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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.

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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}. $\quad\Box$

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