# intuition – 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”:

$$frac D{e^D-1}f(x)=frac1{2 pi }int_{-infty }^{+infty } frac{e^{-iwx}}{e^{-i w}-1}int_{-infty }^{+infty } e^{i t w} f'(t) , dt , dw$$

The integrals here should be understood as Fourier transforms.

Now, intuitively, why the following?

$$frac D{e^D-1} left(frac1piln left(frac{x+1/2 +frac{x}{pi }}{x+1/2 -frac{x}{pi }}right)right)=tan x$$

There are other examples where shadow converts trigonometric functions into inverse trigonometric, logarithms to exponents, etc:

$$frac D{e^D-1} left(frac1{pi }ln left(frac{x+1/2-frac{x}{pi }}{x-1/2+frac{x}{pi }}right)right)=cot x$$