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,
$$\int_{0}^{\infty} \frac{1}{a_{n}x^{n} + … + a_{2}x^{2} + a_{o}}dx = 2 \pi i \bigg( \frac{1}{(n!)}\frac{\log(P) – a_{n}P^{n} + … + a_{2}P^{2} + a_{o}}{(\log(P)^{2})} \bigg)$$
Best Answer
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 $$ \int^{\infty}_{0}\frac{dt}{a(t-\rho_1)(t-\rho_2)\ldots(t-\rho_N)}=-\sum^{N}_{k=1}\frac{\log(-\rho_k)}{H'(\rho_k)} $$