Show that the Cauchy principal value of $ \int_{-\infty}^{\infty}\frac{P(x)}{Q(x)}dx$ exists when $\mbox{deg}(Q)=\mbox{deg}(P)+1$.

Question

Let $R=\frac{P}{Q}$ where $P$ and $Q$ are polynomials such that $Q$ has not zeros in $\mathbb{R}$ and $\mbox{deg}(Q)=\mbox{deg}(P)+1$. Show that the Cauchy principal value of $ \int_{-\infty}^{\infty}R(x)dx$ exists, that is,

$$\lim_{s\rightarrow \infty} \int_{-s}^{s}R(x)dx \qquad \mbox{exists.}$$

Remark: I know that $\lim_{s\rightarrow \infty} \int_{-s}^{s}\frac{\hat{P}(x)}{\hat{Q}(x)}dx $ exists when we have $\mbox{deg}(\hat{Q})\geq\mbox{deg}(\hat{P})+2$ then considering $P(x)=a_{0}+a_{1}x+\cdots+a_{n}x^{n}$ and $Q(x)=b_{0}+b_{1}x+\cdots+b_{n}x^{n}+b_{n+1}x^{n+1}$, we have

$$\lim_{s\rightarrow \infty} \int_{-s}^{s}R(x)dx=\lim_{s\rightarrow \infty} \int_{-s}^{s}\frac{a_{0}+a_{1}x+\cdots+a_{n-1}x^{n-1}}{Q(x)}dx+\lim_{s\rightarrow \infty} \int_{-s}^{s}\frac{a_{n}x^{n}}{Q(x)}dx $$

Note that $\lim_{s\rightarrow \infty} \int_{-s}^{s}\frac{a_{0}+a_{1}x+\cdots+a_{n-1}x^{n-1}}{Q(x)}dx $ exists. So my problem is summarized to show that $ \lim_{s\rightarrow \infty} \int_{-s}^{s}\frac{a_{n}x^{n}}{Q(x)}dx $ exists, for this purpose, I have tried to consider the curve $\gamma=\gamma_{1}+\gamma_{2}$ where $\gamma_{2}(t)=se^{it}$ where $t\in[0,\pi]$ is the semicircle and $\gamma_{1}(t)=t$ is the segment joining $-s$ with $s$ the problem is in show that $$\lim_{s\rightarrow \infty} \int_{\gamma_{2}}\frac{a_{n}z^{n}}{Q(z)}dz=0.$$

Answer 1

Without loss of generality we may assume that the leading coefficient of $P$ and $Q$ is $1$. Let $d$ be the degree of $P$ and $Q=x\,P+p$, where $p$ is a polynomial of degree at most $d$. Then $$ \frac PQ=\frac1x-\frac{p}{x\,Q}. $$ The principal value of $1/x$ exists, and $p/(x\,Q)$ is integrable since the degree of $p$ is at most $d$ and the degree of $x\,Q$ is $d+2$.

Added to answer Diego's comment:

Let $$\begin{align} P(x)&=x^d+a_{d-1}x^{d-1}+\dots+a_1x+a_0\\ Q(x)&=x^{d+1}+b_dx^d+\dots+b_1x+b_0\\ p(x)&=Q(x)-x\,P(x)=(b_d-a_{d-1})x^d+\dots(b_1-a_0)x+b_0. \end{align}$$ Now $$ \frac{P}{Q}=\frac1x\,\frac{x\,P}{Q}=\frac1x\,\frac{x\,P+p-p}{x\,P+p}=\frac1x\Bigl(1-\frac pQ\Bigr). $$ Example: $$ \frac{x+2}{x^2+3\,x+1}=\frac1x-\frac{x+1}{x(x^2+3\,x+1)}. $$