To show that the integral $\int_{-\infty}^{\infty}\frac{(p'(x))^2}{(p(x))^2+(p'(x))^2}dx$ converges and is less or equal than $n^{3/2}\pi$

Question

Consider a polynomial $p \in \mathbb{R}[x]$ of degree $n$ and with no real roots. Prove that$$\int_{-\infty}^{\infty}\frac{(p'(x))^2}{(p(x))^2+(p'(x))^2}dx$$converges, and is less or equal than $n^{3/2}\pi$

My approach

Now let $x_1, x_2, \dots, x_n$ be the roots of $p$. By Cauchy-Schwarz

$$(\sum_{k=1}^{n}{\frac{1}{x-x_k}})^2\leq n\sum_{k=1}^{n}{\frac{1}{|x-x_k|^2}}$$

I don't know what to do next. If I am wrong kindly provide a detailed answer in the answer section. I have shown what I have thought of or what I have done .

Can anyone confirm if my thought process is right?

Just a reminder... This question has been lying unanswered for a long time

Thank you.

Answer 1

First of all, we can define: $$p_n(x)=\sum_{k=0}^na_kx^k\tag{1}$$ $$p_n'(x)=\sum_{k=0}^nka_kx^{k-1}$$

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Now by the multinomial theorem: $$\left(\sum_{i=1}^mx_i\right)^n=\sum_{\sum j_i=n}{n\choose{j_1,j_2...j_m}}\prod_{t=1}^mx_t^{j_t}$$ From this you should be able to come up with an expression for: $p_n^2$ and $p_n'^2$.

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Now also note that from what we know (due to there being no real roots): $$p_n(x_0)=\sum_{k=0}^na_kx_0^k\ne0\,\,\,\,x_0\in\mathbb{R}$$ we know that: $$p_n(x)^2=O(x^{2n})$$ $$p_n'(x)^2=O(x^{2(n-1)})$$ and so it is clear that: $$\frac{p_n'(x)^2}{p_n(x)^2+p_n'(x)^2}=O\left(\frac1{x^2}\right)$$