diff --git a/docs/kakomonn/kyushu_university/ISEE/kyotsu_2018_complex_function_theory.md b/docs/kakomonn/kyushu_university/ISEE/kyotsu_2018_complex_function_theory.md new file mode 100644 index 000000000..52179d437 --- /dev/null +++ b/docs/kakomonn/kyushu_university/ISEE/kyotsu_2018_complex_function_theory.md @@ -0,0 +1,144 @@ +--- +comments: false +title: 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 +tags: + - Kyushu-University +--- +# 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、 + +$$ +-\frac{\pi^3}{4} = 0 + 2\int_0^{\infty}\frac{(\ln x)^2}{x^2 + 1}dx - \frac{\pi^3}{2} + 0 +$$ + +$$ +\therefore \int_0^{\infty}\frac{(\ln x)^2}{x^2 + 1}dx = \frac{\pi^3}{8} +$$ \ No newline at end of file diff --git a/docs/kakomonn/kyushu_university/index.md b/docs/kakomonn/kyushu_university/index.md index 0850a4cae..69074437a 100644 --- a/docs/kakomonn/kyushu_university/index.md +++ b/docs/kakomonn/kyushu_university/index.md @@ -73,6 +73,8 @@ tags: - [解析学・微積分](ISEE/kyotsu_2019_analysis_calculus.md) - [ベクトル解析](ISEE/kyotsu_2019_vector_analysis.md) - [確率・統計](ISEE/kyotsu_2019_prob_stat.md) + - 2018年度: + - [複素関数論](ISEE/kyotsu_2018_complex_function_theory.md) - 2016年度: - [複素関数論](ISEE/kyotsu_2016_complex_function_theory.md) - 情報理工学専攻: diff --git a/mkdocs.yml b/mkdocs.yml index 055ad2494..009d512a6 100644 --- a/mkdocs.yml +++ b/mkdocs.yml @@ -924,6 +924,8 @@ nav: - 解析学・微積分: kakomonn/kyushu_university/ISEE/kyotsu_2019_analysis_calculus.md - ベクトル解析: kakomonn/kyushu_university/ISEE/kyotsu_2019_vector_analysis.md - 確率・統計: kakomonn/kyushu_university/ISEE/kyotsu_2019_prob_stat.md + - 2018年度: + - 複素関数論: kakomonn/kyushu_university/ISEE/kyotsu_2018_complex_function_theory.md - 2016年度: - 複素関数論: kakomonn/kyushu_university/ISEE/kyotsu_2016_complex_function_theory.md - 情報理工学専攻: