= Sine power kernel integrability
{title2=$\int_0^\infty|\sin x|x^{-p}\,dx<\infty\iff1<p<2$}
For $p>0$, the <function> $x\mapsto\sin x/x^p$ is <Lebesgue integrable> on $(0,\infty)$ exactly when $1<p<2$. Near zero, $\sin x/x\to1$, so absolute integrability is equivalent to that of $x^{1-p}$, which requires $p<2$. At infinity, $p>1$ gives the upper bound $x^{-p}$. For $0<p\leq1$, on each <interval> $[k\pi,(k+1)\pi]$,
$$
\int_{k\pi}^{(k+1)\pi}\frac{|\sin x|}{x^p}\,dx\geq\frac{2}{((k+1)\pi)^p},
$$
and the corresponding <P-series> diverges.
The signed <improper integral> nevertheless converges for every $0<p<2$: near zero the same local comparison applies, and at infinity the <Dirichlet test for improper integrals> uses the bounded antiderivative of <sine> and the decreasing factor $x^{-p}\to0$. Thus $0<p\leq1$ gives <conditional convergence>, while $1<p<2$ gives <absolute integrability>. In particular, substituting $t=1/x$ turns $\int_0^1|x^{-1}\sin(1/x)|\,dx$ into the divergent tail $\int_1^\infty|\sin t|/t\,dt$.
\Image[/past-exam-of-the-integration-course-of-the-university-of-oxford/2017-sine-power-integrability.png]
{title=Conditional and absolute convergence have different thresholds for a sine power kernel}
{description=The left panel assumes $p>0$ and excludes the open endpoints of each range. The right panel compares signed and absolute truncated integrals at $p=1$, with a divergent harmonic lower bound.}
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