Given that $f\in L^p(\mathbb{R}),g\in L^q(\mathbb{R})$, $1\le p,q\le\infty.$ Define $F(x)=\int_0^xf(t)dt$.
How can one show $$(\vert x\vert+1)^{-a}F(x)g(x)\in L^1(\mathbb{R})$$ when $a>2-\frac{1}{p}-\frac{1}{q}$
I can show that $F$ is continuous. I cannot use Holder's inequality since $p$ and $q$ are not conjugate. Integration by parts came to mind but I am not familiar with an analogue of this method for Lebesgue integral functions. Even if I assume that $p,q$ are conjugate it doesn't help much since $f$ is not in the integrand.
We know that $|F(x)|\le|x|^{1-1/p}\|f\|_p$ since Jensen (or Hölder) says $$ \begin{align} \left|\int_0^xf(t)\,\mathrm{d}t\right| &\le|x|^{1-1/p}\left(\int_0^x|f(t)|^p\,\mathrm{d}t\right)^{1/p}\\ &\le(|x|+1)^{1-1/p}\left(\int_0^x|f(t)|^p\,\mathrm{d}t\right)^{1/p}\tag{1} \end{align} $$ Therefore, $$ \left|(|x|+1)^{-a}F(x)g(x)\right|\le(|x|+1)^{1-a-1/p}\|f\|_p|g(x)|\tag{2} $$ We know that $(|x|+1)^{1-a-1/p}\in L^r$ for $1-a-\frac1p\lt-\frac1r$.
Hölder says that if $\frac1q=1-\frac1r\gt2-a-\frac1p$, then $(2)\in L^1$. Thus, if $$ a\gt2-\frac1q-\frac1p\tag{3} $$ then $(|x|+1)^{-a}F(x)g(x)\in L^1$.