= Solution
For every fixed $x>0$, $\log(x+n)/n\to0$, so $f_n(x)\to0$. To exchange the <limit> with the <Lebesgue integral>, we need a common integrable bound over the whole half-line.
For $n\geq1$, $x+n\leq n(1+x)$, so
$$
0\leq\frac{\log(x+n)}n
\leq\frac{\log n}{n}+\frac{\log(1+x)}n
\leq1+\log(1+x).
$$
Together with $|\sin x|\leq\min(x,1)$, this gives
$$
|f_n(x)|\leq G(x):=\bigl(1+\log(1+x)\bigr)\frac{\min(x,1)}{x^q}.
$$
For $0<x\leq1$, $\log(1+x)\leq x\leq1$, hence $G(x)\leq2x^{1-q}$, which is integrable because $q<2$. For $x\geq1$,
$$
G(x)\leq\frac{1+\log2+\log x}{x^q}.
$$
Both terms are integrable because $q>1$:
$$
\int_1^\infty x^{-q}\,dx=\frac1{q-1},\qquad
\int_1^\infty\frac{\log x}{x^q}\,dx=\frac1{(q-1)^2}.
$$
The second equality follows from <integration by parts>, or the <substitution> $x=e^t$. Consequently $G$ is a <Lebesgue integrable function>, and every $f_n$ is <Lebesgue integrable>.
The <dominated convergence theorem> states that <almost everywhere> convergence of <measurable functions> under one integrable absolute-value bound implies <L1 convergence>, and in particular convergence of their <integrals>. Here the pointwise <limit> is zero and the dominator is $G$, so
$$
\boxed{\lim_{n\to\infty}\int_0^\infty f_n(x)\,dx=0.}
$$
In fact we obtain the stronger conclusion $\|f_n\|_1\to0$. A quantitative version follows by retaining the factors $1/n$ in the bound:
$$
\|f_n\|_1\leq\frac{A\log n+B}{n},
\quad
A=\int_0^\infty\frac{|\sin x|}{x^q}\,dx,
\quad
B=\int_0^\infty\frac{\log(1+x)|\sin x|}{x^q}\,dx.
$$
The preceding endpoint estimates show $A,B<\infty$, so the <L1 norm> is $O(\log n/n)$ as $n\to\infty$, with $O$ denoting <Big O notation>. This is <logarithmic reweighting under dominated convergence>.
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