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Upon Fourier transformation $x\mapsto k$, this becomes a diagonal operator with matrix elements $\langle k|\ln D|k'\rangle=2\pi \delta(k-k')\ln k$. So to find the matrix elements in the $x$-representation we would need to inverse the Fourier transform of the logarithm $\ln k$. From this MSE answer for the Fourier transform of $\ln |k|$ (with absolute value signs) I would conclude that $$\langle x|\ln D|x'\rangle=\left(\frac{i \pi}{2}-\gamma\right) \delta (x-x')+\text{P.V.}\left(\frac{1}{2 (x-x')}-\frac{1}{2 | x-x'| }\right).$$