= Logarithmic reweighting under dominated convergence
{title2=$\left\|\dfrac{\log(n+x)}n u(x)\right\|_1\leq\dfrac{\log n\,\|u\|_1+\|\log(1+x)u\|_1}{n}$}
Suppose $u$ and $\log(1+x)u$ are <Lebesgue integrable functions> on $(0,\infty)$. For integers $n\geq1$, $x+n\leq n(1+x)$ gives the displayed <L1 norm> bound. It tends to zero because $\log n/n\to0$, proving <L1 convergence> of the logarithmically weighted <functions> to zero. Alternatively, $|\log(n+x)u(x)/n|\leq(1+\log(1+x))|u(x)|$ and pointwise convergence permit the <dominated convergence theorem>.
For $u(x)=\sin x/x^q$ with $1<q<2$, both integrability conditions hold. Near zero, $|\sin x|\leq x$ and $\log(1+x)\leq x$ give integrable bounds by powers. At infinity, $|\sin x|\leq1$ and $\log(1+x)\leq\log2+\log x$, with $\int_1^\infty x^{-q}\log x\,dx=(q-1)^{-2}$. The <L1 norm> consequently decays as $O(\log n/n)$ in <Big O notation>.
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