Continuous Fourier transform
Consider the Fourier transform: \begin{eqnarray} D(x) & = & \frac{1}{4\pi^2} \int_{-\infty}^\infty d\kappa e^{-i\kappa x}\tilde{D}(\kappa) \\ & \approx & \frac{1}{4\pi^2} \int_{-K}^K d\kappa e^{-i\kappa x}\tilde{D}(\kappa) \\ & \approx & \frac{1}{4\pi^2} \sum_{k=0}^{N-1} \frac{2K}{N} \exp\left[-i\left(\frac{2K}{N}k-K\right) x\right]\tilde{D}_k \\ D_m & \propto & e^{iKx}\sum_{k=0}^{N-1}\exp\left[-i\frac{2K}{N}k\left(\frac{2L}{N}m-L\right)\right]\tilde{D}_k \\ & \propto & \exp\frac{i2KLm}{N} \sum_{k=0}^{N-1} \exp\frac{-i 4KLkm}{N^2}\exp\frac{i2KLk}{N} \tilde{D}_k \end{eqnarray} Let $2KL=\pi N$, we get \begin{equation} e^{-i\pi m}D_m\propto\sum_{k=0}^{N-1}\exp\left(-2\pi i\frac{km}{N}\right)e^{i\pi k}\tilde{D}_k \end{equation}
Fourier integral theorem and Dirac δ-function
Dirichlet integral
\[
\int_{-\infty}^\infty dx \frac{\sin(xL)}{x} = \pi
\]
As $L$ goes to infinity, the integrand becomes more and more like a δ-function.