I am studying some system on a ring with $N$ discrete sites, and I obtain some quantity $Q$:
\begin{equation} Q(t,r)=\frac{2}{N}\sum_{k=1}^{N-1}\left(1-(-1)^k\right)\sin\left[\frac{k\pi}{N}\right]\sin\left[\frac{rk\pi}{N}\right]\exp\left\{-2t\left(1-\cos\left[\frac{k\pi}{N}\right]\right)\right\} \end{equation}
Where $r$ represents a distance, it is discrete and can take any value up to $N-1$.
I am interested in the case where $N\to \infty$, so I was thinking of approximating my sum into an integral. There is this post that provides a method, here is what I tried:
First I set $k \to 2k-1 $ to only take into account the odd terms. I obtain:
\begin{equation} Q(t,r)=\frac{2}{N}\sum_{k=1}^{N/2}\sin\left[\frac{(2k-1)\pi}{N}\right]\sin\left[\frac{r(2k-1)\pi}{N}\right]\exp\left\{-2t\left(1-\cos\left[\frac{(2k-1)\pi}{N}\right]\right)\right\} \end{equation} I could then just identify my $a, b$ to then apply this rule (for $f$ continuous on $[a,b]$): \begin{equation} \int_{a}^{b} f(x) d x=\lim _{n \rightarrow \infty} \frac{b-a}{n} \sum_{i=1}^{n} f\left(a+\frac{b-a}{n} i\right) \end{equation} However I am unsure of the following points:
- How does $r$ scales if I defined $x=\frac{(2k-1)\pi}{N}$ and let $N\to \infty$?
- By summing only over the odd terms I can change the indices into $k \to 2k-1 $, but then my sum goes only to $N/2$ (assuming N is even). What happens to the integral?