Convergence of $\int_1^\infty\frac{x^2+kx}{x^4+k^px^2+k^2}dm(x)$; $k\in\mathbb{N},1\le p<\infty$ Consider the integrals
$$I(k,p)=\int_1^\infty\frac{x^2+kx}{x^4+k^px^2+k^2}dm(x),$$ with $k\in\mathbb{N}$ and $1\le p<\infty$.

*

*For which $p$ does the integrand have an integrable majorant?


*For which $p$ do the integrals tend to $0$?
I'm thinking that the integrals have an integrable majorant when $p\ge 1$, but I'm not sure if it's true or how to show it.

Here we obtain an integral majorant ( independent from $k$) and convergence for $p>1$.  For $p=1$ we obtain convergence but no integrable majorant ( independent from $k$)
Case $p>1$
$$ \frac{kx}{x^4+k^px^2+k^2}\leq \frac{kx}{x^4+k^px^2}=\frac{k}{x(x^2+ k^p)}=:\phi(x,k)
$$
$\partial_k\phi = \frac{(x^2+k^p)-pk^p}{x(x^2+k^p)^2}=\frac{x^2-(p-1)k^p}{x(x^2+k^p)^2}=0$ has solution $k_*=\Big(\frac{x^2}{p-1}\Big)^{1/p}$, and $\partial^2_{kk}\phi(k_*)<0$. Since $\phi(x,0)=0=\lim_{k\rightarrow\infty}\phi(x,k)$, we have that $\phi(x,k)\leq \phi(x,k^*)$ for all $k$ and so
$$
\phi(x,k)=\frac{k}{x(x^2+ k^p)}\leq C_p\frac{x^{2/p}}{x^3}=C_px^{\tfrac2p -3}
$$
for all $x\geq1$ and some constant  $C_p$ depending only on $p$. Then, for $p>1$
$$
f_{p,k}(x):=\frac{x^2+kx}{x^4+k^px^2+k^2}\leq \frac{1}{x^2}+C_px^{\tfrac2p -3}\in L_1([1,\infty),m)
$$
For each $x\geq1$
$$
\lim_{k\rightarrow\infty}\frac{x^2+kx}{x^4+k^px^2+k^2}=0
$$
Hence, if $p>1$, $\lim_{k\rightarrow\infty}\int^\infty_1 f_{p,k}\,dm =0$ by dominated convergence.

Case  $p=1$:
Convergence can be addressed by direct computation:
$$\int^\infty_1 f_{1,k}=\int^\infty_1\frac{x^2\,dx}{x^4+kx^2+1} +k\int^\infty_1\frac{x\, dx}{x^4+kx^2+k^2}\,dx$$
Since $g_k(x):=\frac{x^2}{x^4+kx^2+k^2}\leq \frac{1}{x^2}$ and $0<g_k(x)\xrightarrow{k\rightarrow\infty}0$, $$I_k:=\int^\infty_1\frac{x^2\,dx}{x^4+kx^2+1} \xrightarrow{k\rightarrow\infty}0$$
The second intergral can be estimated directly:
\begin{aligned}
J_k=k\int^\infty_1\frac{x}{x^4+kx^2+k^2}\,dx&=k\int^\infty_1\frac{x\,dx}{(x^2+\tfrac{k}{2})^2+\tfrac34k^2}\\
&=\frac{k}{2}\int^\infty_{1+\tfrac{k}{2}}\frac{du}{u^2+\tfrac{3}{4}k^2}\\
&=\frac{1}{\sqrt{3}}\Big(\frac{\pi}{2}-\arctan\big(\tfrac{2(k+\tfrac12)}{\sqrt{3}k}\big)\Big)\xrightarrow{k\rightarrow\infty}\frac{1}{\sqrt{3}}\Big(\frac{\pi}{2}-\arctan\big(\tfrac{2}{\sqrt{3}}\big)\Big)
\end{aligned}
It follow from this that there is not integrable majorant for $f_{1,k}$ (that is independent of $k$)

Alternative method: only $p\geq2$
From
$$\sqrt{x^4k^px^2}=x^3 k^{p/2}\leq\frac{x^4  + k^px^2}{2}$$
we obtain
\begin{aligned}
f_{p,k}(x)&=\frac{x^2+kx}{x^4+k^px^2+k^2}\leq \frac{1}{x^2}+\frac{kx}{x^4+k^px^2}\leq \frac{1}{x^2}+ \frac{1}{2x^2}k^{1-p/2}
\end{aligned}
For $p\geq2$
$$0<f_{p,k}(x)\leq \frac{3}{2x^2}$$
since $x\geq1$ and $k\geq1$.

