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.
Best Answer
For the desired convergence it is enough that the $2\pi$-periodic function $f$ be just locally integrable.
Indeed, let \begin{equation*} g:=f-A,\quad A:=\frac1{2\pi}\int_\pi^{3\pi}f, \end{equation*} \begin{equation*} \int_\pi^{3\pi}g=0. \tag{1}\label{1} \end{equation*} We want to show that \begin{equation*} I_N:=\int_\pi^N dr\,\frac{g(r)}r \end{equation*} converges (as $N\to\infty$) to a real number.
Note that \begin{equation*} I_N=\sum_{k=0}^{k_N}J_k+R_N, \tag{2}\label{2} \end{equation*} where \begin{equation*} k_N:=\Big\lfloor\frac{N-3\pi}{2\pi}\Big\rfloor, \end{equation*} \begin{equation*} J_k:=\int_{\pi+2\pi k}^{3\pi+2\pi k} dr\,\frac{g(r)}r = \int_{\pi}^{3\pi} dr\,\frac{g(r)}{r+2\pi k}, \end{equation*} \begin{equation*} R_N:=\int_{3\pi+2\pi k_N}^N dr\,\frac{g(r)}r = \int_{3\pi}^{N-2\pi k_N} dr\,\frac{g(r)}{r+2\pi k_N}. \end{equation*} Next, \begin{equation*} |R_N|\le\int_{3\pi}^{5\pi} dr\,\frac{|g(r)|}{2\pi k_N}\to0. \tag{3}\label{3} \end{equation*} Further, letting \begin{equation*} G(u):=\int_\pi^u dr\,g(r), \end{equation*} we get \begin{equation*} \begin{aligned} J_k&=\int_{\pi}^{3\pi} dr\,g(r)\int_{r+2\pi k}^\infty\frac{ds}{s^2} \\ &=\int_{\pi+2\pi k}^\infty\frac{ds}{s^2}\,G(\min(s-2\pi k,3\pi)) \\ &=\int_{\pi+2\pi k}^{3\pi+2\pi k}\frac{ds}{s^2}\,G(s-2\pi k) \\ &=\int_{\pi}^{3\pi}\frac{ds}{(s+2\pi k)^2}\,G(s), \end{aligned} \end{equation*} since $G(3\pi)=0$, by \eqref{1}. So, \begin{equation*} |J_k|\le\frac{c}{(\pi+2\pi k)^2} \tag{4}\label{4} \end{equation*} where \begin{equation*} c:=\int_{\pi}^{3\pi}ds\,|G(s)|<\infty. \end{equation*}
Now the convergence of $I_N$ to a real number follows from \eqref{2}, \eqref{3}, and \eqref{4}. $\quad\Box$