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 2017年8月実施 複素関数論

标签：

• Kyushu-University
• Branch-Cut
• Residue-Theorem
• Real-Integral-by-Residues

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。

$$\begin{align} \oint_C f(z)dz = \int_{CR}f(z)dz + \int_{C_1}f(z)dz + \int_{C\varepsilon}f(z)dz + \int_{C_2}f(z)dz \tag{\textcircled{1}} \end{align}$$

ここで、

$$\begin{aligned} \bigg|\int_{C_R}f(z)dz\bigg| &= \bigg|\int_0^{\pi}\frac{(\ln R + i\theta)^2}{R^2e^{2i\theta} + 1} \cdot iRe^{i\theta}d\theta\bigg| \\ &\leqq \int_0^{\pi}\bigg|\frac{(\ln R + i\theta)^2}{R^2e^{2i\theta} + 1}\cdot R d\theta\bigg| \\ &\leqq \int_0^{\pi}\frac{(\ln R)^2 + \pi^2}{R^2 - 1} \cdot R d\theta \\ &= \frac{R}{R^2 - 1}[(\ln R)^2 + \pi^2] \cdot \pi \overset{R \rightarrow \infty}{\longrightarrow}0 \end{aligned}$$

$$\begin{aligned} \bigg|\int_{C\varepsilon}f(z)dz\bigg| &= \bigg|\int_{\pi}^0 \frac{(\ln \varepsilon + i\theta)^2}{\varepsilon^2e^{2i\theta} + 1}i\varepsilon e^{i\theta}d\theta\bigg| \\ &\leqq \int_0^{\pi}\bigg|\frac{(\ln \varepsilon)^2 + \pi^2}{1 - \varepsilon^2}\cdot \varepsilon d\theta\bigg| \\ &= \frac{R}{R^2 - 1}[(\ln R)^2 + \pi^3] \cdot \pi \overset{\varepsilon \rightarrow 0}{\longrightarrow} 0 \end{aligned}$$

$$\begin{align} \lim_{R \rightarrow \infty,\varepsilon \rightarrow 0}\int_{C_1}f(z)dz &= \int_0^{\infty}\frac{(\ln x + \pi i)^2}{x^2 + 1}dx \notag \\ &= \int_0^{\infty} \frac{(\ln x)^2}{x^2 + 1}dx + 2\pi i \int_0^{\infty}\frac{\ln x}{x^2 + 1}dx - \pi^2\int_0^{\infty}\frac{1}{x^2 + 1}dx \notag \\ &= \int_0^{\infty} \frac{(\ln x)^2}{x^2 + 1}dx + 2\pi i \int_0^{\infty}\frac{\ln x}{x^2 + 1}dx - \pi^2[\tan^{-1}x]_0^{\infty} \notag \\ &= \int_0^{\infty}\frac{(\ln x)^2}{x^2 + 1}dx + 2\pi i \int_0^{\infty}\frac{ln x}{x^2 + 1}dx - \frac{\pi^3}{2} \tag{\textcircled{2}} \end{align}$$

$$\begin{align} \lim_{R \rightarrow \infty,\varepsilon \rightarrow 0}\int_{C_1}f(z)dz = \int_0^{\infty} \frac{(\ln x)^2}{x^2 + 1}dx \tag{\textcircled{3}} \end{align}$$

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、

$$\frac{\pi^2}{2}i = 2\int_0^{\infty}\frac{ln x}{x^2 + 1}dx + \frac{\pi^2}{2}i$$

$$\begin{align} \therefore \int_0^{\infty}\frac{ln x}{x^2 + 1}dx = 0 \tag{\textcircled{4}} \end{align}$$

① ~ ④　から、

$$-\frac{\pi^3}{4} = 0 + 2\int_0^{\infty}\frac{(\ln x)^2}{x^2 + 1}dx - \frac{\pi^3}{2} + 0$$

$$\therefore \int_0^{\infty}\frac{(\ln x)^2}{x^2 + 1}dx = \frac{\pi^3}{8}$$