Logarithmic reweighting under dominated convergence (${\Vert \frac{\log (n+x)}{n}u(x)\Vert }_1\le \frac{\log n\Vert u{\Vert }_1+\Vert \log (1+x)u{\Vert }_1}{n}$)

Suppose $u$ and $\log (1+x)u$ are Lebesgue integrable functions on $(0,\infty )$. For integers $n\ge 1$, $x+n\le n(1+x)$ gives the displayed L1 norm bound. It tends to zero because $\log n/n\to 0$, proving L1 convergence of the logarithmically weighted functions to zero. Alternatively, $|\log (n+x)u(x)/n|\le (1+\log (1+x))|u(x)|$ 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, $|\sin x|\le x$ and $\log (1+x)\le x$ give integrable bounds by powers. At infinity, $|\sin x|\le 1$ and $\log (1+x)\le \log 2+\log x$, with ${\int }_1^{\infty }{x}^{-q}\log xdx=(q-1{)}^{-2}$. The L1 norm consequently decays as $O(\log n/n)$ in Big O notation.