & 0\\ \vdots & \vdots & \vdots & \ddots & & \vdots\\ 0 & 0 & 0 & ... & & 2\\ \end{bmatrix} $$ is a $(N-1) \times (N-1)$ matrix.
I found that $\det A=a^{N-1} N$ but I have no idea how to compute the inverse of this matrix for general $N$.
Then when I do the same procedure for the $\bar{u}_n$ integral I get that the solution of the integral (first equation) is
$$ I=\frac{(2 \pi)^{(N-1)/2}}{\det(A)^{1/2}}\frac{(2 \pi)^{(N-1)/2}}{\det(A^{-1}-A)^{1/2}}e^{\textbf{G}^T (A^{-1}-A)^{-1} \textbf{G}}, $$
where $\textbf{G}_i=2 b^2 A^{-1}_{ij} r_j$ and $r_j$ is a a constant vector that I know. Now, even if I know how to deal it $\det A$ I have no idea how to compute $\det(A^{-1}-A)$.
So these are my questions:
i) Do you know if there is a generic way that I can write the inverse of the matrix $A$, $A^{-1}_{ij}$, in terms of N?
ii) Knowing $\det(A)$ can I compute $\det(A^{-1}-A)$?
iii) Do you think that there is a clever way of dealing with this integral?
Thank you all for you time!
