My goal is to show that the function $f(x) = \sin(x)\frac{\ln(x^2+1)}{\lvert x \rvert^n}$ where $n \in (\frac{1}{2},1)$ is in $L^2(\mathbb{R})$.
I know that when evaluating the integral
$$ \int_{-\infty}^{\infty} \lvert \sin(x)\frac{\ln(x^2+1)}{\lvert x \rvert^n} \rvert^2 dx$$
the term $\lvert \sin^2(x) \rvert$ is bounded by 1. But I can't figure out how to go from there.
There are really two things to check here: convergence as $x \to \infty$ and convergence at $x = 0$.
Near $x = 0$, we have $\sin(x) \sim x$ and $\ln(1 + x^2) \sim x^2$ and thus $\lvert f(x) \rvert^2 \sim x^{6-2n}$ and so the integral converges near $x =0$ so long as $n < \tfrac 7 2$.
The behaviour at infinity is dominated by the polynomial decay. Indeed, for any $\epsilon > 0$, there is $C > 0$, such that $\ln(1+x^2) \le C \lvert x\rvert^\epsilon$ for all sufficiently large $x$. Thus $$\lvert f(x) \rvert^2 \le \frac{C}{ \lvert x \rvert^{2n-\epsilon}}$$ and for sufficiently small $\epsilon >0$, we have $2n - \epsilon > 1$ (so long as $n > \tfrac 1 2$) so the integral converges at $x = \infty$.
Thus it seems that $f \in L^2(\mathbb R)$ whenever $n \in (\tfrac 1 2, \tfrac 7 2)$.