Evaluation a this integral If $f$ ia a continuously differentiable function on the unit circle and
$$
g(x)=\frac{1}{2\pi}\int_{-\pi}^{\pi}\frac{f(x+t)-f(x-t)}{2\tan\frac{1}{2}t}dt
$$
evaluate
$$
\frac{1}{2\pi}\int_{-\pi}^{\pi}\frac{g(x+t)-g(x-t)}{2\tan\frac{1}{2}t}dt
$$
 A: Let's call the original transform $f \mapsto g$ by $g = \widetilde{f}$.
The point is that $f \mapsto \widetilde{f}$ acts diagonally in the Fourier basis for $C^1[-\pi, \pi]$; you can see this any one of a number of fancier ways but I'll stick to showing it computationally:
$$\widetilde{e^{i n (\cdot)}}(x) = \frac{1}{2 \pi} \int_{- \pi}^{\pi} \frac{e^{i n (x + t)} - e^{i n (x - t)}}{2 \tan \frac{t}{2}} dt = i e^{i n x} \left( \frac{1}{2 \pi} \int_{- \pi}^{\pi} \frac{\sin n t}{2 \tan \frac{t}{2}} dt \right)
$$
From here we can show that the parenthetical quantity on the far right side is equal to 1 for all $n \neq 0$, and 0 for $n = 0$. To show this directly is an exercise in complex analysis; simply expand the $\sin, \tan$'s into exponential form and then perform the substitution $z = e^{i t}$, integrating counterclockwise around the unit circle, picking up a residue at $z = 0$.
Now it should be clear that $\tilde{f} = i f$ when the zeroth fourier coefficient is zero, i.e. when $\int_{- \pi}^{\pi} f(x) dx = 0$. As $f \mapsto \tilde{f}$ acts as the zero operator on the subspace of constant functions, we now have
$$
\tilde{f}(x) = i \left(f(x) - \int_{-\pi}^{\pi} f(y) dy \right)
$$
A second application of $\mapsto$ gives $- \left(f(x) - \int_{-\pi}^{\pi} f(y) dy \right)$.
