I'm considering integrals of the (Hilbert transform) type
$$p.v.\int_{-\infty}^\infty\frac{f(r)}{r}\,dr$$
where **$f(r)$ is periodic**, say, with period $2\pi$. I'm assuming very little regularity on $f$. To be concrete, let's say that $f(r)$ is **$\alpha$-Holder continuous with $\alpha<1$**. Now I'm wondering if the above expression is necessarily well defined and finite. I don't know if this is true, but below is some work towards (maybe) showing that it's true.

First we have \begin{align} p.v.\int_{-\infty}^\infty\frac{f(r)}{r}\,dr=p.v.\int_{-\pi}^\pi\frac{f(r)}{r}\,dr+\lim_{N\to\infty}\left(\int_{\pi}^N\frac{f(r)}{r}\,dr+\int_{-N}^{-\pi}\frac{f(r)}{r}\,dr\right) \end{align} Holder continuity of $f$ is enough to show that the first integral on the right hand side is finite. Now consider the remaining two. Letting $$A=\frac{1}{2\pi}\int_0^{2\pi}f(r)\,dr$$ and using the fact that $$\int_\pi^N\frac{A}{r}\,dr+\int_{-N}^{-\pi}\frac{A}{r}\,dr=0$$ we write $$\lim_{N\to\infty}\left(\int_{\pi}^N\frac{f(r)}{r}\,dr+\int_{-N}^{-\pi}\frac{f(r)}{r}\,dr\right)=\lim_{N\to\infty}\left(\int_{\pi}^N\frac{f(r)-A}{r}\,dr+\int_{-N}^{-\pi}\frac{f(r)-A}{r}\,dr\right)\quad (*)$$

Point is, $f(r)-A$ is now a periodic function that oscillates about 0 (i.e. takes on both negative and positive values), so maybe (just maybe) we have that the integrals on the right hand side of (*) converge. Of course, one is led to this hopefulness due to the fact that integrals like

$$\int_1^\infty\frac{sin(r)}{r}\,dr,\quad\int_1^\infty\frac{cos(r)}{r}\,dr$$ converge. I concede though that the above are very specific examples, and there's really no reason that convergence should hold when $\sin$ and $\cos$ are replaced by other periodic functions (with quite minimal regularity; although regularity may not even be the issue here). But, who knows, maybe. Any intuition one way or the other would be greatly appreciated.