We know a discrete Fourier transformation with discrete $n$ and continuous $x_1,x_2$: $$ \sum_{n\in\mathbb{Z}} e^{-in(x_1-x_2)\frac{2\pi}{L}}=L\delta(x_1-x_2) $$ with Dirac delta function $\delta$.
and a discrete Fourier transformation with discrete $n$ and discrete $x_1,x_2$: $$ \sum_{n\in\mathbb{Z}} e^{-in(x_1-x_2)\frac{2\pi}{L}}=L\delta_{x_1-x_2,0} $$ with Kronecker delta function $\delta$.
Question: what is the discrete Fourier transformation with discrete $n$ and discrete $x_1,x_2$ below: $$ \sum_{n\in\mathbb{Z}} n \;e^{-in(x_1-x_2)\frac{2\pi}{L}}=? $$
Comment: For continuous $x_1,x_2$ case, I have obtained, and it is easy to see that $$ \sum_{n\in\mathbb{Z}} n e^{-in(x_1-x_2)\frac{2\pi}{L}}=i\frac{L^2}{2\pi} \frac{\partial \delta(x_1-x_2)}{\partial {(x_1-x_2)}} $$
For discrete $x_1,x_2$, I obtained a form of $$\sum_{n\in\mathbb{Z}} n \;e^{-in(x_1-x_2)\frac{2\pi}{L}}=\frac{L\delta_{x_1-x_2,0}}{1-e^{-i(x_1-x_2)\frac{2\pi}{L}}}=\frac{L^2}{2\pi}\frac{\delta_{x_1-x_2,0}}{i(x_1-x_2)+\dots}.$$
I would like to make sure that my answer is really the correct case?
The initial identity is wrong. First, the series does not converge in the classical sense, and second, as a series of distributions the limit is periodic, what is called a Dirac comb. $$ \sum_{n\in\mathbb{Z}} e^{-2\pi i\, n(x_1-x_2)\frac{1}{L}}=L\sum_{k\in\Bbb Z}\delta(x_1-x_2+kL) $$ See Poisson summation.
There is no discrete case, since then the sums are not even interpretable as distributions and thus diverge in all circumstances.
The second formula has the same caveats, the right side should be the derivative of a Dirac comb. The general approach is correct.