For which $p\in\mathbb{R}$ does $f_n(x)=\frac{nx^p}{1+n^2x^2}dx$, where $n\geq 1$, become integrable on $(0,\infty)$ with respect to Lebesgue Measure Question: For which $p\in\mathbb{R}$ does the family of functions $f_n(x)=\frac{nx^p}{1+n^2x^2}$ $(n\geq1)$ have an integrable upper bound on $(0,\infty)$ with respect to the Lebesgue measure?
My Attempt:  First, note that $\frac{nx^p}{1+n^2x^2}=\frac{x^p}{\frac{1}{n}+nx^2}$.  Now, $\frac{d}{dn}(1+nx^2)\approx-\frac{1}{n^2}+x^2=0$ (to minimize).  Therefore, $x^2=\frac{1}{n^2}$ and so $n=\frac{1}{x}$ whenever $x\leq1$ since $n\geq1$ by assumption.  Thus, $\min_n(\frac{1}{1/x}+\frac{1}{x}x^2)=2x$, and so $\sup_n(\frac{nx^p}{1+n^2x^2})\approx\frac{x^p}{2x}$.  Now, we have that $\frac{1}{2}\int_0^1 x^{p-1}dx<\infty$ whenever $p>0$.  Now, as $x\rightarrow\infty$, $\frac{nx^p}{1+n^2x^2}\rightarrow\frac{nx^p}{n^2x^2}=\frac{x^{p-2}}{n}\leq x^{p-2}$, since $n\geq1$.  So, $\int_0^1 x^{p-2}dx<\infty$ if $p<1$.  Hence we have an integrable upper bound whenever $p\in(0,1)$.

The change of variables $u=nx$ gives
$$\int^\infty_0\frac{nx^p}{1+n^2x^2}\,dx=\frac{1}{n^p}\int^\infty_0\frac{u^p}{1+u^2}\,du=\frac{1}{n^p}I_p$$

*

*If $p<-1$, then $I_p\geq\frac12\int^1_0u^p\,du=\infty$

*If $-1<p<0$, the $I_p<\infty$ and so $\frac{1}
{n^p}I_p\xrightarrow{n\rightarrow\infty}\infty$

*If $0<p<1$, $I_p<\infty$ and so $\frac{1}{n^p}I_p\xrightarrow{n\rightarrow\infty}0$

*If $1\leq p$, then $I_p=\infty$

*When $p=0$, you get $\frac{\pi}{2}$
