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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 $\text{Problem (1)}$

$\text{Problem (1)}$

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,

$$\int_{0}^{\infty} \frac{1}{a_{n}x^{n} + ... + a_{2}x^{2} + a_{o}}dx = 2 \pi i \bigg( \frac{1}{(n!)}\frac{\log(P) - a_{n}P^{n} + ... + a_{2}P^{2} + a_{o}}{(\log(P)^{2})} \bigg)$$

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  • $\begingroup$ Now looking at the proof I suspect that if one provides a $\delta-\epsilon$ definitions of $\eta_{R}$ it should clean up some of the analysis made. Could anyone assist with this ? $\endgroup$
    – Zophikel
    Oct 5, 2018 at 19:21
  • $\begingroup$ First, is the term $a_1x$ missing from $p_n(x)$? Second, why do you believe that $f(z)$ has poles of any particular order? $\endgroup$
    – Mark Viola
    Oct 5, 2018 at 20:26
  • $\begingroup$ @MarkViola hmm the fact that our particular choice of $f$ is very erroneous on my part, I'll have to answer with a correct proof sometime later. But diving into the meat of the issue is that our problem in $(1)$ the integral we are considering is correctly written as $(1)$ $$ \int_{-\infty}^{\infty} \frac{1}{p(x)}dx = \int_{-\infty}^{\infty} \frac{1}{a_{n}x^{n} + a_{n-1}x^{n-1}+ \cdot \cdot \cdot + a_{2}x^{2} + a_{1}x + a_{o} }dx $$ Applying the Fundamental Theorem of Algebra, $(1)$ can be rewritten as $\endgroup$
    – Zophikel
    Oct 7, 2018 at 1:21
  • $\begingroup$ $$\int_{-\infty}^{\infty} \frac{1}{k(x-x_{1})(x-x_{2}) \cdot \cdot \cdot (x-x_{n})}dx$$ $\endgroup$
    – Zophikel
    Oct 7, 2018 at 1:22
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    $\begingroup$ What exactly do you expect for an answer? Almost everywhere you say you want $\int_0^\infty\frac{dx}{p(x)}$. And then, you also say you want $\int_{-\infty}^\infty\frac{dx}{p(x)}$. These two integrals cannot be dealt with in the same manner. $\endgroup$
    – user593746
    Oct 14, 2018 at 18:12

4 Answers 4

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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 $$ \int^{\infty}_{0}\frac{dt}{a(t-\rho_1)(t-\rho_2)\ldots(t-\rho_N)}=-\sum^{N}_{k=1}\frac{\log(-\rho_k)}{H'(\rho_k)} $$

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  • $\begingroup$ Isn't the required integral from $0$ to $\infty$, and not from $-\infty$ to $\infty$? How does your work deal with the integral $\int_0^\infty\frac{1}{(x+1)(x+2)(x+3)}dx$ for instance? $\endgroup$
    – user593746
    Oct 14, 2018 at 18:08
  • $\begingroup$ I have add some notes for the case of simple roots polynomial, with degree>1. $\endgroup$ Oct 15, 2018 at 16:24
  • $\begingroup$ It's been a couple of months and I've just gotten back to read this working through your solution and a question that comes to mind. Consider $P(X)$, and $(Q(x)$ being polynomials of degree $N > 1$ with simple roots $\rho_1,\rho_2,\ldots,\rho_N$, also note that $deg(Q(z)) > deg(P(z))$ could one generalize that, $$\sum_{\rho/H / P}\frac{Q'(p)}{P'(\rho)}=0.$$ $\endgroup$
    – Zophikel
    Jul 5, 2019 at 16:39
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    $\begingroup$ @ Zophikel. Yes: If $P(z)$ is polynomial of degree $N>1$ and has distinct roots in $\textbf{C}:,z_1,z_2,\ldots,z_N$, then we have $\sum^{N}_{k=1}\frac{z_k^r}{P'(Z_k)}=0$, $\forall r=0,1,2,\ldots,N-2$. $\endgroup$ Jul 5, 2019 at 17:41
  • $\begingroup$ so a better way I could have written this to make it more clear is if $Q(z)$, and $P(z)$ are polynomials of degree $N>1$ and has distinct roots in $\mathbb{C}:$ (i.e) we have $\textbf{C}:,z_1,z_2,\ldots,z_N$ it trivially follows that, $$\sum^{N}_{k=1}\frac{Q'(Z_k)}{P'(Z_k)}=0, \forall r=0,1,2,\ldots,N-2.$$ Now probing the result and doing some thinking, makes me wonder if one could also say as well that , $$\sum^{N}_{k=1}\frac{Q'(Z_k)a}{P'(Z_k)}=0, \forall r=0,1,2,\ldots,N-2. $$ $\text{Remark}$ $a$ is just some constant $\endgroup$
    – Zophikel
    Jul 5, 2019 at 18:56
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Let $p(X)\in\mathbb{C}[X]$ be a polynomial of degree $n\geq 2$ whose roots are in $\mathbb{C}\setminus\mathbb{R}_{\geq 0}$. The goal is to evaluate $$I:=\int_0^\infty\,\frac{1}{p(x)}\,\text{d}x\,.$$ Let $z_1,z_2,\ldots,z_k\in\mathbb{C}$ be all pairwise distinct roots of $p(X)$, respectively, with multiplicities $m_1,m_2,\ldots,m_k$ (whence $n=m_1+m_2+\ldots+m_k$). Define $f:\left(\mathbb{C}\setminus\mathbb{R}_{\geq 0}\right)\to\mathbb{C}$ to be the holomorphic function $$f(z):=\frac{\ln(z)}{p(z)}\text{ for all }z\in\left(\mathbb{C}\setminus\mathbb{R}_{\geq 0}\right)\,.$$ Here, for a complex number $z\in\left(\mathbb{C}\setminus \mathbb{R}_{\ge0}\right)$, we write $z=r\,\exp(\text{i}\phi)$ with $\phi\in(0,2\pi)$ and $r>0$, and then define $$\ln(z):=\ln(r)+\text{i}\phi\,,$$ so that the branch cut is $\mathbb{R}_{\geq 0}$. For $\epsilon\in(0,1)$, set $$J(\epsilon):=\oint_{\Gamma_{\epsilon}}\,f(z)\,\text{d}z\,,$$ where $\Gamma_{\epsilon}$ is the positively oriented keyhole contour given by $$\begin{align}\Biggr[\epsilon\,\exp(+\text{i}\epsilon),\frac{1}{\epsilon}\,\exp(+\text{i}\epsilon)\Biggl]&\cup\Biggl\{\frac{1}{\epsilon}\,\exp(\text{i}\theta)\,\Big|\,\theta\in[\epsilon,2\pi-\epsilon]\Biggr\}\\&\cup\Biggr[\frac1\epsilon\,\exp(-\text{i}\epsilon),{\epsilon}\,\exp(-\text{i}\epsilon)\Biggl]\cup\Biggl\{{\epsilon}\,\exp(\text{i}\theta)\,\Big|\,\theta\in[2\pi-\epsilon,\epsilon]\Biggr\}\,.\end{align}$$

Observe that $$\lim_{\epsilon\to0^+}\,J(\epsilon)=-2\pi\text{i}\,I\,.$$ By the Residue Theorem, $$\lim_{\epsilon\to0^+}\,J(\epsilon)=2\pi\text{i}\,\sum_{j=1}^k\,\text{Res}_{z=z_j}\big(f(z)\big)\,.$$ Hence, $$I=-\sum_{j=1}^k\,\text{Res}_{z=z_j}\big(f(z)\big)\,.$$ Since $$\begin{align}\text{Res}_{z=z_j}\big(f(z)\big)&=\frac{1}{\left(m_j-1\right)!}\,\lim_{z\to z_j}\,\frac{\text{d}^{m_j-1}}{\text{d}z^{m_j-1}}\,\Big(\left(z-z_j\right)^{m_j}\,f(z)\Big)\\&=\frac{1}{\left(m_j-1\right)!}\,\lim_{z\to z_j}\,\frac{\text{d}^{m_j-1}}{\text{d}z^{m_j-1}}\,\left(\frac{\left(z-z_j\right)^{m_j}\,\ln(z)}{p(z)}\right)\end{align}$$ for $j=1,2,\ldots,k$, we conclude that $$I=-\sum_{j=1}^k\,\frac{1}{\left(m_j-1\right)!}\,\lim_{z\to z_j}\,\frac{\text{d}^{m_j-1}}{\text{d}z^{m_j-1}}\,\left(\frac{\left(z-z_j\right)^{m_j}\,\ln(z)}{p(z)}\right)\,.\tag{*}$$ In particular, if $m_j=1$ for all $j=1,2,\ldots,k$, then $k=n$ and $$I=-\sum_{j=1}^n\,\frac{\ln\left(z_j\right)}{p'\left(z_j\right)}\,.\tag{$\star$}$$

For example, let $p(X):=\left(X^2+1\right)^2\,(X+1)$. You can see that the partial fraction decomposition gives $$\int\,\frac{1}{p(x)}\,\text{d}x=\frac{1}{8}\,\left(\frac{2\,(x+1)}{x^2+1}-\ln\left(x^2+1\right)+2\,\ln(x+1)+4\,\arctan(x)\right)+\text{constant}\,,$$ so that $$I=\int_0^\infty\,\frac{1}{p(x)}\,\text{d}x=\frac{1}{4}\,(\pi-1)\,.$$ You can also use (*) to compute $I$. Let $z_1=-1$, $z_2=+\text{i}$, and $z_3=-\text{i}$, so that $m_1=1$, $m_2=2$, and $m_3=2$. Then, $$\frac{1}{\left(m_1-1\right)!}\,\lim_{z\to z_1}\,\frac{\text{d}^{m_1-1}}{\text{d}z^{m_1-1}}\,\left(\frac{\left(z-z_1\right)^{m_1}\,\ln(z)}{p(z)}\right)=\lim_{z\to -1}\,\frac{\ln(z)}{\left(z^2+1\right)^2}=\frac{\pi\text{i}}{4}\,,$$ $$\begin{align}\frac{1}{\left(m_2-1\right)!}\,\lim_{z\to z_2}\,\frac{\text{d}^{m_2-1}}{\text{d}z^{m_2-1}}\,\left(\frac{\left(z-z_2\right)^{m_2}\,\ln(z)}{p(z)}\right)&=\lim_{z\to +\text{i}}\,\frac{\text{d}}{\text{d}z}\,\frac{\ln(z)}{(z+\text{i})^2\,(z+1)}\\&=\frac{1+\pi}{8}-\text{i}\,\left(\frac{\pi-2}{16}\right)\,,\end{align}$$ and $$\begin{align}\frac{1}{\left(m_3-1\right)!}\,\lim_{z\to z_3}\,\frac{\text{d}^{m_3-1}}{\text{d}z^{m_3-1}}\,\left(\frac{\left(z-z_3\right)^{m_3}\,\ln(z)}{p(z)}\right)&=\lim_{z\to -\text{i}}\,\frac{\text{d}}{\text{d}z}\,\frac{\ln(z)}{(z-\text{i})^2\,(z+1)}\\&=\frac{1-3\pi}{8}-\text{i}\,\left(\frac{3\pi+2}{16}\right)\,.\end{align}$$ Hence, $$I=-\left(\frac{\pi\text{i}}{4}+\left(\frac{1+\pi}{8}-\text{i}\,\left(\frac{\pi-2}{16}\right)^{\vphantom{a^a}}\right)+\left(\frac{1-3\pi}{8}-\text{i}\,\left(\frac{3\pi+2}{16}\right)^{\vphantom{a^a}}\right)^{\vphantom{a^a}}\right)=\frac{\pi-1}{4}\,.$$


In fact, if $m_j=1$ for all $j=1,2,\ldots,k$ (so $k=n$), then it follows that $$\frac{1}{p(x)}=\sum_{j=1}^n\,\frac{1}{p'\left(z_j\right)\,\left(x-z_j\right)}\,.$$ Without loss of generality, we assume that $$\text{arg}\left(z_1\right)\geq \text{arg}\left(z_2\right) \geq \ldots \geq \text{arg}\left(z_n\right)\,,$$ wher $\text{arg}(z)$ is the argument of $z\in\left(\mathbb{C}\setminus\mathbb{R}_{\geq 0}\right)$ in the interval $(0,2\pi)$. Because $$\sum_{j=1}^n\,\frac{1}{p'\left(z_j\right)}=\lim_{x\to\infty}\,\frac{x}{p(x)}=0\,,$$ we can write $$\frac{1}{p(x)}=\sum_{j=1}^{n-1}\,\frac{1}{p'\left(z_j\right)}\,\left(\frac{1}{x-z_j}-\frac{1}{x-z_n}\right)\,.$$ That is, $$\int\,\frac{1}{p(x)}\,\text{d}x=\sum_{j=1}^{n-1}\,\frac{1}{p'\left(z_j\right)}\,\ln\left(\frac{x-z_j}{x-z_n}\right)+\text{constant}=\sum_{j=1}^n\,\frac{1}{p'\left(z_j\right)}\,\ln\left(x-z_j\right)+\text{constant}\,,$$ whence $$\int_0^\infty\,\frac{1}{p(x)}\,\text{d}x=-\sum_{j=1}^{n-1}\,\frac{1}{p'\left(z_j\right)}\,\ln\left(\frac{z_j}{z_n}\right)=-\sum_{j=1}^{n}\,\frac{\ln\left(z_j\right)}{p'\left(z_j\right)}\,.$$ For example, one can evaluate $$\int_0^\infty\,\frac{1}{(x+2)(x+3)}\,\text{d}x$$ to be $$-\left(\frac{\ln(-2)}{(-2+3)}+\frac{\ln(-3)}{(-3+2)}\right)=\ln\left(\frac{3}{2}\right)\,.$$


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 share a common factor, then $$\int_0^\infty\,\frac{s(x)}{p(x)}\,\text{d}x$$ can be evaluated in the same way. That is, we have $$\int_0^\infty\,\frac{s(x)}{p(x)}\,\text{d}x=-\sum_{j=1}^k\,\frac{1}{\left(m_j-1\right)!}\,\lim_{z\to z_j}\,\frac{\text{d}^{m_j-1}}{\text{d}z^{m_j-1}}\,\left(\frac{\left(z-z_j\right)^{m_j}\,\ln(z)\,s(z)}{p(z)}\right)\,.$$ In particular, if $m_j=1$ for all $j=1,2,\ldots,k$ (so $k=n$), then $$\frac{s(x)}{p(x)}=\sum_{j=1}^n\,\frac{s\left(z_j\right)}{p'\left(z_j\right)\,\left(x-z_j\right)}$$ and so $$\int_0^\infty\,\frac{s(x)}{p(x)}\,\text{d}x=-\sum_{j=1}^n\,\frac{\ln\left(z_j\right)\,s\left(z_j\right)}{p'\left(z_j\right)}\,.$$

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  • $\begingroup$ It looks like you posted nearly the same answer twice. Was this intended? $\endgroup$
    – Alex R.
    Oct 12, 2018 at 22:13
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    $\begingroup$ @AlexR. They are not the same answers (one has the extra assumption that $p(X)$ is an even polynomial, and the other deals with a more general setting). And I intended to do this, because MathJax is very slow for long answers. $\endgroup$ Oct 12, 2018 at 22:14
  • $\begingroup$ I've been reworking through your proof I apologize for the late response(it's been a couple of months), but why isn't okay to end the proof by noticing that, $$ \int_0^\infty\,\frac{1}{p(x)}dx= -\sum_{j=1}^k\,\frac{1}{\left(m_j-1\right)!}\,\lim_{z\to z_j}\,\frac{\text{d}^{m_j-1}}{\text{d}z^{m_j-1}}\,\left(\frac{\left(z-z_j\right)^{m_j}\,\ln(z)}{p(z)}\right)\,.\tag{*} $$ $\endgroup$
    – Zophikel
    Jul 5, 2019 at 16:49
  • $\begingroup$ I initially ask because it seems $(*)$ would work for any order pole while $\tag{$\star$}$ seems to work for the cause of first order poles ? $\endgroup$
    – Zophikel
    Jul 5, 2019 at 17:31
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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 $$\int_0^\infty\,\frac{s(x)}{p(x)}\,\text{d}x=\pi\text{i}\,\sum_{j=1}^{\frac{n}{2}}\,\frac{s\left(z_j\right)}{p'\left(z_j\right)}=\frac{\pi\text{i}}{2}\,\sum_{j=1}^{\frac{n}{2}}\,\frac{u\left(z_j^2\right)}{z_j\,q'\left(z_j^2\right)}\,.$$

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    $\begingroup$ Let $d>1$ be an integer. Suppose that $p(X)=q\left(X^d\right)$ and $s(X)=u\left(X^d\right)$ 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 $z_1,z_2,\ldots,z_l$ be all the distinct roots of $p(X)$ with arguments in the interval $\left(0,\dfrac{2\pi}{d}\right)$, respectively, with multiplicities $m_1,m_2,\ldots,m_l$. $\endgroup$ Oct 12, 2018 at 23:08
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    $\begingroup$ We can then show that $$\int_0^\infty\,\frac{s(x)}{p(x)}\,\text{d}x=\small\frac{2\pi\text{i}}{1-\exp\left(\frac{2\pi\text{i}}{d}\right)}\,\sum_{j=1}^l\,\frac{1}{\left(m_j-1\right)!}\,\lim_{z\to z_j}\,\frac{\text{d}^{m_j-1}}{\text{d}z^{m_j-1}}\,\frac{\left(z-z_j\right)^{m_j}\,s(z)}{p(z)}\,.$$ $\endgroup$ Oct 12, 2018 at 23:10
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    $\begingroup$ In particular, if $m_j=1$ for every $j=1,2\ldots,l$, then $$\int_0^\infty\,\frac{s(x)}{p(x)}\,\text{d}x=\small\frac{2\pi\text{i}}{1-\exp\left(\frac{2\pi\text{i}}{d}\right)}\,\sum_{j=1}^l\,\frac{s\left(z_j\right)}{p'\left(z_j\right)}=\frac{\frac{2\pi\text{i}}{d}}{1-\exp\left(\frac{2\pi\text{i}}{d}\right)}\,\sum_{j=1}^l\,\frac{u\left(z_j^d\right)}{z_j^{d-1}\,q'\left(z_j^d\right)}\,.$$ $\endgroup$ Oct 12, 2018 at 23:16
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$\text{Batominovski's First Idenity}:$

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,

$$\lim_{R \rightarrow \infty} \oint_{\gamma_{R_{\mathcal{H}}}} \frac{p(z)}{q(z)}dz = \lim_{R \rightarrow \infty}\oint_{\gamma_{R^{1}_{\mathcal{H}}}} \frac{p(z)}{q(z)}dz + \oint_{\gamma_{R^{2}_{\mathcal{H}}}} \frac{p(z)}{q(z)}dz $$

$$\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, = \frac{2 \pi i}{1 - \exp( \frac{2 \pi i}{d})} \sum_{j}^{l} \frac{1}{(m_{j} -1)!}\lim_{z \rightarrow z_{j}} \big( \partial_{z} \big)^{m_{j}-1} \frac{(z-z_{j})^{m_{j}}s(z)}{p(z)}.$$

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