Derive the Hilbert transform for periodic function

Question

I know that the Hilbert transform of a function $f$ is given by \begin{equation} \mathcal{H}f(t)=P.V\int_{-\infty}^{\infty}\frac{f(x)}{t-x}dx \end{equation} But there is another formula for periodic function \begin{equation} \mathcal{H}f(t)=P.V\frac{1}{2\pi}\int_{0}^{2\pi}f(x)\cot(\frac{t-x}{2})dx \end{equation} How to derive the last formula from the first? I was confused for weeks.

Answer 1

I think the chapter about the periodic functions should work for you http://liu.diva-portal.org/smash/get/diva2:872439/FULLTEXT02.pdf .

Answer 2

The Hilbert transform is $\phi \mapsto \phi \ast PV(\frac{1}{x})$ which in the Fourier domain means $\hat{\phi} \mapsto \frac{1}{i \pi} \hat{\phi}(\xi) \text{ sign}(\xi)$. This is well-defined whenever $\hat{\phi}(\xi) \text{ sign}(\xi)$ is well-defined as a (tempered) distribution.

If $\phi$ is $T$ periodic then $\hat{\phi}(\xi) = \sum_{n=-\infty}^\infty c_n \delta(\xi - n/T)$ where $c_n = \int_0^T \phi(x)e^{-2i \pi n x/T}dx$ are the Fourier series coefficients.

Iff $c_0 = 0$ then $\widehat{H \phi}(\xi) = \frac{1}{i \pi}\sum_{n =1}^\infty c_n \delta(\xi - n/T)-c_{-n} \delta(\xi + n/T)$ is a tempered distribution, and it is the Fourier transform of the $T$-periodic distribution $H \phi(x) = \phi \star h(x) = \int_{-T/2}^{T/2} \phi(x-y) h(y)dy$ (this integral being in the sense of distributions, if $\phi$ is $C^1$ then there is a principal value)

where $h$ is the $T$-periodic distribution whose Fourier transform is $\sum_{n =1}^\infty \delta(\xi - n/T)-\delta(\xi + n/T)$.

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Thus, it is a good idea to look at $\phi_r(x) = \sum_{n=-\infty}^\infty r^{|n|} c_n e^{2i \pi n x /T}$ and $h_r(x) = \sum_{n=1}^\infty r^n (e^{2i \pi n x /T}-e^{2i \pi n x /T})$ so that $$\phi \star h(x) = \lim_{r \to 1^-} \int_{-T/2}^{T/2} \phi_r(x-y) h_r(y)dy$$ the integral being convergent, and the limit $r \to 1^-$ being in the sense of distributions.