Convergence of oscillatory integrals 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.
 A: 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}. $\quad\Box$
