May I know how I can approximate the following sum for large $n$:
$$\frac{1}{n^2 xy}\left[\sum_{i=0}^{nx}\sum_{j=0}^{ny}\frac{1}{1 + \left(\frac{ |i-j|\wedge (n-|i-j|)}{n}\right)^2}\right]$$
where $0 \le x < y \le 1$, such that $nx,ny$ are integers and $\wedge$ is the min function i.e. $a\wedge b = \min\{a,b\}$? I think I should get an integral approximation but is not occurring to me right now. Thanks.
I couldn't find a closed-form or an approximation, but was able to make it into $3$ single sums instead. The objective is to find the following:
$$\frac{1}{n^2 xy}\left[\sum_{i=0}^{nx}\sum_{j=0}^{ny}\frac{1}{1 + \left(\frac{ |i-j|\wedge (n-|i-j|)}{n}\right)^2}\right]$$
By splitting this into $i-j < 0$ and $i-j > 0$, we get
$$\frac{1}{n^2 xy}\left[{\sum_{i=0}^{nx}\sum_{j=0}^{i}\frac{1}{1 + \left(\frac{ (i-j)\wedge (n-i+j)}{n}\right)^2}}+\sum_{i=0}^{nx}\sum_{j=i+1}^{ny}\frac{1}{1 + \left(\frac{ (j-i)\wedge (n-j+i)}{n}\right)^2}\right]$$
Then since $i-0 = i, i-1 = i-1..., i-(i) = 0$ ranges from $0$ to $i$ also, the first sum can be rewritten. Also the second sum is reindexed so that it goes over the difference.
$$\frac{1}{n^2 xy}\left[\sum_{i=0}^{nx}\sum_{d=0}^{i}\frac{1}{1 + \left(\frac{ (d)\wedge (n-d)}{n}\right)^2}+\sum_{i=0}^{nx}\sum_{d=1}^{ny-i}\frac{1}{1 + \left(\frac{ (d)\wedge (n-d)}{n}\right)^2}\right]$$
Finally, the frequency of each difference is found
$$\frac{1}{n^2 xy}\left[\sum_{d=0}^{nx}\frac{nx+1-d}{1 + \left(\frac{ (d)\wedge (n-d)}{n}\right)^2}+\sum_{d=1}^{ny-nx} \frac{nx+1}{1 + \left(\frac{ (d)\wedge (n-d)}{n}\right)^2} + \sum_{d = ny-nx+1}^{ny} \frac{ny+1-d}{1 + \left(\frac{ (d)\wedge (n-d)}{n}\right)^2}\right]$$
That brings it from one double sum to three single sums.