コーシーの積分公式

コーシーの積分公式

\(K\)を単連結な(穴のない)領域とし，\(f(z)\)は\(K\)上で正則である(微分可能な)複素凾数とする． \(C\)を\(K\)内のある単純(自己交点を持たない)閉曲線であるとし，\(D\)を\(C\)の内部の領域とする． \(z\)を\(D\)内の任意の一点とすると，この時以下が成り立つ． $$\begin{eqnarray} \frac{1}{2\pi i}\oint_C\frac{f(\zeta)}{\zeta-z}\mathrm{d}\zeta&=&f(z)&\;\ldots\;z\in D \end{eqnarray}$$ \(f(z)\)の係数となる\(\frac{1}{\zeta-z}\)が\(\zeta=z\)で微分不可(特異点)となる．

\(z\)を\(D\)の外の場合

$$\begin{eqnarray} \href{https://shikitenkai.blogspot.com/2021/08/blog-post.html}{\frac{1}{2\pi i}\oint_C\frac{f(\zeta)}{\zeta-z}\mathrm{d}\zeta=0\;\ldots\;z\notin D} \end{eqnarray}$$ 積分対象全域で正則のままなので，\(\frac{f(\zeta)}{\zeta-z}\)を改めて\(f(z)\)と考えてコーシーの積分定理が成り立つ．

特異点を囲む微小半径の円(単純閉曲線)を考える

積分経路を\(C\)は\(z\)を含むかどうかが問題なので，\(z\)を中心とした半径\(\varepsilon\)の経路\(C^\prime\)に置き換えて考える． $$\begin{eqnarray} \oint_C\frac{f(\zeta)}{\zeta-z}\mathrm{d}\zeta &=&\oint_{C\prime}\frac{f(\zeta)}{\zeta-z}\mathrm{d}\zeta\;\ldots\;C\prime:|\zeta-z|=\varepsilon \\\zeta&=&z+\varepsilon e^{i\theta}　(0\le\theta<2\pi) \\\frac{\mathrm{d}\zeta}{\mathrm{d}\theta}&=&\varepsilon \cdot ie^{i\theta} \\\mathrm{d}\zeta&=&i\varepsilon e^{i\theta}d\theta \\\oint_{C\prime}\frac{f(\zeta)}{\zeta-z}\mathrm{d}\zeta &=&\lim_{\varepsilon\rightarrow0}\int_0^{2\pi}\frac{f\left(z+\varepsilon e^{i\theta}\right)}{z+\varepsilon e^{i\theta}-z}i\varepsilon e^{i\theta}d\theta \\ \\&=&\lim_{\varepsilon\rightarrow0}\int_0^{2\pi}\frac{f\left(z+\varepsilon e^{i\theta}\right)}{\varepsilon e^{i\theta}}i\varepsilon e^{i\theta}d\theta \\ \\&=&\lim_{\varepsilon\rightarrow0}i\int_0^{2\pi}f\left(z+\varepsilon e^{i\theta}\right)d\theta \\&=&\lim_{\varepsilon\rightarrow0}i\int_0^{2\pi}f\left(z+\varepsilon e^{i\theta}\right)d\theta \\&=&i\int_0^{2\pi}f\left(z\right)d\theta \\&=&if\left(z\right)\int_0^{2\pi}d\theta \\&=&if\left(z\right)2\pi \\&=&2\pi if\left(z\right) \\\oint_C\frac{f(\zeta)}{\zeta-z}\mathrm{d}\zeta&=&2\pi if\left(z\right) \\f\left(z\right)&=&\frac{1}{2\pi i}\oint_C\frac{f(\zeta)}{\zeta-z}\mathrm{d}\zeta \end{eqnarray}$$

コーシーの積分定理

コーシーの積分定理(Cauchy's Integral Theorem)

\(K\)を単連結な(穴のない)領域とし，\(f(z)\)は\(K\)上で正則である(微分可能な)複素凾数とする． \(C\)を\(K\)内のある単純(自己交点を持たない)閉曲線であるとし，\(D\)を\(C\)の内部の領域とすると，この時以下が成り立つ． $$\begin{eqnarray} \oint_C f(\zeta)\mathrm{d}\zeta&=&0 \end{eqnarray}$$

グリーンの定理(Green's theorem)

閉曲線\(C\)で囲まれた領域\(D\)を考える場合，\(C^1\)級凾数\(P(x, y), Q(x, y)\)について以下が成り立つ。 $$ \begin{eqnarray} \iint_D \left(\frac{\partial Q(x, y)}{\partial x}-\frac{\partial P(x, y)}{\partial y}\right)\mathrm{d}x\mathrm{d}y &=& \oint_C P(x, y)\mathrm{d}x+Q(x,y)\mathrm{d}y \end{eqnarray} $$

グリーンの定理とコーシー・リーマンの関係式より

$$\begin{eqnarray} \oint_C f(\zeta)\mathrm{d}\zeta &=&\oint_C \left\{u(x, y) + i v(x, y)\right\}(\mathrm{d}x+i\mathrm{d}y) \\&=&\oint_C \left\{u(x, y)\mathrm{d}x + i v(x, y)\mathrm{d}x+u(x, y)i\mathrm{d}y + i v(x, y)i\mathrm{d}y\right\} \\&=&\oint_C \left\{u(x, y)\mathrm{d}x - v(x, y)\mathrm{d}y\right\}+ i \left\{v(x, y)\mathrm{d}x+u(x, y)\mathrm{d}y \right\} \\&=&\iint_D \left\{-\frac{\partial v(x,y)}{\partial x}-\frac{\partial u(x,y)}{\partial y}\right\}\mathrm{d}x\mathrm{d}y + i \iint_D \left\{\frac{\partial u(x,y)}{\partial x}-\frac{\partial v(x,y)}{\partial y}\right\}\mathrm{d}x\mathrm{d}y \\&&\;\ldots\;グリーンの定理:\oint_C\left(P\mathrm{d}x+Q\mathrm{d}y\right) =\iint_D\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right) \\&=&-\iint_D \left\{\frac{\partial v(x,y)}{\partial x}+\frac{\partial u(x,y)}{\partial y}\right\}\mathrm{d}x\mathrm{d}y + i \iint_D \left\{\frac{\partial u(x,y)}{\partial x}-\frac{\partial v(x,y)}{\partial y}\right\}\mathrm{d}x\mathrm{d}y \\&=&-\iint_D 0\mathrm{d}x\mathrm{d}y+ i \iint_D 0\mathrm{d}x\mathrm{d}y \\&&\;\ldots\;K上で正則凾数なのでコーシー・リーマンの関係式がなりたつ \href{https://shikitenkai.blogspot.com/2021/07/blog-post_19.html}{\left\{\begin{array} \\\frac{\partial u(x,y)}{\partial x}=\frac{\partial v(x,y)}{\partial y} \\\frac{\partial v(x,y)}{\partial x}=-\frac{\partial u(x,y)}{\partial y} \end{array}\right.} \\&=&0+i0=0 \end{eqnarray}$$

バーゼル問題(ディガンマ凾数の相反公式から)

平方数の逆数全ての和はいくつかという問題(バーゼル問題)

$$\begin{eqnarray} \sum_{k=1}^{\infty}\frac{1}{k^2}&=&\frac{\pi}{6} \end{eqnarray}$$

ディガンマ凾数の相反公式

$$\begin{eqnarray} \psi\left(1-z\right)-\psi\left(z\right) &=&\href{https://shikitenkai.blogspot.com/2021/07/blog-post_22.html}{\pi \cot{\left(\pi z\right)}} \end{eqnarray}$$
その微分 $$\begin{eqnarray} \frac{\partial}{\partial z}\pi \cot{\left(\pi z\right)} &=& \pi\frac{\partial}{\partial z} \cot{\left(\pi z\right)} \\&=&\pi\left\{-\frac{\pi}{\sin^2{\left(\pi z\right)}}\right\} \;\ldots\;\href{https://shikitenkai.blogspot.com/2021/07/cotaz.html}{\frac{\partial}{\partial w} \cot{\left(a w\right)}=-\frac{a}{\sin^2{\left(aw\right)}}} \\&=&-\frac{\pi^2}{\sin^2{\left(\pi z\right)}} \\&=&f(z) \end{eqnarray}$$

\(f(z)\)の\(z=\frac{1}{3}\)の値

$$\begin{eqnarray} -\frac{\pi^2}{\sin^2{\left(\pi \frac{1}{3}\right)}} &=&-\frac{\pi^2}{\left(\frac{\sqrt{3}}{2}\right)^2} \;\ldots\;\sin{\left(\frac{\pi}{3}\right)}=\frac{\sqrt{3}}{2} \\&=&-\frac{\pi^2}{\frac{3}{4}} \\&=&\frac{-4\pi^2}{3} \end{eqnarray}$$

ディガンマ凾数の相反公式の積分表示

$$\begin{eqnarray} \psi\left(1-z\right)-\psi\left(z\right) &=&\int_0^1\frac{u^{(z)-1}-u^{(1-z)-1}}{1-u}\mathrm{d}u \;\ldots\;\href{https://shikitenkai.blogspot.com/2021/07/blog-post_13.html}{\psi\left(y\right)-\psi\left(x\right)=\int_{0}^1 \frac{u^{x-1}-u^{y-1}}{1-u}\mathrm{d}u} \\&=&\int_0^1\frac{u^{z-1}-u^{-z}}{1-u}\mathrm{d}u \\&=&\int_\infty^0\frac{(e^{-x})^{z-1} - (e^{-x})^{-z}}{ 1-e^{-x} } (-e^{-x})\mathrm{d}x \\&&\;\ldots\;u=e^{-x},\;\frac{\mathrm{d}u}{\mathrm{d}x}=-e^{-x} \\&&\;\ldots\;u=e^{-x},\;u:0\rightarrow1, x:\infty\rightarrow0 \\&=&\int_0^\infty\frac{e^{-x}\left\{e^{-x(z-1)} - e^{-x(-z)}\right\}}{ 1-e^{-x} } \mathrm{d}x \\&=&\int_0^\infty\frac{e^{-x}e^{-xz+x} - e^{-x}e^{xz}}{ 1-e^{-x} } \mathrm{d}x \\&=&\int_0^\infty\frac{e^{-x-xz+x} - e^{-x+xz}}{ 1-e^{-x} } \mathrm{d}x \\&=&\int_0^\infty\frac{e^{-xz} - e^{-x(1-z)}}{ 1-e^{-x} } \mathrm{d}x \end{eqnarray}$$
その微分 $$\begin{eqnarray} \frac{\partial}{\partial z}\int_0^\infty\frac{e^{-xz} - e^{-x(1-z)}}{ 1-e^{-x} } \mathrm{d}x &=&\int_0^\infty\frac{1}{ 1-e^{-x} } \left(\frac{\partial}{\partial z}e^{-xz} - \frac{\partial}{\partial z}e^{-x(1-z)}\right) \mathrm{d}x \\&=&\int_0^\infty\frac{1}{ 1-e^{-x} } \left(-xe^{-xz} - (-x)(-1)e^{-x(1-z)}\right) \mathrm{d}x \\&=&\int_0^\infty\frac{1}{ 1-e^{-x} } \left(-xe^{-xz} - xe^{-x(1-z)}\right) \mathrm{d}x \\&=&\int_0^\infty\frac{1}{ 1-e^{-x} } \left\{-x\left(e^{-xz} + e^{-x(1-z)}\right)\right\} \mathrm{d}x \\&=&-\int_0^\infty\frac{1}{ 1-e^{-x} } \left\{x\left(e^{-xz} + e^{-x(1-z)}\right)\right\} \mathrm{d}x \\&=&g(z) \end{eqnarray}$$

\(g(z)\)の\(z=\frac{1}{3}\)の値

$$\begin{eqnarray} g\left(\frac{1}{3}\right) &=&-\int_0^\infty\frac{1}{ 1-e^{-x} } \left\{x\left(e^{-x\frac{1}{3}} + e^{-x(1-\frac{1}{3})}\right)\right\} \mathrm{d}x \\&=&-\int_0^\infty\frac{1}{ 1-e^{-x} } \left\{x\left(e^{-\frac{x}{3}} + e^{\frac{-2x}{3}}\right)\right\} \mathrm{d}x \\&=&-\int_0^\infty \frac{e^{x}}{e^{x}}\frac{1}{ 1-e^{-x} } \left\{ x\left( e^{-\frac{x}{3}} + e^{\frac{-2x}{3}} \right) \right\} \mathrm{d}x \\&=&-\int_0^\infty \frac{e^{x}}{ e^{x}\left(1-e^{-x}\right) } \left\{ x\left( e^{-\frac{x}{3}} + e^{\frac{-2x}{3}} \right) \right\} \mathrm{d}x \\&=&-\int_0^\infty \frac{e^{x}}{ e^{x}-e^{x}e^{-x} } \left\{ x\left( e^{-\frac{x}{3}} + e^{\frac{-2x}{3}} \right) \right\} \mathrm{d}x \\&=&-\int_0^\infty \frac{e^{x}}{ e^{x}-1 } \left\{ x\left( e^{-\frac{x}{3}} + e^{\frac{-2x}{3}} \right) \right\} \mathrm{d}x \\&=&-\int_0^\infty \left(\sum_{k=0}^{\infty}e^{-kx}\right) \left\{ x\left( e^{-\frac{x}{3}} + e^{\frac{-2x}{3}} \right) \right\} \mathrm{d}x \;\ldots\;\sum_{k=0}^{\infty}e^{-kx}=\frac{e^x}{e^x-1} ,\;\href{https://shikitenkai.blogspot.com/2021/07/a-kx.html}{\sum_{k=0}^{\infty}a^{-kx}=\frac{a^x}{a^x-1}} \\&=&-\sum_{k=0}^{\infty}\int_0^\infty e^{-kx} \left\{ x\left( e^{-\frac{x}{3}} + e^{\frac{-2x}{3}} \right) \right\} \mathrm{d}x \\&=&-\sum_{k=0}^{\infty}\int_0^\infty x\left( e^{-kx}e^{-\frac{x}{3}} + e^{-kx}e^{\frac{-2x}{3}} \right) \mathrm{d}x \\&=&-\sum_{k=0}^{\infty}\int_0^\infty x\left( e^{-x(k+\frac{1}{3})} + e^{-x(k+\frac{2}{3})} \right) \mathrm{d}x \\&=&-\sum_{k=0}^{\infty} \left( \int_0^\infty xe^{-x(k+\frac{1}{3})} \mathrm{d}x + \int_0^\infty xe^{-x(k+\frac{2}{3})} \mathrm{d}x \right) \\&=&-\sum_{k=0}^{\infty} \left( \frac{1}{\left(k+\frac{1}{3}\right)^2} + \frac{1}{\left(k+\frac{2}{3}\right)^2} \right) \;\ldots\;\href{https://shikitenkai.blogspot.com/2021/07/xe-ax0.html}{\int_0^\infty xe^{-ax} \mathrm{d}x=\frac{1}{a^2}} \\&=&-\sum_{k=0}^{\infty} \left( \frac{1}{\left(\frac{3k+1}{3}\right)^2} + \frac{1}{\left(\frac{3k+2}{3}\right)^2} \right) \\&=&-\sum_{k=0}^{\infty} \left( \frac{3^2}{\left(3k+1\right)^2} + \frac{3^2}{\left(3k+2\right)^2} \right) \\&=&-\sum_{k=0}^{\infty} \left( \frac{3^2}{\left(3k+1\right)^2} + \frac{3^2}{\left(3k+2\right)^2} \color{red}{+ \frac{3^2}{\left(3k+3\right)^2} - \frac{3^2}{\left(3k+3\right)^2}} \color{black}{} \right) \\&=&-\sum_{k=0}^{\infty} \left( \frac{3^2}{\left(3k+1\right)^2} + \frac{3^2}{\left(3k+2\right)^2} + \frac{3^2}{\left(3k+3\right)^2} \right) - \sum_{k=0}^{\infty} \left( - \frac{3^2}{\left(3k+3\right)^2} \right) \\&=&-\sum_{k=0}^{\infty} \left( \frac{3^2}{\left(3k+1\right)^2} + \frac{3^2}{\left(3k+2\right)^2} + \frac{3^2}{\left(3k+3\right)^2} \right) + \sum_{k=0}^{\infty} \frac{3^2}{\left(3k+3\right)^2} \\&=&-\sum_{k=0}^{\infty} \left( \frac{3^2}{\left(3k+1\right)^2} + \frac{3^2}{\left(3k+2\right)^2} + \frac{3^2}{\left(3k+3\right)^2} \right) +\sum_{k=0}^{\infty}\frac{3^2}{3^2\left(k+1\right)^2} \\&=&-\sum_{k=0}^{\infty} \left( \frac{3^2}{\left(3k+1\right)^2} + \frac{3^2}{\left(3k+2\right)^2} + \frac{3^2}{\left(3k+3\right)^2} \right) +\sum_{k=0}^{\infty}\frac{1}{\left(k+1\right)^2} \\&=&-\sum_{k=0}^{\infty} \left( \frac{3^2}{\left(3k+1\right)^2} + \frac{3^2}{\left(3k+2\right)^2} + \frac{3^2}{\left(3k+3\right)^2} \right) +\sum_{k=1}^{\infty}\frac{1}{k^2} \\&&\;\ldots\;\sum_{k=0}^{\infty}\frac{1}{\left(k+1\right)^2} =\frac{1}{\left(0+1\right)^2}+\frac{1}{\left(1+1\right)^2}+\frac{1}{\left(2+1\right)^2}+\cdots =\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\cdots =\sum_{k=1}^{\infty}\frac{1}{k^2} \\&=&-3^2\sum_{k=0}^{\infty} \left( \frac{1}{\left(3k+1\right)^2} + \frac{1}{\left(3k+2\right)^2} + \frac{1}{\left(3k+3\right)^2} \right) +\sum_{k=1}^{\infty}\frac{1}{k^2} \\&=&-9\sum_{k=1}^{\infty}\frac{1}{k^2} +\sum_{k=1}^{\infty}\frac{1}{k^2} \\&&\;\ldots\; \sum_{k=0}^{\infty} \left( \frac{1}{\left(3k+1\right)^2} + \frac{1}{\left(3k+2\right)^2} + \frac{1}{\left(3k+3\right)^2} \right) = \left( \frac{1}{\left(3\cdot0+1\right)^2} + \frac{1}{\left(3\cdot0+2\right)^2} + \frac{1}{\left(3\cdot0+3\right)^2} \right) + \left( \frac{1}{\left(3\cdot1+1\right)^2} + \frac{1}{\left(3\cdot1+2\right)^2} + \frac{1}{\left(3\cdot1+3\right)^2} \right) +\cdots = \left( \frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} \right) + \left( \frac{1}{4^2} + \frac{1}{5^2} + \frac{1}{6^2} \right) +\cdots =\sum_{k=1}^{\infty}\frac{1}{k^2} \\&=&-8\sum_{k=1}^{\infty}\frac{1}{k^2} \end{eqnarray}$$

fとgが等しいことを利用

$$\begin{eqnarray} g\left(\frac{1}{3}\right) &=&f\left(\frac{1}{3}\right) \\-8\sum_{k=1}^{\infty}\frac{1}{k^2} &=&\frac{-4\pi^2}{3} \\\sum_{k=1}^{\infty}\frac{1}{k^2} &=&\frac{-1}{8}\frac{-4\pi^2}{3} \\&=&\frac{\pi^2}{6} \end{eqnarray}$$