留数定理の利用例

MathJaxのアップデートにともない動作を確認したかったため，典型的な複素積分の問題とその解答を載せる．

問 次の積分の値を求めよ．

1. \(\displaystyle \int_{-\infty}^\infty \frac{1}{1 + x^2} dx\)
2. \(\displaystyle \int_{0}^\infty \frac{1}{1 + x^3} dx\)
3. \(\displaystyle \int_{-\infty}^\infty \frac{1}{1 + x^4} dx\)

解答 (a) \( \displaystyle f(z) = \frac{1}{z^2 + 1}\ (z \in \mathbb{C}) \)とおく． この関数は\( z = \pm i \)を孤立特異点にもつ． また曲線\( C \)を\( C \colon z = Re^{i\theta} \) \( (R \gt 1,\ 0 \leq \theta \leq \pi) \)で定義する．このとき，留数定理により \begin{align} \int_{-R}^R f(x) dx + \int_C f(z) dz = 2 \pi i \res(f, i) \label{a} \end{align} が成り立つ．

\(f\)について \begin{align*} f(z) &= \frac{1}{(z + i)(z - i)} \\ &= \frac{1}{z - i} \cdot \frac{1}{z - i + 2i} \\ &= \frac{1}{2i} \cdot \frac{1}{z - i} \cdot \frac{1}{\displaystyle 1 + \frac{z - i}{2i}} \end{align*} であるから，\( 0 \lt \abs{z - i} \lt 1 \)の範囲で\( \displaystyle \abs{\frac{z - i}{2i}} \leq \frac{\abs{z - i}}{\abs{2i}} \leq \frac{1}{2} \)となり，級数展開できて \begin{align*} f(z) &= \frac{1}{2i} \frac{1}{z - i} \parentheses{ 1 - \frac{z - i}{2i} + \parentheses{\frac{z - i}{2i}}^2 - \parentheses{\frac{z - i}{2i}}^3 + \cdots } \\ &= - \frac{i}{2} \frac{1}{z - i} + \frac{1}{4} + \cdots \end{align*} となる．したがって\( \displaystyle \res(f, i) = - \frac{i}{2} \)である．

したがって\eqref{a}は \begin{align*} \int_{-R}^R f(x) dx + \int_C f(z) dz = \pi \end{align*} となる．

ここで \begin{align*} \abs{\int_C f(z) dz} &= \abs{\int_0^{\pi} \frac{i R e^{i \theta}}{R^2 e^{2i\theta} + 1} d\theta} \\ &= \int_0^{\pi} \frac{\abs{i R e^{i \theta}}}{\abs{R^2 e^{2i\theta} + 1}} d\theta \\ &\leq \int_0^{\pi} \frac{R}{\abs{R^2 e^{2i\theta}} - 1} d\theta \\ &= \frac{R}{R^2 - 1} \int_0^{\pi} d\theta \\ &= \frac{1}{\displaystyle R - \frac{1}{R}} \pi. \end{align*}

最終的に \begin{align*} \int_{-\infty}^\infty f(x) dx &= \lim_{R \to \infty} \int_{-R}^R f(z) dz \\ &= \lim_{R \to \infty} \parentheses{\pi - \int_C f(z) dz} \\ &= \pi \end{align*} を得る．

(b) \( \displaystyle f(z) = \frac{1}{z^3 + 1} = \frac{1}{(z + 1)(z^2 - z + 1)}\ (z \in \mathbb{C}) \)とおく． この関数は\( \displaystyle \alpha = \frac{1}{2} + \frac{\sqrt{3}}{2}i \) とすると\( \displaystyle z = - 1, \alpha, \bar{\alpha} \)を孤立特異点としてもつ．曲線\( C_1,\, C_2 \)を\( C_1\colon z = Re^{i\theta}\) \(\displaystyle \parentheses{R \gt 0, 0 \leq \theta \leq \frac{2 \pi}{3}}, C_2\colon z = r e^{2\pi i /3} \) \((0 \leq r \leq R) \)で定める．留数定理より \begin{align} \int_{0}^R f(x) dx + \int_{C_1} f(z) dz + \int_{-C_2} f(z) dz = 2 \pi i \res(f, \alpha) \label{b} \end{align} が成り立つ．

\(f\)について \begin{align*} f(z) &= \frac{1}{(z + 1)(z - \alpha)(z - \bar{\alpha})} \\ &= \frac{1}{(z - \alpha + \alpha + 1)(z - \alpha)(z - \alpha + \alpha - \bar{\alpha})} \\ &= \frac{1}{\displaystyle \parentheses{1 + \frac{z - \alpha}{\alpha + 1}} (z - \alpha)\parentheses{1 + \frac{z - \alpha}{\alpha - \bar{\alpha}}}} \\ &= \frac{1}{\alpha + 1}\cdot \frac{1}{\alpha - \bar{\alpha}} \cdot \frac{1}{z - \alpha} \parentheses{1 - \frac{z - \alpha}{\alpha - 1} + \parentheses{\frac{z - \alpha}{\alpha - 1}}^2 - \cdots} \\ &\quad\,\,\parentheses{1 - \frac{z - \alpha}{\alpha - \bar{\alpha}} + \parentheses{\frac{z - \alpha}{\alpha - \bar{\alpha}}}^2 - \cdots}. \end{align*} であるから\( \displaystyle \res (f, \alpha) = \frac{1}{(\alpha + 1)(\alpha - \bar{\alpha})} = \frac{1}{(\alpha + 1)\sqrt{3}i} \)を得る．

一方， \begin{align*} \abs{\int_{C_1} f(z) dz} &= \abs{\int_0^{\pi/2} \frac{i R e^{i \theta}}{R^3 e^{3i\theta} + 1} d\theta} \\ &= \int_0^{2 \pi/3} \frac{\abs{i R e^{i \theta}}}{\abs{R^3 e^{3i\theta} + 1}} d\theta \\ &\leq \int_0^{2 \pi/3} \frac{dr}{\abs{r^3 e^{3i\theta}} - 1}d\theta \\ &\leq \int_0^{2 \pi/3} \frac{dr}{\abs{r^3 e^{3i\theta}} - 1}d\theta \\ &= \frac{2\pi}{3} \frac{1}{\displaystyle R^2 - \frac{1}{R}}. \end{align*} また \begin{align*} \int_{-C_2} f(z) dz &= - \int_0^R f(z) \frac{dz}{dr} dr \\ &= - \int_0^R \frac{e^{2\pi i /3}}{r^3 e^{2\pi i} + 1} dr \\ &= \bar{\alpha} \int_0^R \frac{1}{r^3 + 1} dr \end{align*} であるから，\eqref{b}は \begin{gather*} \int_{0}^R f(z) dz + \int_{C_1} f(z) dz + \int_{-C_2} f(z) dz = \frac{2 \pi i}{(\alpha + 1)\sqrt{3}i} \\ (1 + \bar{\alpha}) \int_{0}^R \frac{dx}{x^3 + 1} + \int_{C_1} f(z) dz = \frac{2 \pi}{(\alpha + 1)\sqrt{3}} \end{gather*} となり，\( R \to \infty \)として \begin{align*} \int_{0}^\infty \frac{dx}{x^3 + 1} & = \frac{2 \pi}{\sqrt{3}}\cdot\frac{1}{(\alpha + 1)(\bar{\alpha} + 1)} \\ & = \frac{2 \pi}{3 \sqrt{3}} \\ \end{align*} を得る．

(c) \( \displaystyle f(z) = \frac{1}{z^4 + 1}\) \((z \in \mathbb{C}) \)とおく．この関数は\( \displaystyle \alpha = \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}i \) とすると\( z = \alpha, \bar{\alpha}, -\alpha, -\bar{\alpha} \)を孤立特異点としてもつ．曲線\( C \)を\( C\colon z = Re^{i\theta} \) \( (R \gt 0, 0 \leq \theta \leq \pi) \) で定める．留数定理より \begin{align} \int_{-R}^R f(x) dx + \int_C f(z) dz = 2 \pi i \sum_{z \in \{\alpha, -\bar{\alpha}\}} \res(f, z) \label{c} \end{align} が成り立つ．

\(f\)について\( \displaystyle 0 \lt \abs{z - \alpha} \lt \frac{\sqrt{2}}{2} \)の範囲で \begin{align*} f(z) &= \frac{1}{(z - \alpha)(z + \bar{\alpha})(z + \alpha)(z - \bar{\alpha})} \\ &= \frac{1}{(z - \alpha)(z - \alpha + \alpha + \bar{\alpha})(z - \alpha + 2 \alpha)(z - \alpha + \alpha - \bar{\alpha})} \\ &= \frac{\displaystyle \frac{1}{\alpha + \bar{\alpha}} \cdot \frac{1}{2\alpha} \cdot \frac{1}{\alpha - \bar{\alpha}}}{\displaystyle (z - \alpha)\parentheses{1 + \frac{z - \alpha}{\alpha + \bar{\alpha}}} \parentheses{1 + \frac{z - \alpha}{2\alpha}} \parentheses{1 + \frac{z - \alpha}{\alpha - \bar{\alpha}}} } \\ &= \frac{1}{\alpha + \bar{\alpha}} \cdot \frac{1}{2\alpha} \cdot \frac{1}{\alpha - \bar{\alpha}} \cdot \frac{1}{z - \alpha} \cdot \parentheses{1 - \frac{z - \alpha}{\alpha + \bar{\alpha}} + \parentheses{\frac{z - \alpha}{\alpha + \bar{\alpha}}}^2 - \cdots} \\ &\quad\,\, \parentheses{1 - \frac{z - \alpha}{2\alpha} + \parentheses{\frac{z - \alpha}{2\alpha}}^2 - \cdots} \parentheses{1 - \frac{z - \alpha}{\alpha - \bar{\alpha}} + \parentheses{\frac{z - \alpha}{\alpha - \bar{\alpha}}}^2 - \cdots}. \end{align*} また\( \displaystyle 0 \lt \abs{z + \bar{\alpha}} \lt \frac{\sqrt{2}}{2} \)の範囲で \begin{align*} f(z) &= \frac{1}{(z + \bar{\alpha} - \bar{\alpha} - \alpha)(z + \bar{\alpha})(z + \bar{\alpha} - \bar{\alpha} + \alpha)(z + \bar{\alpha} - 2\bar{\alpha})} \\ &= \frac{\displaystyle \frac{-1}{\alpha + \bar{\alpha}} \cdot \frac{1}{\alpha - \bar{\alpha}} \cdot \frac{-1}{2\bar{\alpha}} }{\displaystyle \parentheses{1 - \frac{z + \bar{\alpha}}{\alpha + \bar{\alpha}}} (z + \bar{\alpha}) \parentheses{1 + \frac{z + \bar{\alpha}}{\alpha - \bar{\alpha}}} \parentheses{1 - \frac{z + \bar{\alpha}}{2\bar{\alpha}}} } \\ &= \frac{1}{\alpha + \bar{\alpha}} \cdot \frac{1}{2\bar{\alpha}} \cdot \frac{1}{\alpha - \bar{\alpha}} \cdot \frac{1}{z + \bar{\alpha}} \cdot \parentheses{1 + \frac{z + \alpha}{\alpha + \bar{\alpha}} + \parentheses{\frac{z + \bar{\alpha}}{\alpha + \bar{\alpha}}}^2 + \cdots} \\ &\quad\,\, \parentheses{1 - \frac{z + \bar{\alpha}}{\alpha - \bar{\alpha}} + \parentheses{\frac{z + \bar{\alpha}}{\alpha - \bar{\alpha}}}^2 - \cdots} \parentheses{1 + \frac{z + \bar{\alpha}}{2\bar{\alpha}} + \parentheses{\frac{z + \bar{\alpha}}{2\bar{\alpha}}}^2 + \cdots}. \end{align*} 以上より \begin{align*} 2 \pi i \sum_{z \in \{\alpha, -\bar{\alpha}\}} \res(f, z) &= 2\pi i \parentheses{\frac{1}{2\alpha(\alpha + \bar{\alpha})(\alpha - \bar{\alpha})} + \frac{1}{2\bar{\alpha}(\alpha + \bar{\alpha})(\alpha - \bar{\alpha})}} \\ &= 2 \pi i \cdot \frac{2 \real \alpha}{2 \abs{\alpha}^2 \cdot 2 \real \alpha \cdot 2 \imaginary \alpha} \\ &= \frac{\pi i}{2 \imaginary \alpha} \\ &= \frac{\sqrt{2}}{2}\pi. \end{align*}

一方 \begin{align*} \abs{\int_C f(z) dz} &= \abs{\int_0^{\pi} \frac{i R e^{i \theta}}{R^4 e^{4i\theta} + 1} d\theta} \\ &= \int_0^{\pi} \frac{\abs{i R e^{i \theta}}}{\abs{R^4 e^{4i\theta} + 1}} d\theta \\ &\leq \int_0^{\pi} \frac{R}{\abs{R^4 e^{4i\theta}} - 1}d\theta \\ &= \frac{R}{R^4 - 1} \pi \\ &= \frac{1}{\displaystyle R^3 - \frac{1}{R}} \pi \end{align*} である．よって\eqref{c}より \begin{align*} \int_{-\infty}^\infty f(x) dx &= \lim_{R \to \infty} \int_{-R}^R f(z) dz \\ &= \lim_{R \to \infty} \parentheses{\frac{\sqrt{2}}{2}\pi - \int_C f(z) dz} \\ &= \frac{\sqrt{2}}{2}\pi \end{align*} を得る．