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In the text "Functions of a Complex Variable" by Robert E.Greene and Steven G.Krantz I'm having trouble verifying my solution to $\text{Problem (1)}$

$\text{Problem (1)}$

$$\int_{0}^{\infty} \frac{1}{a_{n}x^{n} + ... + a_{2}x^{2} + a_{o}}dx \, \, \, $$

$\text{Remark}$

$p(x)$ 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,

$(*)$
$$\oint_{\eta_{R}} \frac{\log(z)}{a_{n}z^{n} + ... + a_{2}z^{2} + a_{o}}dz  = 2 \pi i \bigg(  \frac{1}{(n!)}\frac{\log(P) - a_{n}P^{n} + ... + a_{2}P^{2} + a_{o}}{(\log(P)^{2})} \bigg) \cdot 1$$

Alternately, to compute residues for our choice of $f$ consider that

$$\oint_{\eta_{R}} \frac{\log(z)}{a_{n}z^{n} + ... + a_{2}z^{2} + a_{o}}dz = \sum_{j} \operatorname{Res_{\eta_{R}}}(P_{j})\cdot \operatorname{Ind_{\eta_{R}}}(P_{j})$$

Taking the series expansion around our target $P_{j}$, so we note that our the principle part for our choice of $f$ takes the form

$$\sum_{j=n}^{-1} \frac{1}{(k+j)!} \big( \partial_{z} \big)^{k+j} \Big( ( z-P)^{k} \cdot \frac{\log(z)}{a_{n}z^{n} + ... + a_{2}z^{2} + a_{o}} \Big) \Bigg|_{z = P}.$$

Calculating our principle part yields that,

$(**)$

Applications, of the Residue Theorem unfortunately aren'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 claims 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 can be done for $\eta_{R}^{4}$, hence
\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 the precise estimate over $\eta_{4}$,

$$\bigg| \oint_{\eta_{R}^{4}} \frac{\log(e^{it}/\sqrt{R})}{a_{n}(e^{it}/\sqrt{R})^{n} + ... + a_{2}(e^{it}/ \sqrt{R})^{2} + a_{o}}  dz\bigg|  \leq  \text{length}(\eta_{R}^{4})  \cdot \sup_{\eta_{R}^{4}}(g) \leq \pi R \frac{O(\log(R))}{\sqrt{R}} \, \text{as} \, R \rightarrow \infty.$$

Fri, 05 Oct 2018 18:30 GMT