There are a finite number of poles of
\(\frac{P}{Q}\) that lie in the upper half-plane, so we can find a real number
\(R\) such that the poles all lie inside the contour
\(C\text{,}\) which consists of the segment
\(-R \le x \le R\) of the
\(x\)-axis and the upper semicircle
\(C_R\) of radius
\(R\) shown in
Figure 8.3.5. By properties of integrals,
\begin{equation*}
\int_{-R}^{R}\frac{P(x)}{Q(x)}\,dx = \int_{C}\frac{P(z)}{Q(z)}\,dz - \int_{C_R}\frac{P(z)}{Q(z)}\,dz\text{.}
\end{equation*}
Using the residue theorem, we rewrite this equation as
\begin{equation}
\int_{-R}^{R}\frac{P(x)}{Q(x)}\,dx = 2\pi i\sum_{j=1}^k\mathrm{Res}\left[\frac{P}{Q},z_j\right] -\int_{C_R}\frac{P(z)}{Q(z)}\,dz\text{.}\tag{8.3.3}
\end{equation}
Our proof will be complete if we can show that \(\int_{C_R}\frac{P(z)}{Q(z)}\,dz\) tends to zero as \(R \to \infty\text{.}\) Suppose that
\begin{align*}
P(z) \amp = a_{m}z^{m}+a_{m-1}z^{m-1} + \cdots +a_1z + a_0, \text{ and }\\
Q(z) \amp = b_nz^n+b_{n-1}z^{n-1} + \cdots + b_1z + b_0\text{.}
\end{align*}
Then
\begin{equation*}
\frac{zP(z)}{Q(z)} = \frac{z^{m+1}(a_{m} + a_{m-1}z^{-1}+\cdots + a_1z^{-m+1}+a_0z^{-m})}{z^n(b_n + b_{n-1}z^{-1} + \cdots + b_1z^{-n+1}+b_0z^{-n})}\text{,}
\end{equation*}
so
\begin{align*}
\lim_{|z| \to \infty}\frac{zP(z)}{Q(z)} \amp = \lim_{|z| \to \infty}\frac{z^{m+1}(a_{m}+a_{m-1}z^{-1} + \cdots + a_1z^{-m+1}+a_0z^{-m})}{z^n(b_n+b_{n-1}z^{-1} + \cdots + b_1z^{-n+1}+b_0z^{-n})}\\
\amp = \left(\lim_{|z| \to \infty}\frac{z^{m+1}}{z^n}\right) \left(\lim_{|z| \to \infty}\frac{a_{m}+a_{m-1}z^{-1} + \cdots + a_1z^{-m+1}+a_0z^{-m}}{b_n+b_{n-1}z^{-1} + \cdots +b_1z^{-n+1}+b_0z^{-n}}\right)\text{.}
\end{align*}
Since \(n \ge m+2\text{,}\) this limit reduces to \(0(\frac{a_{m}}{b_n}) =0\text{.}\) Therefore, for any \(\varepsilon >0\text{,}\) we may choose \(R\) large enough so that \(|\frac{zP(z)}{Q(z)}| \lt \frac{\varepsilon}{\pi}\) whenever \(z\) lies on \(C_R\text{.}\) But this means that
\begin{equation}
\left|\frac{P(z)}{Q(z)}\right| \lt \frac{\varepsilon}{\pi |z|} = \frac{\varepsilon}{\pi R}\tag{8.3.4}
\end{equation}
whenever
\(z\) lies on
\(C_R\text{.}\) Using the ML inequality (
Theorem 6.2.19) and Inequality
(8.3.4), we get
\begin{equation*}
\left|\int_{C_R}\frac{P(z)}{Q(z)}\,dz\right| \le \int_{C_R}\frac{\varepsilon}{\pi R}\,|dz| = \frac{\varepsilon}{\pi R}\pi R=\varepsilon\text{.}
\end{equation*}
Since \(\varepsilon >0\) was arbitrary, we conclude that
\begin{equation}
\lim_{R \to \infty}\int_{C_R}\frac{P(z)}{Q(z)}\,dz=0\text{.}\tag{8.3.5}
\end{equation}
If we let
\(R \to \infty\) and combine Equations
(8.3.3) and
(8.3.5), we arrive at the desired conclusion.
