All Questions
1 question
16
votes
2
answers
2k
views
Intuitive explanation why "shadow operator" $\frac D{e^D-1}$ connects logarithms with trigonometric functions?
Consider the operator $\frac D{e^D-1}$ which we will call "shadow":
$$\frac {D}{e^D-1}f(x)=\frac1{2 \pi }\int_{-\infty }^{+\infty } e^{-iwx}\frac{-iw}{e^{-i w}-1}\int_{-\infty }^{+\infty } e^...