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Let $P, Q$, be polynomials, with deg $Q\geq 2+$ deg $P$, $Q\neq 0$, on $\{z\in\mathbb{R} \hspace{0.1cm} | \hspace{0.1cm} z\geq 0\}$. Choose holomorphic branch of $\log z$ on $\mathbb{C}\backslash \{z\in\mathbb{R} \hspace{0.1cm} | \hspace{0.1cm} z\geq 0\}$ such that $\log(-1)=i\pi$. $\Gamma_{\epsilon, R} = $ 2 segments, semi-circle radius $\epsilon$, and arc of a circle radius $R$. Compute $$ \int_{\Gamma_{\epsilon,R}} \dfrac{P(z)}{Q(z)}\log z\,dz $$ with $\epsilon$ small, and $R$ large, by

  1. finding the limit of $\int \frac{P(z)}{Q(z)}\log z$ along the semi-circle of radius $\epsilon$, and then along the arc of the circle of radius $R$, as $\epsilon \to 0$.
  2. finding the double limit of $\int \frac{P(z)}{Q(z)}\log z$ along the 2 segments, respectively, as $\epsilon \to 0$, and $R\to \infty$.

Answer: Recall that the holomorphic branch of $\log z=\ln|z|+i \cdot arg(z)$, where $|arg(z)|\leq \pi$. Since $\log(z)$ has no singularities in our domain, $\frac{P(z)}{Q(z)}\log z$ is analytic in our domain except at the zeros of $Q(z)$. Recall from class that $|P(z)/Q(z)|\leq c|z|^{-2}=cR^{-2}$ when $|x|>x_0$.

First, we need to show that this integral indeed exists. $$ \left| \dfrac{P(z)}{Q(z)}\log z\right| \leq \dfrac{c}{|z|^2}\cdot \left|\ln |z|+i\cdot arg(z)\right| < \dfrac{c}{R^2}\cdot (\ln R+i\pi) $$ which implies our function is bounded and therefore the integrand exists.

Denote $\Gamma_\epsilon$ be the semi-circle of radius $\epsilon$, $\Gamma_R$ be the circle of radius $R$, and $c_1, c_2$ be the lines $arg(z)=-\pi$ and $arg(z)=\pi$, respectively. Note that $$ \int\limits_{\Gamma_{\epsilon, R}} = \int\limits_{\Gamma_R} + \int\limits_{c_2} + \int\limits_{\Gamma_\epsilon} + \int\limits_{c_1} $$ First, we will work with $\Gamma_R$. \begin{equation*} \begin{aligned} \left| \int\limits_{\Gamma_R} \dfrac{P(z)}{Q(z)} \log(z)\,dz \right| & \leq 2\pi R \left|\dfrac{P(z)}{Q(z)}\right| \cdot |\ln R +i \cdot arg(z)| \\ & \leq 2\pi R \dfrac{c}{R^2} \cdot (|\ln R| + |i \cdot arg(z)|) \\ & < \dfrac{2\pi c}{R} \cdot (|\ln R| + \pi) \to 0\\ \end{aligned} \end{equation*} as $R\to \infty$, since $R^{-1}\cdot |\ln R|\to \infty$ as $R\to \infty$.

Similarly, for $\Gamma_\epsilon$ \begin{equation*} \begin{aligned} \left| \int\limits_{\Gamma_\epsilon} \dfrac{P(z)}{Q(z)} \log(z)\,dz \right| & \leq \pi \epsilon \left|\dfrac{P(z)}{Q(z)}\right| \cdot |\ln \epsilon +i \cdot arg(z)| \\ & \leq \pi \epsilon \dfrac{c}{R^2} \cdot (|\ln \epsilon| + |i \cdot arg(z)|) \\ & < \dfrac{\pi c \epsilon}{R^2} \cdot (|\ln \epsilon| + \pi) \to 0\\ \end{aligned} \end{equation*}

Next, we will work with both segments $c_1$, and $c_2$. Let $z=x+i\epsilon$ on $c_1$, and $z=x-i\epsilon$ on $c_2$, with $x\leq 0$. Then \begin{equation*} \begin{aligned} \int\limits_{c_1} \dfrac{P(z)}{Q(z)} \log(z)\,dz & = \int\limits_R^\epsilon \dfrac{P(x)}{Q(x)} \log(x+i\epsilon)\,dx = \int\limits_R^\epsilon \dfrac{P(x)}{Q(x)} \cdot (\ln(x)+i\pi)\,dx = -\int\limits_\epsilon^R \dfrac{P(x)}{Q(x)} \cdot (\ln(x)+i\pi)\,dx \\ & = -\int\limits_\epsilon^R \dfrac{P(x)}{Q(x)} \cdot \ln(x)\,dx - i\pi \int\limits_\epsilon^R \dfrac{P(x)}{Q(x)} \,dx \\ \\ \int\limits_{c_2} \dfrac{P(z)}{Q(z)} \log(z)\,dz & = \int\limits_\epsilon^R \dfrac{P(x)}{Q(x)} \log(x-i\epsilon)\,dx = \int\limits_\epsilon^R \dfrac{P(x)}{Q(x)} \cdot (\ln(x)-i\pi)\,dx \\ & = \int\limits_\epsilon^R \dfrac{P(x)}{Q(x)} \cdot \ln(x)\,dx - i\pi\int\limits_\epsilon^R \dfrac{P(x)}{Q(x)} \,dx \end{aligned} \end{equation*} Hence the sum of these two integrands yields: $$ \int\limits_{c_1} \dfrac{P(z)}{Q(z)} \log(z)\,dz + \int\limits_{c_2} \dfrac{P(z)}{Q(z)} \log(z)\,dz = -2\pi i\int\limits_\epsilon^R \dfrac{P(z)}{Q(z)} \,dz $$

Furthermore, as $\epsilon\to 0$ and $R\to\infty$, $$ \int_{\Gamma_{R,\epsilon}} \frac{P(z)}{Q(z)}\log z\,dz = -2\pi i\int\limits_0^\infty \dfrac{P(z)}{Q(z)} \,dz $$ Recall in class, we showed that $$ \int\limits_{-\infty}^\infty \dfrac{P(z)}{Q(z)} \,dz = 2\pi i \sum_{\substack{z_j \text{ sing.} \\ Im(z_j)>0}} \underset{z_j}{\text{ Res }} \dfrac{P(z)}{Q(z)} $$ Hence \begin{equation*} \begin{aligned} \int_{\Gamma_{R,\epsilon}} \frac{P(z)}{Q(z)}\log z\,dz & = -\pi i\int\limits_{-\infty}^\infty \dfrac{P(z)}{Q(z)} \,dz \\ & = (-\pi i)(2\pi i) \sum_{\substack{z_j \text{ sing.} \\ Im(z_j)>0}} \underset{z_j}{\text{ Res }} \dfrac{P(z)}{Q(z)} = 2(\pi)^2 \sum_{\substack{z_j \text{ sing.} \\ Im(z_j)>0}} \underset{z_j}{\text{ Res }} \dfrac{P(z)}{Q(z)} \end{aligned} \end{equation*}

Referring to Problem 2: What if $\log z$ is replaced by $(\log z)^2$?

Answer: First, we need to show that this integral indeed exists. $$ \left| \dfrac{P(z)}{Q(z)}(\log z)^2\right| \leq \dfrac{c}{|z|^2}\cdot \left|\ln |z|+i\cdot arg(z)\right|^2 < \dfrac{c}{R^2}\cdot (\ln R+i\pi)^2 $$ which implies our function is bounded and therefore the integrand exists.

First, we will work with $\Gamma_R$. \begin{equation*} \begin{aligned} \left| \int\limits_{\Gamma_R} \dfrac{P(z)}{Q(z)} (\log(z))^2\,dz \right| & \leq 2\pi R \left|\dfrac{P(z)}{Q(z)}\right| \cdot |\ln R +i \cdot arg(z)|^2 \\ & \leq 2\pi R \dfrac{c}{R^2} \cdot (|\ln R| + |i \cdot arg(z)|)^2 \\ & < \dfrac{2\pi c}{R} \cdot (|\ln R| + \pi)^2 \\ & = \dfrac{2\pi c}{R} \cdot (|\ln R|^2 + 2\pi|\ln R| + (\pi)^2) \end{aligned} \end{equation*}

Following the procedure: I see already one different. If I let $R\to \infty$, the value above doesn't go to $0$, because I have $(\ln R)^2/R \to \infty$ as $R\to \infty$. How can I fix this?

Please don't give alternative proofs for the first part of the question. i.e. $ \int_{\Gamma_{\epsilon,R}} \dfrac{P(z)}{Q(z)}(\log z)\,dz $

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  • $\begingroup$ "$\Gamma_{\epsilon, R} = $ 2 segments, semi-circle radius $\epsilon$, and arc of a circle radius $R$." This seems sloppy to me. Please me more precise on this. $\endgroup$ – zhw. Mar 5 '18 at 23:57
  • $\begingroup$ you most certainly do not have $\dfrac{\ln^2 R}{R}\to\infty$ when $R\to\infty.$ In fact, the fraction easily approaches $0.$ $\endgroup$ – dezdichado Mar 6 '18 at 5:50
  • $\begingroup$ "Recall that the holomorphic branch of $\log z=\ln|z|+i \cdot \arg(z)$, where $|\arg(z)|\leq \pi$." You need $|\arg(z)|<\pi$ $\endgroup$ – zhw. Mar 6 '18 at 17:56

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