Evaluating $\int_{0}^{\infty} \frac{1}{a_{n}x^{n} + ... + a_{2}x^{2} + a_{o}}dx$ via Residue Theory? In the text "Functions of a Complex Variable" by Robert E. Greene and Steven G.Krantz I'm having trouble verifying my solution to $\text{Problem (1)}$

$\text{Problem (1)}$
Using Calculus of Residue evaluate the following 
$$\int_{0}^{\infty} \frac{1}{a_{n}x^{n} + ... + a_{2}x^{2} + a_{o}}dx \, \, \, $$
$\text{Remark}$
$p(x)$ is any polynomial with no zero's on the nonnegative real axis

$\text{Solution}$
For $(1)$ real variable methods would be fruitless we have to take the, 
$$\oint_{\eta_{R}} \frac{\log(z)}{a_{n}z^{n} + ... + a_{2}z^{2} + a_{o}}dz.$$
For our choice $f$, we initially let
$$\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \eta_{R}^{1}(t)  =  t + i/\sqrt{2R},  \, \, \, \,   1/\sqrt{2R} \leq t \leq R,$$
$$\eta_{R}^{2}(t)= Re^{it}, \, \, \, \,  \theta_{0} \leq t \leq 2 \pi - \theta_{0},$$
where $\theta_{0} = \theta_{0}(R) = \sin^{-1}(1/(R \sqrt{2R}))$
$$\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \eta_{R}^{3}(t)  =  R -t -i/\sqrt{2R},  \, \, \, \, 0 \leq t \leq R-1/\sqrt{2R}.$$
$$\eta_{R}^{4}(t)  =  e^{it}/\sqrt{R}, \, \, \, \, \,  \, \, \, \, \, \, \, \, \, \, \, \, \, \pi/4 \leq t \leq 7 \pi /4.$$
It's important to consider that, 
$$\oint_{\eta_{R}} \frac{\log(z)}{a_{n}z^{n} + ... + a_{2}z^{2} + a_{o}}dz = 2 \pi i \bigg( \sum_{j} \operatorname{Ind_{\eta_{R}}}(P_{j}) \cdot \operatorname{Res_{\eta_{R}}}(P_{j}) \bigg) $$
Clearly our choice of $f$ has a pole of the order of $P$ and a pole of the order $n$. Clearly, 
\begin{align*} 
\operatorname{Res_{f}(P)} &=  \frac{1}{(n-1)!} \bigg( \partial_{z} \bigg)^{n-1} \bigg( (z-n)^{n} \frac{\log(z)}{a_{n}z^{n} + ... + a_{2}z^{2} + a_{o}}\bigg) \bigg|_{z=P}\\ 
\, \, \,  &=  \frac{1}{(n)!} \bigg( \partial_{z} \frac{\log(z)}{a_{n}x^{n} + ... + a_{n}z^{2} + a_{o}}\bigg|_{z = P} \bigg) \\
 &= \frac{1}{(n!)}\frac{\log(z) - a_{n}z^{n} + ... + a_{2}P^{2} + a_{o}}{(\log(x)^{2})}\\   &= \frac{1}{(n!)}\frac{\log(P) - a_{n}P^{n} + ... + a_{2}P^{2} + a_{o}}{(\log(P)^{2})}.
\end{align*}
Putting the pieces together,
$(*)$
$$\oint_{\eta_{R}} \frac{\log(z)}{a_{n}z^{n} + ... + a_{2}z^{2} + a_{o}}dz  = 2 \pi i \bigg(  \frac{1}{(n!)}\frac{\log(P) - a_{n}P^{n} + ... + a_{2}P^{2} + a_{o}}{(\log(P)^{2})} \bigg) \cdot 1$$ 
Applying the  Residue Theorem unfortunately isn't enough to finish our game so it becomes imperative to claim that
$(**)$
$$ \Bigg| \lim_{R \rightarrow \infty}\oint_{\eta^{2}_{R}} f(z)dz \Bigg| \rightarrow 0 $$
and that, 
$(***)$
$$ \Bigg| \lim_{R \rightarrow \infty}\oint_{\eta^{4}_{R}} f(z)dz\Bigg| \rightarrow 0.$$
A particular device used to justify convergence over $\eta_{4}$ and $\eta_{2}$ is the fact that 
$$\bigg(\log \bigg( \frac{x + i \sqrt{2R}}{(x-i/\sqrt{2R}} \bigg) \bigg)\rightarrow -2 \pi i \text{.}$$
We will return to this particular device after dealing with our analysis of convergence over $\eta_{4}$ and $\eta_{2}$. First we take that, 
$$\sum_{\psi}^{4} \bigg(\oint_{\eta_{R}^{\psi}} \frac{\log(z)}{a_{n}z^{n} + ... + a_{2}z^{2} + a_{o}}dz \bigg). $$
Now over $\eta_{2}$ one can see that, 
\begin{align*}
\bigg| \oint_{\eta_{R}^{2}}\frac{\log(z)}{a_{n}z^{n} + ... + a_{2}z^{2} + a_{o}}dz\bigg|& = \bigg| \int_{-R}^{+Ri} \frac{\log(Re^{it})}{a_{n}(Re^{it})^{n} + ... + a_{2}(Re^{it})^{2} + a_{o}} iRe^{i \theta} d \theta\bigg|\\&=  \int_{-R}^{+Ri} \bigg|\frac{\log(Re^{it})}{{a_{n}(Re^{it})^{n} + ... + a_{2}(Re^{it})^{2} + a_{o}}} \bigg| \big| iRe^{i \theta} d \theta \big|\\&= \int_{-R}^{+Ri} \frac{\bigg|\log(Re^{it}) \bigg|}{\bigg| {a_{n}(Re^{it})^{n} + ... + a_{2}(Re^{it})^{2} + a_{o}} \bigg|}  \bigg|iRe^{i \theta} \bigg| d \theta  \bigg| \\& = \int_{\theta_{0}}^{2 \pi - \theta_{0}} \frac{\bigg|\log(Re^{it}) \bigg|}{\bigg|{a_{n}(Re^{it})^{n} + ... + a_{2}(Re^{it})^{2} + a_{o}} \bigg|}  \bigg|iRe^{i \theta} \bigg| \bigg|d \theta  \bigg|.
\end{align*}
Now we can establish a precise estimate for $\eta_{2}$
$$\bigg| \oint_{\eta_{R}^{2}} \frac{\log(z)}{a_{n}z^{n} + ... + a_{2}z^{2} + a_{o}}dz\bigg| \leq  \frac{\ln(R) + \pi }{R^{n} - a_{o}} \pi R \, \, \text{as} \, \, \, R \rightarrow \infty.$$
A similar process for $\eta_{4}$ says that, 
\begin{align*}
\bigg| \oint_{\eta_{R}^{4}} \frac{\log(e^{it}/\sqrt{R})}{{a_{n}z^{n} + ... + a_{2}z^{2} + a_{o}}}  dz\bigg|& =  \oint_{\eta_{R}^{4}} \bigg| \frac{\log(e^{it}/\sqrt{R})}{{a_{n}(e^{it}/\sqrt{R})^{n} + ... + a_{2}(e^{it}/ \sqrt{R})^{2} + a_{o}}}  iRe^{i \theta} d \theta\bigg|\\&= \oint_{\eta_{R}^{4}}  \frac{\bigg|\log(e^{it}/\sqrt{R}) \bigg|}{\bigg|a_{n}(e^{it}/\sqrt{R})^{n} + ... + a_{2}(e^{it}/ \sqrt{R})^{2} + a_{o} \bigg|}  iRe^{i \theta} d \theta \\&= \oint_{\eta_{R}^{4}}  \frac{\bigg| \log(e^{it})- \frac{1}{2}\log(R^{}) \bigg|}{ \bigg|a_{n}(e^{it}/\sqrt{R})^{n} + ... + a_{2}(e^{it}/ \sqrt{R})^{2} + a_{o} \bigg|} \bigg|  iRe^{i \theta} d \theta \bigg|\\& =\oint_{\frac{\pi}{4}}^{\frac{7 \pi}{4}}  \frac{\bigg| it\log(e^{})- \frac{1}{2}\log(R^{}) \bigg|}{ \bigg|a_{n}(e^{it}/\sqrt{R})^{n} + ... + a_{2}(e^{it}/ \sqrt{R})^{2} + a_{o}\bigg|} \bigg|  iRe^{i \theta}\bigg| d \theta \bigg|.  \end{align*}
Now we can establish a precise estimate for $\eta_{4}$ hence, 
$$\bigg| \oint_{\eta_{R}^{4}} \frac{\log(e^{it}/\sqrt{R})}{{a_{n}z^{n} + ... + a_{2}z^{2} + a_{o}}}  dz\bigg|   \leq  \text{length}(\eta_{R}^{4})  \cdot \sup_{\eta_{R}^{4}}(g) \leq \pi R \frac{O(\log(R))}{\sqrt{R}} \, \text{as} \, R \rightarrow \infty.$$
By taking care to provide estimates over $\eta_{2}$ and $\eta_{4}$ we have proved $(***)$ and $(**)$. 
Applying Szeto's Lemma it becomes apparent that, 
$(****)$
$$\oint_{\eta^{1}_{R}}g(z) dz + \oint_{\eta^{3}_{R}}g(z) dz \rightarrow  - 2 \pi i \int_{0}^{\infty} \frac{1}{a_{n}t^{n} + ... + a_{2}t^{2} + a_{o}}dx \, \, \,$$ 
Now taking $(*)$, $(**)$, $(***)$, $(****)$ taken together yield, 
$$\int_{0}^{\infty} \frac{1}{a_{n}x^{n} + ... + a_{2}x^{2} + a_{o}}dx = 2 \pi i \bigg(  \frac{1}{(n!)}\frac{\log(P) - a_{n}P^{n} + ... + a_{2}P^{2} + a_{o}}{(\log(P)^{2})} \bigg)$$
 A: Let $p(X)\in\mathbb{C}[X]$ be a nonconstant polynomial whose roots are in $\mathbb{C}\setminus\mathbb{R}_{\geq 0}$.  If $p(X)$ has only even powers of $X$ (i.e., $p(X)=q\left(X^2\right)$ for some nonconstant polynomial $q(X)\in\mathbb{C}[X]$), then the answer can be made simpler.  Let $z_1,z_2,\ldots,z_l$ be the roots of $p(X)$ in the upper half plane $$\mathbb{H}^+:=\big\{z\in\mathbb{C}\,|\,\text{Im}(z)>0\big\}\,,$$
respectively, with multiplicities $m_1,m_2,\ldots,m_l$.  (Thus, $p(X)$ also has $-z_1,-z_2,\ldots,-z_l$ as roots, respectively, with multiplicities $m_1,m_2,\ldots,m_l$.  Ergo, $n=2\,\left(m_1+m_2+\ldots+m_l\right)$, where $n$ is the degree of $p(X)$, which must be an even positive integer.) 
For $R>0$, consider the contour positively oriented contour $C_R$ given by
$$[-R,+R]\cup\big\{R\,\exp(\text{i}t)\,\big|\,t\in[0,\pi]\big\}\,.$$
Let
$$I:=\int_0^\infty\,\frac{1}{p(x)}\,\text{d}x\text{ and }K(R):=\oint_{C_R}\,\frac{1}{p(z)}\,\text{d}z\,.$$
Thus,
$$2I=\lim_{R\to\infty}\,K(R)=2\pi\text{i}\,\sum_{j=1}^{l}\,\text{Res}_{z=z_j}\left(\frac{1}{p(z)}\right)\,.$$
Therefore,
$$I=\pi\text{i}\,\sum_{j=1}^l\,\frac{1}{\left(m_l-1\right)!}\,\lim_{z\to z_j}\,\frac{\text{d}^{m_j-1}}{\text{d}z^{m_j-1}}\,\frac{\left(z-z_j\right)^{m_j}}{p(z)}\,.\tag{$\square$}$$
In particular, if $m_j=1$ for every $j=1,2,\ldots,l$, then we get $l=\dfrac{n}{2}$ and
$$I=\pi\text{i}\,\sum_{j=1}^{\frac{n}{2}}\,\frac{1}{p'\left(z_j\right)}=\frac{\pi\text{i}}{2}\,\sum_{j=1}^{\frac{n}{2}}\,\frac{1}{z_j\,q'\left(z_j^2\right)}\,.\tag{#}$$
Note that (#) can be proven by ($\star$) from my other answer.  Similarly, ($\square$) also follows from (*).
For example, if $p(X)=\left(X^2+1\right)\,\left(X^2+4\right)$, then $q(X)=(X+1)\,(X+4)$.  You can use the partial fraction decomposition to get
$$\int\,\frac{1}{p(x)}\,\text{d}x=\frac{1}{6}\,\Biggl(2\,\arctan(x)-\arctan\left(\frac{x}{2}\right)\Biggr)+\text{constant}\,,$$
so that
$$I=\int_0^\infty\,\frac{1}{p(x)}\,\text{d}x=\frac{\pi}{12}\,.$$
However, using (#), we get 
$$I=\frac{\pi\text{i}}{2}\,\left(\frac{1}{\text{i}\cdot 3}+\frac{1}{2\text{i}\cdot(-3)}\right)=\frac{\pi}{12}\,.$$
On the other hand, if $p(X)=\left(X^2+1\right)^3$, then we need to use ($\square$).  Note that $l=1$, with $z_1=\text{i}$ and $m_1=3$.  Hence,
$$I=\int_0^\infty\,\frac{1}{\left(x^2+1\right)^3}\,\text{d}x$$
equals
$$\frac{\pi\text{i}}{2!}\,\lim_{z\to\text{i}}\,\frac{\text{d}^2}{\text{d}z^2}\,\frac{1}{(z+\text{i})^3}=\frac{\pi\text{i}}{2}\,\left(\frac{12}{(2\text{i})^5}\right)=\frac{3\pi}{16}\,.$$
Using the partial fraction decomposition yields
$$\int\,\frac{1}{p(x)}\,\text{d}x=\frac{1}{8}\,\left(\frac{x\,\left(3\,x^2+5\right)}{\left(x^2+1\right)^2}+3\,\arctan(x)\right)+\text{constant}\,,$$
so we get the same answer $I=\dfrac{3\pi}{16}$.

Indeed, if $m_j=1$ for every $j=1,2,\ldots,l$ (so $l=\dfrac{n}{2}$), then 
$$\frac{1}{p(x)}=\sum_{j=1}^{\frac{n}{2}}\,\frac{2\,z_j}{p'\left(z_j\right)\,\left(x^2-z_j^2\right)}=\sum_{j=1}^{\frac{n}{2}}\,\frac{1}{q'\left(z_j^2\right)\,\left(x^2-z_j^2\right)}\,.$$  This provides an alternative proof of (#).

If $s(X)\in\mathbb{C}[X]$ is a polynomial of degree at most $n-2$ such that $p(X)$ and $s(X)$ do not have a common factor, and if $s(X)$ only has even-degree terms (namely, $s(X)=u\left(X^2\right)$ for some $u(X)\in\mathbb{C}[X]$), then we also have
$$\int_0^\infty\,\frac{s(x)}{p(x)}\,\text{d}x=\pi\text{i}\,\sum_{j=1}^l\,\frac{1}{\left(m_l-1\right)!}\,\lim_{z\to z_j}\,\frac{\text{d}^{m_j-1}}{\text{d}z^{m_j-1}}\,\frac{\left(z-z_j\right)^{m_j}\,s(z)}{p(z)}\,.$$
In particular, if $m_j=1$ for every $j=1,2,\ldots,l$, then $l=\dfrac{n}{2}$ and
$$\int_0^\infty\,\frac{s(x)}{p(x)}\,\text{d}x=\pi\text{i}\,\sum_{j=1}^{\frac{n}{2}}\,\frac{s\left(z_j\right)}{p'\left(z_j\right)}=\frac{\pi\text{i}}{2}\,\sum_{j=1}^{\frac{n}{2}}\,\frac{u\left(z_j^2\right)}{z_j\,q'\left(z_j^2\right)}\,.$$
A: Assume $P(x)$ and $Q(x)$ are polynomials with real coefficients and
I.
$$
deg(P(x))\leq deg(Q(x))-2
$$
and
II. $Q(x)$ have no roots $z_j$ in $\textbf{R}=(-\infty,+\infty)$.
Assume that $c$ is a simple closed curve that contains all roots in the upper plane and $\gamma_R$ is the sigment $[-R,R]$, $R>0$, along with 
$$
\delta(R):=\left\{z\in\textbf{C}:|z|=R\textrm{ and }0\leq arg(z)\leq \pi \right\},
$$ 
then if $\gamma_R$ encloses $c$, we can write:
$$
\oint_c\frac{P(z)}{Q(z)}dz=\int_{\gamma_R}\frac{P(z)}{Q(z)}dz=\int^{R}_{-R}\frac{P(x)}{Q(x)}dx+\int_{\delta(R)}\frac{P(z)}{Q(z)}dz
$$
From (I) exist $M>0$ and $z_0\in \textbf{C}$ such that
$$
\left|\frac{P(z)}{Q(z)}\right|\leq \frac{M}{|z|^2}\textrm{, }\forall |z|>|z_0|.
$$
Hence
$$
\lim_{z\rightarrow \infty}\left|\int_{\gamma_R}\frac{P(z)}{Q(z)}dz\right|\leq \lim_{z\rightarrow \infty}\int_{\gamma_R}\left|\frac{P(z)}{Q(z)}\right||dz|\leq
\lim_{R\rightarrow \infty}\frac{M}{R^2}\int_{\gamma}|dz|=
$$
$$
=\lim_{R\rightarrow \infty}\frac{M}{R^2}\pi R=0.
$$
Hence
$$
\int^{R}_{-R}\frac{P(x)}{Q(x)}dx+\int_{\delta(R)}\frac{P(z)}{Q(z)}dz=\oint_{c}\frac{P(z)}{Q(z)}dz=2\pi i\sum^{n}_{j=1}Res\left[\frac{P(z)}{Q(z)},z_j\right],
$$
where $z_j$ are the roots of $Q(z)=0$ in the upper plane. Taking the limit $R\rightarrow +\infty$, we arive to
$$
\int^{\infty}_{-\infty}\frac{P(x)}{Q(x)}dx=2\pi i\sum^{n}_{j=1}Res\left[\frac{P(z)}{Q(z)},z_j\right],
$$
which is the desired result.
Note that $n$ are the number of distinct roots (without counting multiplicity) in the upper plane. 
If we set $R(z):=\frac{P(z)}{Q(z)}$, then 
i) If $z_0$ is a pole of first class, we have
$$
Res\left[R(z),z_0\right]=\lim_{z\rightarrow z_0}\left((z-z_0)R(z)\right).
$$
ii) If $z_0$ is a pole of higher class$-k$, where $k$ integer greater than 1, then
$$
Res\left[R(z),z_0\right]=\frac{1}{(k-1)!}\lim_{z\rightarrow z_0}\left(\frac{d^{k-1}}{dz^{k-1}}(z-z_0)^k R(z)\right).
$$
CONTINUING NOTE.
Assume now the differential equation
$$
y'(x)=\sum^{N}_{n=1}a_ny(x)^n=H(y(x))
$$
This differential equation have solution
$$
x+C=\sum_{\rho/H}\frac{\log(y(x)-\rho)}{H'(\rho)},
$$
where the summation is taken over all roots of $H(x)=0$, (here $H$ is a simple polynomial function). If we invert $y$ we get
$$
y^{(-1)}(x)=\int\frac{1}{H(x)}dx=\sum_{\rho/H}\frac{\log(x-\rho)}{H'(\rho)}.
$$
Hence given a polynomial $H(x)$, with simple roots$-\rho$, then 
$$
\int\frac{1}{H(x)}dx=\sum_{\rho/H}\frac{\log(x-\rho)}{H'(\rho)}+C_1
$$
Now I use a lemma
Lemma (Mathematical Olympiad, Poland 1979)
Let $H(x)$ be a polynomial of degree $N>1$ with simple distinct roots $\rho_1,\rho_2,\ldots,\rho_N$. Then
$$
\sum_{\rho/H}\frac{1}{H'(\rho)}=0.
$$ 
From the above lemma we have
$$
S(h):=\sum_{\rho/H}\frac{\log(h-\rho)}{H'(\rho)}=\sum^{N-1}_{k=1}\frac{\log(h-\rho_k)}{H'(\rho_k)}+\frac{1}{H'(\rho_N)}\log(h-\rho_N)=
$$
$$
=\sum^{N-1}_{k=1}\frac{\log(h-\rho_k)}{H'(\rho_k)}-\sum^{N-1}_{k=1}\frac{1}{H'(\rho_k)}\log(h-\rho_N)=\sum^{N-1}_{k=1}\frac{\log(h-\rho_k)-\log(h-\rho_N)}{H'(\rho_k)}
$$
From which (easily) one can see that
$$
\lim_{h\rightarrow+\infty}S(h)=0.
$$
Hence
$$
\int^{\infty}_{0}\frac{dt}{H(t)}=-\sum_{\rho/H}\frac{\log(-\rho)}{H'(\rho)}.
$$
$qed$
Hence knowing that $H(x)=a(x-\rho_1)(x-\rho_2)\ldots (x-\rho_N)$ is a polynomial with simple roots, then the following formula (1) give us the value of 
$$
\int^{\infty}_{0}\frac{dt}{a(t-\rho_1)(t-\rho_2)\ldots(t-\rho_N)}=-\sum^{N}_{k=1}\frac{\log(-\rho_k)}{H'(\rho_k)}
$$
A: 
$\text{Batominovski's First Idenity}:$
Let $d > 1$ be an integer. Suppose that $p(X) = q(X^{d})$ and $s(X) = u(X^{d})$ for some $q(X)$, $u(X) \in \mathbb{C}[X]$ with $deg(q) > deg(u)$, and that $p(X)$ and $s(X)$ share no common roots. Assume further that $p(X)$ has no nonnegative real roots. Let $z_{1}, z_{2}, …, z_{n}$ be all the distinct roots of $p(X)$ with arguments in the interval $(0, \frac{2 \pi}{d})$, respectively, with multiplicities $m_{1}, m_{2}, …, m_{n}$ one can then show that 
$(*)$
$$\int_{0}^{\infty} \frac{s(x)}{p(x)}dx = \frac{2 \pi i}{1 - \exp( \frac{2 \pi  i}{d})} \sum_{j}^{l} \frac{1}{(m_{j} -1)!} \! \! \! \! \, \, \lim_{z \rightarrow z_{j}} \big( \partial_{z} \big)^{m_{j}-1} \frac{(z-z_{j})^{m_{j}}s(z)}{p(z)}.$$

$\text{Proof}$
Before embarking on our journey to conquer the Conjecture, first we must make some adjustments to our integral on the RHS side of $(*)$, so notice that
$$ \int_{0}^{\infty} \frac{s(x)}{p(x)}dx =  \int_{0}^{\infty} \frac{b(x-x_{1})(x-x_{2})(x-x_{3}) \cdot \cdot \cdot (x-x_{n})}{a(z-z_{1})(z-z_{2})(z-z_{3}) \cdot \cdot \cdot (z-z_{n})}dx$$
$$ \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \,  = \frac{2 \pi i}{1 - \exp( \frac{2 \pi  i}{d})} \sum_{j}^{l} \frac{1}{(m_{j} -1)!}\lim_{z \rightarrow z_{j}} \big( \partial_{z} \big)^{m_{j}-1} \frac{(z-z_{j})^{m_{j}}s(z)}{p(z)} $$
Now it's important to note that our choice of $f$ is analytic on an open set $\psi$ containing the closed upper half plane, 
$$ \mathcal{H} = \Big\{ z \in \mathcal{C} | \operatorname{Im(z)} \geq 0 \Big\}.$$
Furthermore consider that $R >0$ the positively oriented contour positively oriented contour $\phi_{R} \subset \mathcal{H}$ given by
$$\big[-R, +R \big] \bigcup \big\{R \, \exp(it)| t \in [0, \pi] \big\}.$$
Now to begin on our quest we make a note that $\phi_{R} \subset \mathcal{H}$, and consider 
$$\oint_{\phi_{R_{\mathcal{H}}}} \frac{s(z)}{q(z)}dz$$
Now notice that, 
$$\oint_{\phi_{R_{\mathcal{H}}}} \frac{s(z)}{q(z)}dz= 2 \pi i \bigg( \sum  \mathcal{Res_{f}(P_{j}}) \cdot \operatorname{Ind_{\gamma}}(P_{j}) \bigg)$$
$$\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \,   = 2 \pi i \bigg[ \sum_{j=1}^l\,\frac{1}{\left(m_j-1\right)!}\,\lim_{z\to z_j}\,\frac{\text{d}^{m_j-1}}{\text{d}z^{m_j-1}}\,\left(\frac{\left(z-z_j\right)^{m_j}\,p(z)}{p(z)}\right) \bigg].$$
Now for every if $m_{j} = 1$ for every $j=1,2,\ldots,l$ then we get $l=\dfrac{2 \pi}{d}$ so we now have that,
$\text{#}$
$$ \oint_{\phi_{R_{\mathcal{H}}}} \frac{s(z)}{q(z)}dz = \frac{2 \pi i}{1 - \exp( \frac{2 \pi  i}{d})} \sum_{j}^{l} \frac{1}{(m_{j} -1)!}\lim_{z \rightarrow z_{j}} \big( \partial_{z} \big)^{m_{j}-1} \frac{(z-z_{j})^{m_{j}}s(z)}{p(z)}.$$
To prove $\text{#}$ notice that if $m_{j} = 1$ for every $j =1,2,..l$ we have that
$$ \frac{s(x)}{p(x)}=\sum_{j=1}^{\frac{2 \pi}{d}}\,\frac{d(s'(x_{j})) \, \big(x^{d}-x_{j}^{d})}{p'\left(z_j\right)\,\left(z^d-z_j^d\right)} =  \sum_{j=1}^{\frac{2 \pi}{d}}\,\frac{u'\left(z_j^d\right)\,\left(x^d-z_j^d\right)}{q'\left(z_j^d\right)\,\left(x^d-z_j^d\right)}.$$
Now unfortunately what we done at this point isn't enough to realize a full proof of the conjecture we still need to make a road-trip around $\phi_{R}.$ We can begin making this journey by noticing through Cauchy's Theorem that,
$$\oint_{\phi_{R_{\mathcal{H}}}} \frac{s(z)}{p(z)}dz = \oint_{\phi_{R^1_{\mathcal{H}}}}\frac{s(z)}{p(z)}dz  + \oint_{\phi_{R^{2}_{\mathcal{H}}}}\frac{s(z)}{p(z)}dz.$$
Now it's worthwhile to claim that, 
$$\oint_{\gamma_{R^{1}_{\mathcal{H}}}}\frac{s(z)}{p(z)}dz \rightarrow \int_{-\infty}^{c} \frac{p(x)}{q(x)}dx + \int_{c}^{\infty} \frac{p(x)}{q(x)}dx  \, \, \text{as} \, R \rightarrow \infty.$$
Before one start's making any qualms make note using the tringle inequality for our choice of $p(z)$, since $|z| > R$ we have that, 
$$|p(z)| \leq |a_{m}z^{m}| + \cdot \cdot \cdot |a_{1}z| + |a_{o}| \leq |a_{m}|R^{m} + \cdot \cdot \cdot + |a_{1}|R + |a_{o}|.$$
On a similar note, 
$$|s(z)| \leq |b_{m}x^{m}| + \cdot \cdot \cdot |b_{1}x| + |b_{o}| \leq |b_{m}|R^{m} + \cdot \cdot \cdot + |b_{1}|R + |b_{o}|.$$
After taking care of $p(z)$, and $s(z)$ respectively we have the estimate, 
$$|f(z)| = \frac{|f(z)|}{|q(z)|} \leq \frac{|b_{m}|R^{m} + \cdot \cdot \cdot + |b_{1}|R + |b_{o}|}{|a_{n}|R^{n}/2} \leq \frac{M}{R^{n-m}}.$$
Now we have the following, 
$$\bigg| \oint_{\phi_{R^{2}_{\mathcal{H}}}} \frac{p(z)}{q(z)}dz \bigg| \leq \frac{M(2 \pi R)}{R^{n-m}} = \frac{2M \pi}{R^{n-m-1}}.$$
It's to safe to say that, 
$$\lim_{R \rightarrow \infty} \bigg| \oint_{\phi_{R^{2}_{\mathcal{H}}}} \frac{p(z)}{q(z)} \bigg| \rightarrow 0.$$
Putting everything together it's easy to note that, 
$$\lim_{R \rightarrow \infty} \oint_{\gamma_{R_{\mathcal{H}}}} \frac{p(z)}{q(z)}dz =  \lim_{R \rightarrow \infty}\oint_{\gamma_{R^{1}_{\mathcal{H}}}} \frac{p(z)}{q(z)}dz + \oint_{\gamma_{R^{2}_{\mathcal{H}}}} \frac{p(z)}{q(z)}dz  $$
$$\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, = \frac{2 \pi i}{1 - \exp( \frac{2 \pi  i}{d})} \sum_{j}^{l} \frac{1}{(m_{j} -1)!}\lim_{z \rightarrow z_{j}} \big( \partial_{z} \big)^{m_{j}-1} \frac{(z-z_{j})^{m_{j}}s(z)}{p(z)}.$$
