Let $p \in [1,\infty[$ and show that $f_{n} \in \mathcal{L}^{p}[0,\infty[$
where $f_{n}(x):=\frac{n\sqrt{x} \sin{(x)}}{1+nx^{2}}$ for $x \in [0,\infty[$
My idea:
Let $n \in \mathbb N$,
$\vert \vert f_{n} \vert \vert_{p}^{p}=\int_{0}^{\infty}\vert \frac{n\sqrt{x} \sin{(x)}}{1+nx^{2}}\vert^{p} dx=\int_{0}^{\infty}\vert \frac{\sqrt{x} \sin{(x)}}{\frac{1}{n}+x^{2}}\vert^{p} dx=\int_{0}^{\infty} \frac{x^{\frac{p}{2}} \vert\sin^{p}{(x)}\vert}{(\frac{1}{n}+x^{2})^{p}} dx$
Now I know that
$\frac{x^{\frac{p}{2}} \vert\sin^{p}{(x)}\vert}{(\frac{1}{n}+x^{2})^{p}} 1_{[0,\infty[}\leq x^{\frac{p}{2}}1_{[0,1[}+\frac{\vert\sin^{p}{(x)}\vert}{x^{\frac{3}{2}p}}1_{[1,\infty[}$. Note that $\frac{\vert\sin^{p}{(x)}\vert}{x^{\frac{3}{2}p}}1_{[1,\infty[}\leq \frac{\vert\sin^{p}{(x)}\vert}{x^{p}}1_{[1,\infty[}$
Now I know that $x^{\frac{p}{2}}1_{[0,1[} \in L^{p}$
And I remember that $\int_{0}^{\infty}\frac{\sin{(x)}}{x}dx=\frac{\pi}{2}$ but $\int_{0}^{\infty}\frac{\vert\sin{(x)}\vert}{x}dx$ is divergent, so obviously my bounds do not help me further. Any tips are appreciated.
We have
$\left | \frac{n\sqrt{x} \sin{(x)}}{1+nx^{2}}\right|\le \left | \frac{\sqrt{x}}{x^{2}}\right|=|x^{-3/2}|$.
On the other hand, if $0<x\le 1,$ then
$\left | \frac{n\sqrt{x} \sin{(x)}}{1+nx^{2}}\right|=\left|\frac{n\sqrt x\cdot (x+O(x^3))}{1+nx^2}\right|=\left|\frac{nx^{5/2}+O(x^{7/2}))}{1+nx^2}\right|$,
and these two facts combine to prove the claim.