= Integrating a mean-zero periodic function divided by x
{title2=$\int_L^\infty h(x)/x\,dx$}
Let $h$ be measurable, integrable over a positive period $L$, and have mean zero there. Its positive and negative parts have equal integrals $H$. On the $k$-th period the signed integral of $h(x)/x$ has absolute value at most
$$
\frac H L\left(\frac1k-\frac1{k+1}\right).
$$
This follows by bounding the positive and negative weighted integrals between $H/((k+1)L)$ and $H/(kL)$. The bounds are summable, so the integrals at period endpoints converge. A leftover partial period has absolute integral at most $\|h\|_{L^1(0,L)}/(kL)$, tending to zero, which proves convergence for arbitrary upper endpoints too.
In contrast, each full-period integral of $|h(x)|/x$ is at least $2H/((k+1)L)$. It diverges in total unless $H=0$, equivalently $h=0$ almost everywhere. Thus a nonzero mean-zero periodic function gives a convergent improper integral here but no <Lebesgue integrability>.
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