九州大学 システム情報科学府 情報理工学専攻・電気電子工学専攻 2018年度 複素関数論
Author
Zero
Description
図に示す曲線 \(C\) に沿った複素積分 \(\oint_C\frac{(\ln z)^2}{z^2 + 1}dz\) を考える．ただし，\(R > 1,\varepsilon < 1\) とする．次 の各問に答えよ．
(1) \(\oint_C\frac{(\ln z)^2}{z^2 + 1}dz\) の値を求めよ．
(2) \(\oint_C\frac{(\ln z)^2}{z^2 + 1}dz\) の値を用いて，\(\int_{0}^{\infty}\frac{(\ln x)^2}{x^2 + 1}dx = \frac{\pi^3}{8}\) を示せ．
Kai
(1)
\(f(z) = \frac{(\ln z)^2}{z^2 + 1}\) とおくと、\(f(z)\) は \(C\) 内部に \(1\) 位の極 \(i\) をもつ. よって、留数定理より、
\[
\begin{aligned}
\oint f(z)dz &= 2\pi i \cdot \text{Res}_{z = i} \frac{(\ln z)^2}{z^2 + 1} \\
&= 2\pi i \cdot \frac{(\ln i)^2}{2i} \\
&= (\ln i)^2 \cdot \pi \\
&= -\frac{\pi^3}{4}
\end{aligned}
\]
(2)
曲線 \(C\) を - 区間 \([Re^{io},Re^{i\pi}]\) を \(C_R\) - 区間 \([-R,-\varepsilon]\) を \(C_1\) - 区間 \([\varepsilon e^{i\pi},\varepsilon e^{io}]\) を \(C_\varepsilon\) - 区間 \([\varepsilon, R]\) を \(C_2\) として。\(C = C_R + C_1 + C_{\varepsilon} + C_2\) と表す。
\[
\begin{align}
\oint_C f(z)dz = \int_{CR}f(z)dz + \int_{C_1}f(z)dz + \int_{C\varepsilon}f(z)dz + \int_{C_2}f(z)dz \tag{①}
\end{align}
\]
ここで、
\[
\begin{aligned}
\bigg|\int_{C_R}f(z)dz\bigg| &= \bigg|\int_0^{\pi}\frac{(\ln R + i\theta)^2}{R^2e^{2i\theta} + 1} \cdot iRe^{i\theta}d\theta\bigg| \\
&\leqq \int_0^{\pi}\bigg|\frac{(\ln R + i\theta)^2}{R^2e^{2i\theta} + 1}\cdot R d\theta\bigg| \\
&\leqq \int_0^{\pi}\frac{(\ln R)^2 + \pi^2}{R^2 - 1} \cdot R d\theta \\
&= \frac{R}{R^2 - 1}[(\ln R)^2 + \pi^2] \cdot \pi \overset{R \rightarrow \infty}{\longrightarrow}0
\end{aligned}
\]
\[
\begin{aligned}
\bigg|\int_{C\varepsilon}f(z)dz\bigg| &= \bigg|\int_{\pi}^0 \frac{(\ln \varepsilon + i\theta)^2}{\varepsilon^2e^{2i\theta} + 1}i\varepsilon e^{i\theta}d\theta\bigg| \\
&\leqq \int_0^{\pi}\bigg|\frac{(\ln \varepsilon)^2 + \pi^2}{1 - \varepsilon^2}\cdot \varepsilon d\theta\bigg| \\
&= \frac{R}{R^2 - 1}[(\ln R)^2 + \pi^3] \cdot \pi \overset{\varepsilon \rightarrow 0}{\longrightarrow} 0
\end{aligned}
\]
\[
\begin{align}
\lim_{R \rightarrow \infty,\varepsilon \rightarrow 0}\int_{C_1}f(z)dz &= \int_0^{\infty}\frac{(\ln x + \pi i)^2}{x^2 + 1}dx \notag \\
&= \int_0^{\infty} \frac{(\ln x)^2}{x^2 + 1}dx + 2\pi i \int_0^{\infty}\frac{\ln x}{x^2 + 1}dx - \pi^2\int_0^{\infty}\frac{1}{x^2 + 1}dx \notag \\
&= \int_0^{\infty} \frac{(\ln x)^2}{x^2 + 1}dx + 2\pi i \int_0^{\infty}\frac{\ln x}{x^2 + 1}dx - \pi^2[\tan^{-1}x]_0^{\infty} \notag \\
&= \int_0^{\infty}\frac{(\ln x)^2}{x^2 + 1}dx + 2\pi i \int_0^{\infty}\frac{ln x}{x^2 + 1}dx - \frac{\pi^3}{2} \tag{②}
\end{align}
\]
\[
\begin{align}
\lim_{R \rightarrow \infty,\varepsilon \rightarrow 0}\int_{C_1}f(z)dz = \int_0^{\infty} \frac{(\ln x)^2}{x^2 + 1}dx \tag{③}
\end{align}
\]
② について、\(\int_0^{\infty}\frac{\ln x}{x^2 + 1}dx\) を求める。\(q(z) = \frac{\ln z}{z^2 + 1}\) とおくと。
\[
\begin{aligned}
\oint_C q(z)dz = 2\pi i \cdot \lim_{z \rightarrow i} \frac{\ln z}{z \rightarrow i} = 2\pi i \cdot \frac{ln i}{2i} = \pi \cdot i \cdot \frac{\pi}{2} = \frac{\pi^2}{2}i
\end{aligned}
\]
\(f(z)\) と同様に \(q(z)\) について、
\[
\begin{aligned}
\bigg|\int_{CR}q(z)dz\bigg| &= \bigg|\int_0^{\pi}\frac{ln R + i\theta}{R^2e^{2i\theta} + 1}i Re^{i\theta}d\theta\bigg| \\
&\leqq \int_0^{\pi} \frac{R}{R^2 - 1} \cdot \sqrt{(ln R)^2 + \pi^2} d\theta \\
&= \frac{R}{R^2 - 1}\sqrt{(\ln R)^2 + \pi^2} \cdot \pi \overset{R \rightarrow \infty}{\longrightarrow} 0
\end{aligned}
\]
\[
\begin{aligned}
\bigg|\int_{C\varepsilon}q(z)dz\bigg| &= \bigg|\int_{\pi}^0\frac{\ln \varepsilon + i\theta}{\varepsilon^2 e^{2i\theta} + 1}i\varepsilon e^{i\theta}d\theta\bigg| \\
&\leqq \int_0^{\pi}\frac{\varepsilon}{1 - \varepsilon^2}\sqrt{(\ln \varepsilon)^2 + \pi^2}d\theta \\
&= \frac{\varepsilon}{1 - \varepsilon^2}\sqrt{(\ln \varepsilon)^2 + \pi^2} \cdot \pi \overset{\varepsilon \rightarrow 0}{\longrightarrow} 0
\end{aligned}
\]
\[
\begin{aligned}
\lim_{R \rightarrow \infty,\varepsilon \rightarrow 0}\int_{C_1}q(z)dz &= \int_0^{\infty}\frac{ln x}{x^2 + 1}dx
\end{aligned}
\]
より、
\[
\frac{\pi^2}{2}i = 2\int_0^{\infty}\frac{ln x}{x^2 + 1}dx
\]
より、
\[
\frac{\pi^2}{2}i = 2\int_0^{\infty}\frac{ln x}{x^2 + 1}dx + \frac{\pi^2}{2}i
\]
\[
\begin{align}
\therefore \int_0^{\infty}\frac{ln x}{x^2 + 1}dx = 0 \tag{④}
\end{align}
\]
① ~ ④ から、
\[
-\frac{\pi^3}{4} = 0 + 2\int_0^{\infty}\frac{(\ln x)^2}{x^2 + 1}dx - \frac{\pi^3}{2} + 0
\]
\[
\therefore \int_0^{\infty}\frac{(\ln x)^2}{x^2 + 1}dx = \frac{\pi^3}{8}
\]