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

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

a^(-k)及びa^(-kx)の無限級数

\(a^{-k}\)及び\(a^{-kx}\)の無限級数

\(a^{-k}\)の無限級数

$$\begin{eqnarray} S=\sum_{k=0}^{\infty}a^{-k} &=&\sum_{k=0}^{\infty}\left(a^{-1}\right)^k \;\ldots\;a\in\mathbb{R},\;\left|a\right|^{-1}=\frac{1}{\left|a\right|}\lt1 \\&=&\sum_{k=0}^{\infty}\left(\frac{1}{a}\right)^k \\&=&\sum_{k=0}^{\infty}\frac{1}{a^{k}} \\&=&\frac{1}{a^0}+\frac{1}{a}+\frac{1}{a^{2}}+\frac{1}{a^{3}}+\cdots \\&=&1+\frac{1}{a}+\frac{1}{a\cdot a}+\frac{1}{a\cdot a^{2}}+\cdots \\&=&1+\frac{1}{a}+\frac{1}{a}\frac{1}{a}+\frac{1}{a}\frac{1}{a^{2}}+\cdots \\&=&1+\frac{1}{a}\left(1+\frac{1}{a}+\frac{1}{a^{2}}+\cdots\right) \\&=&1+\frac{1}{a}S \\S-\frac{1}{a}S&=&1 \\S\left(1-\frac{1}{a}\right)&=&1 \\S&=&\frac{1}{1-\frac{1}{a}} \\&=&\frac{1}{\frac{a-1}{a}} \\&=&\frac{a}{a-1} \end{eqnarray}$$

\(a^{-kx}\)の無限級数

$$\begin{eqnarray} S=\sum_{k=0}^{\infty}a^{-kx}&=&\sum_{k=0}^{\infty}\left(a^{-x}\right)^k \;\ldots\;a,x\in\mathbb{R},\;|a|^{-x}=\frac{1}{|a|^{x}}\lt1 \\&=&\sum_{k=0}^{\infty}\left(\frac{1}{a^{x}}\right)^k \\&=&\sum_{k=0}^{\infty}\frac{1}{a^{kx}} \\&=&\frac{1}{a^{0x}}+\frac{1}{a^x}+\frac{1}{a^{2x}}+\frac{1}{a^{3x}}+\cdots \\&=&1+\frac{1}{a^{x}}+\frac{1}{a^{x}a^{x}}+\frac{1}{a^{x}a^{2x}}+\cdots \\&=&1+\frac{1}{a^{x}}+\frac{1}{a^{x}}\frac{1}{a^{x}}+\frac{1}{a^{x}}\frac{1}{a^{2x}}+\cdots \\&=&1+\frac{1}{a^{x}}\left(1+\frac{1}{a^{x}}+\frac{1}{a^{2x}}+\cdots\right) \\&=&1+\frac{1}{a^{x}}S \\S-\frac{1}{a^x}S&=&1 \\S\left(1-\frac{1}{a^x}\right)&=&1 \\S&=&\frac{1}{1-\frac{1}{a^x}} \\&=&\frac{1}{\frac{a^x-1}{a^x}} \\&=&\frac{a^x}{a^x-1} \end{eqnarray}$$

ガンマ凾数の相反公式の積分表示

ガンマ凾数の相反公式の積分表示

ベータ凾数のガンマ凾数での表示

$$\begin{eqnarray} \left.B\left(p,q\right)\right|_{p=1-q}&=&\href{https://shikitenkai.blogspot.com/2020/05/blog-post_22.html}{\left.\frac{\Gamma\left(p\right)\Gamma\left(q\right)}{\Gamma\left(p+q\right)}\right|_{p=1-q}} \;\ldots\;p,q\in\mathbb{C},\;\Re\left(p\right),\Re\left(q\right)\gt0 \\&=&\frac{\Gamma\left(1-q\right)\Gamma\left(q\right)}{\Gamma\left(1-q+q\right)} \\&=&\frac{\Gamma\left(1-q\right)\Gamma\left(q\right)}{\Gamma\left(1\right)} \\&=&\frac{\Gamma\left(1-q\right)\Gamma\left(q\right)}{1} \\&&\;\ldots\;\Gamma\left(1\right)=\int_0^\infty t^{1-1}e^{-t}\mathrm{d}t=\int_0^\infty t^{0}e^{-t}\mathrm{d}t=\int_0^\infty 1\cdot e^{-t}\mathrm{d}t=\int_0^\infty e^{-t}\mathrm{d}t \\&&\;\ldots\;\href{https://shikitenkai.blogspot.com/2021/07/e-ax0.html}{\int_0^\infty e^{-t}\mathrm{d}t=\frac{1}{1}}=1 \\&=&\Gamma\left(1-q\right)\Gamma\left(q\right) \end{eqnarray}$$

ベータ凾数の定義での表示

$$\begin{eqnarray} \left.B\left(p,q\right)\right|_{p=1-q} &=&\int_0^1 t^{p-1}\left(1-t\right)^{q-1}\mathrm{d}t \\&=&\int_\infty^0 \left(\frac{1}{x+1}\right)^{p-1}\left(1-\frac{1}{x+1}\right)^{q-1}\cdot\frac{-1}{\left(x+1\right)^2}\mathrm{d}x \\&&\;\ldots\;t=\frac{1}{x+1},\;x=\frac{1}{t}-1,\;\frac{\mathrm{d}t}{\mathrm{d}x}=\frac{-1}{\left(x+1\right)^2} \\&&\;\ldots\;t:0\rightarrow1,\;x:\infty\rightarrow0 \\&=&\int_0^\infty \left(\frac{1}{x+1}\right)^{p-1}\left(1-\frac{1}{x+1}\right)^{q-1}\cdot\frac{1}{\left(x+1\right)^2}\mathrm{d}x \\&=&\int_0^\infty \left(\frac{1}{x+1}\right)^{p-1}\left(\frac{x}{x+1}\right)^{q-1}\cdot\frac{1}{\left(x+1\right)^2}\mathrm{d}x \\&=&\int_0^\infty \frac{1^{p-1}\cdot x^{q-1}\cdot 1}{\left(x+1\right)^{p-1}\left(x+1\right)^{q-1}\left(x+1\right)^{2}}\mathrm{d}x \\&=&\int_0^\infty \frac{x^{q-1}}{\left(x+1\right)^{(p-1)+(q-1)+2}}\mathrm{d}x \\&=&\int_0^\infty \frac{x^{q-1}}{\left(x+1\right)^{p+q}}\mathrm{d}x \\&=&\int_0^\infty \frac{x^{q-1}}{\left(x+1\right)^{1-q+q}}\mathrm{d}x \\&=&\int_0^\infty \frac{x^{q-1}}{x+1}\mathrm{d}x \end{eqnarray}$$

ガンマ凾数の相反公式の積分表示

$$\begin{eqnarray} \Gamma\left(1-q\right)\Gamma\left(q\right)&=&\int_0^\infty \frac{x^{q-1}}{x+1}\mathrm{d}x \end{eqnarray}$$

ガンマ凾数の相反公式

ガンマ凾数の相反公式

$$\begin{eqnarray} \Gamma\left(z\right)\Gamma\left(1-z\right) &=&\Gamma\left(z\right)\Gamma\left(-z+1\right) \;\ldots\;\Re\left(z\right)\gt0かつ\Re\left(1-z\right)\gt0なので0\lt\Re\left(z\right)\lt1 \\&=&\Gamma\left(z\right)\left(-z\Gamma\left(-z\right)\right) \;\ldots\;\href{https://shikitenkai.blogspot.com/2020/08/s1ss.html}{\Gamma\left(z+1\right)=z\Gamma\left(z\right)} \\&=&-z\Gamma\left(z\right)\Gamma\left(-z\right) \\&=&-z \left(\lim_{n\rightarrow\infty}\frac{n^zn!}{\prod_{k=0}^n (z+k)}\right) \left(\lim_{n\rightarrow\infty}\frac{n^{-z}n!}{\prod_{k=0}^n (-z+k)}\right) \;\ldots\;\href{https://shikitenkai.blogspot.com/2021/07/blog-post_16.html}{\Gamma\left(z\right)=\lim_{n\rightarrow\infty}\frac{n^zn!}{\prod_{k=0}^n (z+k)}} \\&=&-z \left(\lim_{n\rightarrow\infty}\frac{n^zn! \cdot n^{-z}n!}{\prod_{k=0}^n (z+k)(-z+k)}\right) \\&=&-z\left(\lim_{n\rightarrow\infty}\frac{n!n!}{\prod_{k=0}^n (k^2-z^2)}\right) \;\ldots\;A^BA^{-B}=A^{B-B}=A^{0}=1 \\&=&-z\left(\lim_{n\rightarrow\infty}\frac{\left(\prod_{k=1}^nk\right)\left(\prod_{k=1}^nk\right)}{\prod_{k=0}^n (k^2-z^2)}\right) \;\ldots\;n!=\prod_{k=1}^nk \\&=&-z\left(\lim_{n\rightarrow\infty}\frac{\prod_{k=1}^nk^2}{\prod_{k=0}^n (k^2-z^2)}\right) \;\ldots\;\left(\prod_{k=m}^nA_k\right)\left(\prod_{k=m}^nB_k\right)=\prod_{k=m}^nA_kB_k \\&=&-z\left(\lim_{n\rightarrow\infty}\frac{\prod_{k=1}^nk^2}{(0^2-z^2)\prod_{k=1}^n (k^2-z^2)}\right) \\&=&-z\frac{1}{-z^2}\left(\lim_{n\rightarrow\infty}\prod_{k=1}^n\frac{k^2}{k^2-z^2}\right) \\&=&\frac{1}{z}\prod_{k=1}^\infty\frac{k^2}{k^2-z^2} \\&=&\frac{1}{\frac{1}{\frac{1}{z}\prod_{k=0}^\infty\frac{k^2}{k^2-z^2}}}\;\ldots\;A=\frac{1}{\frac{1}{A}} \\&=&\frac{1}{z\prod_{k=1}^\infty\frac{k^2-z^2}{k^2}} \\&=&\frac{\pi}{\pi}\frac{1}{z\prod_{k=0}^\infty\frac{k^2-z^2}{k^2}} \\&=&\frac{\pi}{\pi z\prod_{k=0}^\infty \left(1-\frac{z^2}{k^2}\right)} \\&=&\frac{\pi}{\sin{\left(\pi z\right)}} \\&&\;\ldots\;\href{https://ja.wikipedia.org/wiki/%E4%B8%89%E8%A7%92%E9%96%A2%E6%95%B0%E3%81%AE%E7%84%A1%E9%99%90%E4%B9%97%E7%A9%8D%E5%B1%95%E9%96%8B}{\sin{\left(\pi z\right)} =\pi z\prod_{k=1}^\infty\left(1-\left(\frac{z}{k}\right)^2\right)\;(三角凾数の無限乗積展開 wikipedia)} \end{eqnarray}$$

xe^(-ax)の広義積分[0,∞]

\(xe^{-ax}\)の広義積分\([0,\infty]\)

\begin{eqnarray} \int_0^\infty xe^{-ax} \mathrm{d}x &=&\lim_{n\rightarrow\infty} \int_0^n xe^{-ax} \mathrm{d}x \\&=&\lim_{n\rightarrow\infty} \left[ \left[x \cdot \frac{-1}{a}e^{-ax} \right]_0^n - \int_0^n \frac{-1}{a} \cdot e^{-ax} \mathrm{d}x \right] \\&=&\lim_{n\rightarrow\infty} \left[ \left[n \frac{-1}{a}e^{-an} - 0 \cdot \frac{-1}{a}e^{-a 0}\right] + \frac{1}{a} \int_0^n e^{-ax} \mathrm{d}x \right] \\&=&\lim_{n\rightarrow\infty} \left[ n \frac{-1}{a}e^{-an} + \frac{1}{a} \int_0^n \cdot e^{-ax} \mathrm{d}x \right] \\&=&\frac{-1}{a}\lim_{n\rightarrow\infty} ne^{-an} + \frac{1}{a} \lim_{n\rightarrow\infty} \int_0^n \cdot e^{-ax} \mathrm{d}x \\&=&0 + \frac{1}{a}\lim_{n\rightarrow\infty} \int_0^n e^{-ax} \mathrm{d}x \\&&指数凾数\left(e^{-x}\right)の方が線形凾数\left(x\right)より早く漸近化するので\lim_{n\rightarrow\infty}e^{-ax}=0が収束先となる． \\&=&\frac{1}{a}\lim_{n\rightarrow\infty} \int_0^n e^{-ax} \mathrm{d}x \\&=&\frac{1}{a}\cdot\frac{1}{a}\;\ldots\;\int_0^\infty e^{-ax} \mathrm{d}x=\frac{1}{a} \\&=&\frac{1}{a^2} \end{eqnarray}

e^(-ax)の広義積分[0,∞]

\(e^{-ax}\)の広義積分\([0,\infty]\)

\begin{eqnarray} \int_0^\infty e^{-ax} \mathrm{d}x &=&\lim_{n\rightarrow\infty} \int_0^n e^{-ax} \mathrm{d}x \\&=&\lim_{n\rightarrow\infty} \int_0^{-\infty} e^{u} \left(\frac{-1}{a}\right)\mathrm{d}u \\&&\;\ldots\;u=-ax,\;\frac{\mathrm{d}u}{\mathrm{d}x}=-a,\;\mathrm{d}x=\frac{-1}{a}\mathrm{d}u \\&&\;\ldots\;x:0\rightarrow n,\;u:0\rightarrow -an \\&=&\lim_{n\rightarrow\infty} \frac{-1}{a}\int_0^{-an} e^{u}\mathrm{d}u \\&=&\lim_{n\rightarrow\infty} \frac{-1}{a}\left[e^{u} \right]_0^{-an} \\&=&\lim_{n\rightarrow\infty} \frac{-1}{a}\left[e^{-an} - e^{0}\right] \\&=&\frac{-1}{a}\left[0 - 1\right] \\&=&\frac{-1}{a}\cdot(-1) \\&=&\frac{1}{a} \end{eqnarray}

cos(z)の微分

\(\cos{\left(z\right)}\)の微分

\(u+iv\)で表す

$$\begin{eqnarray} \cos{\left(z\right)} &=&\cos{\left(x\right)}\cos{\left(iy\right)}-\sin{\left(x\right)}\sin{\left(iy\right)} \\&=&\cos{\left(x\right)}\cosh{\left(y\right)}-\sin{\left(x\right)}i\sinh{\left(y\right)} \\&&\;\ldots\;\href{https://shikitenkai.blogspot.com/2021/07/cosi-x-sini-x-cos-sin.html}{\cos\left(iy\right)=i\cosh\left(y\right),\;\sin\left(iy\right)=i\sinh\left(y\right)} \\&=&\cos{\left(x\right)}\cosh{\left(y\right)}-i\sin{\left(x\right)}\sinh{\left(y\right)} \\&=&u(x,y)+iv(x,y) \end{eqnarray}$$ $$\left\{\begin{eqnarray} u(x,y)&=&\cos{\left(x\right)}\cosh{\left(y\right)} \\v(x,y)&=&-\sin{\left(x\right)}\sinh{\left(y\right)} \end{eqnarray}\;\ldots\;x,y\in\mathbb{R}\right.$$

\(u,v\)を\(x,y\)で偏微分する

$$\begin{eqnarray} \frac{\partial u(x,y)}{\partial x} &=&\frac{\partial}{\partial x}\cos{\left(x\right)}\cosh{\left(y\right)} \;\ldots\;x,y\in\mathbb{R} \\&=&\cosh{\left(y\right)}\frac{\partial}{\partial x}\cos{\left(x\right)} \\&=&\cosh{\left(y\right)}\left(-\sin{\left(x\right)}\right) \;\ldots\;\frac{\mathrm{d}}{\mathrm{d}\theta}\cos{\left(\theta\right)}=-\sin{\left(\theta\right)} \\&=&-\sin{\left(x\right)}\cosh{\left(y\right)} \end{eqnarray}$$ $$\begin{eqnarray} \frac{\partial u(x,y)}{\partial y} &=&\frac{\partial}{\partial y}\cos{\left(x\right)}\cosh{\left(y\right)} \;\ldots\;x,y\in\mathbb{R} \\&=&\cos{\left(x\right)}\frac{\partial}{\partial y}\cosh{\left(y\right)} \\&=&\cos{\left(x\right)}\sinh{\left(y\right)} \;\ldots\;\frac{\mathrm{d}}{\mathrm{d}\theta}\cosh{\left(\theta\right)}=\sinh{\left(\theta\right)} \end{eqnarray}$$ $$\begin{eqnarray} \frac{\partial v(x,y)}{\partial x} &=&\frac{\partial}{\partial x}\left(-\sin{\left(x\right)}\sinh{\left(y\right)}\right) \;\ldots\;x,y\in\mathbb{R} \\&=&-\sinh{\left(y\right)}\frac{\partial}{\partial x}\sin{\left(x\right)} \\&=&-\sinh{\left(y\right)}\cos{\left(x\right)} \;\ldots\;\frac{\mathrm{d}}{\mathrm{d}\theta}\sin{\left(\theta\right)}=\cos{\left(\theta\right)} \\&=&-\cos{\left(x\right)}\sinh{\left(y\right)} \end{eqnarray}$$ $$\begin{eqnarray} \frac{\partial v(x,y)}{\partial y} &=&\frac{\partial}{\partial y}\left(-\sin{\left(x\right)}\sinh{\left(y\right)}\right) \;\ldots\;x,y\in\mathbb{R} \\&=&-\sin{\left(x\right)}\frac{\partial}{\partial y}\sinh{\left(y\right)} \\&=&-\sin{\left(x\right)}\cosh{\left(y\right)} \;\ldots\;\frac{\mathrm{d}}{\mathrm{d}\theta}\sinh{\left(\theta\right)}=\cosh{\left(\theta\right)} \end{eqnarray}$$

コーシー・リーマンの関係式を満たす

$$\href{https://shikitenkai.blogspot.com/2021/07/blog-post_19.html}{\left\{ \begin{eqnarray} \frac{\partial u}{\partial x}&=&\frac{\partial v}{\partial y} \\\frac{\partial v}{\partial x}&=&-\frac{\partial u}{\partial y} \end{eqnarray} \right.}$$

実軸方向での微分

$$\begin{eqnarray} \frac{\mathrm{d}}{\mathrm{d}z}\cos{\left(z\right)} &=&\href{https://shikitenkai.blogspot.com/2021/07/blog-post_19.html}{\frac{\partial u(x,y)}{\partial x}+i\frac{\partial v(x,y)}{\partial x}} \\&=&-\sin{\left(x\right)}\cosh{\left(y\right)}+i\left(-\cos{\left(x\right)}\sinh{\left(y\right)}\right) \\&=&-\sin{\left(x\right)}\cosh{\left(y\right)}-i\cos{\left(x\right)}\sinh{\left(y\right)} \\&=&-\left(\sin{\left(x\right)}\cos{\left(iy\right)}+\cos{\left(x\right)}\sin{\left(iy\right)}\right) \\&&\;\ldots\;\href{https://shikitenkai.blogspot.com/2021/07/cosi-x-sini-x-cos-sin.html}{\cos\left(iy\right)=i\cosh\left(y\right),\;\sin\left(iy\right)=i\sinh\left(y\right)} \\&=&-\sin{\left(x+iy\right)} \\&=&-\sin{\left(z\right)} \end{eqnarray}$$

cos(i x), sin(i x) (純虚数に対するcos, sin)

\(\cos{(i x)}, \sin{(i x)}\) (純虚数に対する\(\cos, \sin\))

\(\cos{\left(i x\right)}\)

$$\begin{eqnarray} \cos{\left(i x\right)} &=&\frac{e^{i\left(ix\right)}+e^{-i\left(ix\right)}}{2 }\;\ldots\;x\in\mathbb{R} \\&=&\frac{e^{-x}+e^{x}}{2 } \\&=&\cosh{\left(x\right)} \end{eqnarray}$$

\(\sin{\left(i x\right)}\)

$$\begin{eqnarray} \sin{\left(ix\right)} &=&\frac{e^{i\left(ix\right)}-e^{-i\left(ix\right)}}{2i}\;\ldots\;x\in\mathbb{R} \\&=&\frac{e^{-x}-e^{x}}{2i} \\&=&\frac{1}{i}\frac{-\left(e^{x}-e^{-x}\right)}{2} \\&=&\frac{i}{i}\frac{-1}{i}\sinh{\left(x\right)} \\&=&i\sinh{\left(x\right)} \end{eqnarray}$$

\(cosh{(i x)}, sinh{(i x)}\) (純虚数に対する\(\cosh, \sinh\))

\(cosh{(i x)}, sinh{(i x)}\) (純虚数に対する\(\cosh, \sinh\))

ディガンマ凾数の相反公式

ディガンマ凾数の相反公式

ディガンマ凾数の定義

$$\begin{eqnarray} \psi\left(z\right)&=&\frac{\mathrm{d}}{\mathrm{d}z} \log{\left(\Gamma\left(z\right)\right)} \\&=&\frac{\Gamma^\prime\left(z\right)}{\Gamma\left(z\right)} \end{eqnarray}$$

\(1-z\)のディガンマ凾数

$$\begin{eqnarray} \psi\left(1-z\right) &=&\frac{\mathrm{d}}{\mathrm{d}z} \log{\left(\Gamma\left(1-z\right)\right)} \\&=&\frac{\mathrm{d}}{\mathrm{d}u} \log{\left(\Gamma\left(u\right)\right)}\frac{\mathrm{d}u}{\mathrm{d}z} \;\ldots\;u=1-z,\;\frac{\mathrm{d}u}{\mathrm{d}z}=-1 \\&=&-\frac{\mathrm{d}}{\mathrm{d}z}\log\left(\Gamma\left(1-z\right)\right) \\&=&-\frac{\mathrm{d}}{\mathrm{d}z}\log\left(\frac{\Gamma\left(z\right)}{\Gamma\left(z\right)}\Gamma\left(1-z\right)\right) \\&=&-\frac{\mathrm{d}}{\mathrm{d}z}\log\left(\frac{1}{\Gamma\left(z\right)}\frac{\pi}{\sin{\left(\pi z\right)}}\right) \;\ldots\;\href{https://shikitenkai.blogspot.com/2021/07/blog-post_26.html}{\Gamma\left(z\right)\Gamma\left(1-z\right)=\frac{\pi}{\sin{\left(\pi z\right)}}} \\&=&-\left\{ \frac{\mathrm{d}}{\mathrm{d}z} \log \left(\pi\right) -\frac{\mathrm{d}}{\mathrm{d}z} \log \left(\sin{\left(\pi z\right)}\right) -\frac{\mathrm{d}}{\mathrm{d}z} \log \left(\Gamma\left(z\right)\right) \right\} \\&=&-\left\{0-\pi\cot{\left(\pi z\right)}-\psi\left(z\right)\right\} \\&&\;\ldots\;\href{https://shikitenkai.blogspot.com/2021/07/logsinz.html}{\frac{\mathrm{d}}{\mathrm{d}z} \log \sin{\left(\pi z\right)}=\pi\cot{\left(\pi z\right)}} \\&&\;\ldots\;\frac{\mathrm{d}}{\mathrm{d}z} \log \Gamma\left(z\right)=\psi\left(z\right),\;(ディガンマ凾数の定義) \\&=&\pi\cot{\left(\pi z\right)}+\psi\left(z\right) \end{eqnarray}$$

ディガンマ凾数の相反公式

$$\begin{eqnarray} \psi\left(1-z\right) &=&\pi\cot{\left(\pi z\right)}+\psi\left(z\right) \\\psi\left(1-z\right)-\psi\left(z\right)&=&\pi\cot{\left(\pi z\right)} \end{eqnarray}$$

log(sin(πz))の微分

\(\log{\left(\sin{\left(\pi z\right)}\right)}\)の微分

$$\begin{eqnarray} \frac{\mathrm{d}}{\mathrm{d}z}\log{\left(\sin{\left(\pi z\right)}\right)} &=&\frac{\mathrm{d}}{\mathrm{d}f}\log{\left(\sin{\left(f\right)}\right)}\frac{\mathrm{d}f}{\mathrm{d}z} \;\ldots\;f=\pi z,\;f\in\mathbb{C} \\&=&\frac{\mathrm{d}}{\mathrm{d}g}\log{\left(g\right)}\frac{\mathrm{d}g}{\mathrm{d}f}\frac{\mathrm{d}f}{\mathrm{d}z} \;\ldots\;g=\sin{\left(f\right)},\;g\in\mathbb{C} \\&=&\frac{1}{g}\;\cos{\left(f\right)}\;\pi \\&&\;\ldots\;\href{https://shikitenkai.blogspot.com/2021/07/logz.html}{\frac{\mathrm{d}}{\mathrm{d}g}\log{\left(g\right)}=\frac{1}{g}} \\&&\;\ldots\;\href{https://shikitenkai.blogspot.com/2021/07/sinz.html}{\frac{\mathrm{d}}{\mathrm{d}f}\sin{\left(f\right)}=\cos{\left(f\right)}} \\&&\;\ldots\;\href{https://shikitenkai.blogspot.com/2021/07/az.html}{\frac{\mathrm{d}}{\mathrm{d}z}\pi z=\pi} \\&=&\frac{1}{\sin\left(f\right)}\;\cos{\left(f\right)}\;\pi \\&=&\frac{1}{\sin\left(\pi z\right)}\;\cos{\left(\pi z\right)}\;\pi \\&=&\pi\frac{\cos{\left(\pi z\right)}}{\sin\left(\pi z\right)} \\&=&\pi\frac{1}{\tan\left(\pi z\right)} \;\ldots\;\tan\left(\pi z\right)=\frac{\sin\left(\pi z\right)}{\cos{\left(\pi z\right)}} \\&=&\pi\cot{\left(\pi z\right)} \end{eqnarray}$$

sin(z)の微分

\(\sin{\left(z\right)}\)の微分

\(u+iv\)で表す

$$\begin{eqnarray} \sin{\left(z\right)} &=&\sin{\left(x\right)}\cos{\left(iy\right)}+\cos{\left(x\right)}\sin{\left(iy\right)} \\&=&\sin{\left(x\right)}\cosh{\left(y\right)}+\cos{\left(x\right)}i\sinh{\left(y\right)} \\&&\;\ldots\;\href{https://shikitenkai.blogspot.com/2021/07/cosi-x-sini-x-cos-sin.html}{\cos\left(iy\right)=i\cosh\left(y\right),\;\sin\left(iy\right)=i\sinh\left(y\right)} \\&=&\sin{\left(x\right)}\cosh{\left(y\right)}+i\cos{\left(x\right)}\sinh{\left(y\right)} \\&=&u(x,y)+iv(x,y) \end{eqnarray}$$ $$\left\{\begin{eqnarray} u(x,y)&=&\sin{\left(x\right)}\cosh{\left(y\right)} \\v(x,y)&=&\cos{\left(x\right)}\sinh{\left(y\right)} \end{eqnarray}\;\ldots\;x,y\in\mathbb{R}\right.$$

\(u,v\)を\(x,y\)で偏微分する

$$\begin{eqnarray} \frac{\partial u(x,y)}{\partial x} &=&\frac{\partial}{\partial x}\sin{\left(x\right)}\cosh{\left(y\right)} \;\ldots\;x,y\in\mathbb{R} \\&=&\cosh{\left(y\right)}\frac{\partial}{\partial x}\sin{\left(x\right)} \\&=&\cosh{\left(y\right)}\cos{\left(x\right)} \;\ldots\;\frac{\mathrm{d}}{\mathrm{d}\theta}\sin{\left(\theta\right)}=\cos{\left(\theta\right)} \\&=&\cos{\left(x\right)}\cosh{\left(y\right)} \end{eqnarray}$$ $$\begin{eqnarray} \frac{\partial u(x,y)}{\partial y} &=&\frac{\partial}{\partial y}\sin{\left(x\right)}\cosh{\left(y\right)} \;\ldots\;x,y\in\mathbb{R} \\&=&\sin{\left(x\right)}\frac{\partial}{\partial y}\cosh{\left(y\right)} \\&=&\sin{\left(x\right)}\sinh{\left(y\right)} \;\ldots\;\frac{\mathrm{d}}{\mathrm{d}\theta}\cosh{\left(\theta\right)}=\sinh{\left(\theta\right)} \end{eqnarray}$$ $$\begin{eqnarray} \frac{\partial v(x,y)}{\partial x} &=&\frac{\partial}{\partial x}\cos{\left(x\right)}\sinh{\left(y\right)} \;\ldots\;x,y\in\mathbb{R} \\&=&\sinh{\left(y\right)}\frac{\partial}{\partial x}\cos{\left(x\right)} \\&=&\sinh{\left(y\right)}\left(-\sin{\left(x\right)}\right) \;\ldots\;\frac{\mathrm{d}}{\mathrm{d}\theta}\cos{\left(\theta\right)}=-\sin{\left(\theta\right)} \\&=&-\sin{\left(x\right)}\sinh{\left(y\right)} \end{eqnarray}$$ $$\begin{eqnarray} \frac{\partial v(x,y)}{\partial y} &=&\frac{\partial}{\partial y}\cos{\left(x\right)}\sinh{\left(y\right)} \;\ldots\;x,y\in\mathbb{R} \\&=&\cos{\left(x\right)}\frac{\partial}{\partial y}\sinh{\left(y\right)} \\&=&\cos{\left(x\right)}\cosh{\left(y\right)} \;\ldots\;\frac{\mathrm{d}}{\mathrm{d}\theta}\sinh{\left(\theta\right)}=\cosh{\left(\theta\right)} \end{eqnarray}$$

コーシー・リーマンの関係式を満たす

$$\href{https://shikitenkai.blogspot.com/2021/07/blog-post_19.html}{\left\{ \begin{eqnarray} \frac{\partial u}{\partial x}&=&\frac{\partial v}{\partial y} \\\frac{\partial v}{\partial x}&=&-\frac{\partial u}{\partial y} \end{eqnarray} \right.}$$

実軸方向での微分

$$\begin{eqnarray} \frac{\mathrm{d}}{\mathrm{d}z}\sin{\left(z\right)} &=&\href{https://shikitenkai.blogspot.com/2021/07/blog-post_19.html}{\frac{\partial u(x,y)}{\partial x}+i\frac{\partial v(x,y)}{\partial x}} \\&=&\cos{\left(x\right)}\cosh{\left(y\right)}+i\left(-\sin{\left(x\right)}\sinh{\left(y\right)}\right) \\&=&\cos{\left(x\right)}\cosh{\left(y\right)}-i\sin{\left(x\right)}\sinh{\left(y\right)} \\&=&\cos{\left(x\right)}\cos{\left(iy\right)}-\sin{\left(x\right)}\sin{\left(iy\right)} \\&&\;\ldots\;\href{https://shikitenkai.blogspot.com/2021/07/cosi-x-sini-x-cos-sin.html}{\cos\left(iy\right)=\cosh\left(y\right),\;\sin\left(iy\right)=i\sinh\left(y\right)} \\&=&\cos{\left(x+iy\right)} \\&=&\cos{\left(z\right)} \end{eqnarray}$$

log(z)の微分

\(\log{\left(z\right)}\)の微分

\(u+iv\)で表す

$$\begin{eqnarray} \log{\left(z\right)} &=&\log{\left(x+iy\right)} \;\ldots\;z=x+iy,\;z\in\mathbb{C},\;x,y\in\mathbb{R} \\&=&\log{\left( |z| e^{i\arg{\left(z\right)}} \right)} \\&&\;\ldots\;z=|z| e^{i\arg{\left(z\right)}},\;|z|=|x+iy|=\sqrt{x^2+y^2}は実数,\:\arg{\left(z\right)}は実数で多価(集合) \\&&\;\ldots\;\arg{\left(z\right)}=\mathrm{Arg}{\left(z\right)}+2n\pi,\;n\in\mathbb{Z} \\&&\;\ldots\;-\pi\lt\mathrm{Arg}{\left(z\right)}\leq\pi,\;\mathrm{Arg}{\left(z\right)}は実数で一価 \\&=&\log{\left(|z|\right)}+\log{\left(e^{i\arg{\left(z\right)}}\right)} \\&=&\log{\left(|z|\right)}+\log{\left(e^{i\left(\mathrm{Arg}{\left(z\right)}+2n\pi\right)}\right)} \\&=&\log{\left(|z|\right)}+i\left(\mathrm{Arg}{\left(z\right)}+2n\pi\right) \\&=&u(x,y)+iv(x,y)\;\ldots\;v(x,y)は実数で多価(集合) \end{eqnarray}$$ $$\left\{\begin{eqnarray} u(x,y)&=&\log{\left(\sqrt{x^2+y^2}\right)} \\v(x,y)&=&\mathrm{Arg}{\left(z\right)}+2n\pi=\begin{cases} \arctan{\left(\frac{y}{x}\right)}&+2n\pi & (x\gt0) \\\arctan{\left(\frac{y}{x}\right)}+\pi&+2n\pi & (x\lt0\;かつ\;y\geq0) \\\arctan{\left(\frac{y}{x}\right)}-\pi&+2n\pi & (x\lt0\;かつ\;y\lt0) \\\frac{\pi}{2}&+2n\pi & (x=0\;かつ\;y\gt0) \\-\frac{\pi}{2}&+2n\pi & (x=0\;かつ\;y\lt0) \\\mathrm{identerminate} && (x=0\;かつ\;y=0) \end{cases} \end{eqnarray}\;\ldots\;x,y\in\mathbb{R},\;n\in\mathbb{Z}\right.$$

\(u,v\)を\(x,y\)で偏微分する

$$\begin{eqnarray} \frac{\partial u(x,y)}{\partial x} &=&\frac{\partial}{\partial x}\log{\left(\sqrt{x^2+y^2}\right)} \\&=&\frac{\partial}{\partial f}\log{\left(f\right)}\frac{\partial f}{\partial x} \;\ldots\;f=\sqrt{x^2+y^2}=\left(x^2+y^2\right)^{\frac{1}{2}},\;f\in\mathbb{R} \\&=&\frac{\partial}{\partial f}\log{\left(f\right)} \frac{\partial f}{\partial g}\frac{\partial g}{\partial x} \;\ldots\;g=x^2+y^2,\;g\in\mathbb{R} \\&=&\frac{1}{f}\;\frac{1}{2}\left(g\right)^{-\frac{1}{2}}\;2x \\&=&\frac{1}{\sqrt{x^2+y^2}}\;\frac{1}{2\sqrt{x^2+y^2}}\;2x \\&=&\frac{x}{x^2+y^2} \end{eqnarray}$$ $$\begin{eqnarray} \frac{\partial u(x,y)}{\partial y} &=&\frac{\partial}{\partial y}\log{\left(\sqrt{x^2+y^2}\right)} \\&=&\frac{\partial}{\partial f}\log{\left(f\right)}\frac{\partial f}{\partial y} \;\ldots\;f=\sqrt{x^2+y^2}=\left(x^2+y^2\right)^{\frac{1}{2}},\;f\in\mathbb{R} \\&=&\frac{\partial}{\partial f}\log{\left(f\right)} \frac{\partial f}{\partial g}\frac{\partial g}{\partial y} \;\ldots\;g=x^2+y^2,\;g\in\mathbb{R} \\&=&\frac{1}{f}\;\frac{1}{2}\left(g\right)^{-\frac{1}{2}}\;2y \\&=&\frac{1}{\sqrt{x^2+y^2}}\;\frac{1}{2\sqrt{x^2+y^2}}\;2y \\&=&\frac{y}{x^2+y^2} \end{eqnarray}$$ $$\begin{eqnarray} \frac{\partial v(x,y)}{\partial x} &=&\frac{\partial}{\partial x}\mathrm{Arg}{\left(x+iy\right)}+2n\pi \\&=&\begin{cases} \frac{\partial}{\partial x}\left(\arctan{\left(\frac{y}{x}\right)}+2n\pi\right) & (x\gt0) \\\frac{\partial}{\partial x}\left(\arctan{\left(\frac{y}{x}\right)}+\pi+2n\pi\right) & (x\lt0\;かつ\;y\geq0) \\\frac{\partial}{\partial x}\left(\arctan{\left(\frac{y}{x}\right)}-\pi+2n\pi\right) & (x\lt0\;かつ\;y\lt0) \\\frac{\partial}{\partial x}\left(\frac{\pi}{2}+2n\pi\right) & (x=0\;かつ\;y\gt0) \\\frac{\partial}{\partial x}\left(-\frac{\pi}{2}+2n\pi\right) & (x=0\;かつ\;y\lt0) \\\mathrm{identerminate} & (x=0\;かつ\;y=0) \end{cases} \\&=&\begin{cases} \href{https://shikitenkai.blogspot.com/2021/07/blog-post_9.html}{\frac{-y}{x^2+y^2}} & (x\gt0) \\\href{https://shikitenkai.blogspot.com/2021/07/blog-post_9.html}{\frac{-y}{x^2+y^2}} & (x\lt0\;かつ\;y\geq0) \\\href{https://shikitenkai.blogspot.com/2021/07/blog-post_9.html}{\frac{-y}{x^2+y^2}} & (x\lt0\;かつ\;y\lt0) \\\href{https://shikitenkai.blogspot.com/2021/07/blog-post_9.html}{0} & (x=0\;かつ\;y\gt0) \\\href{https://shikitenkai.blogspot.com/2021/07/blog-post_9.html}{0} & (x=0\;かつ\;y\lt0) \\\mathrm{identerminate}& (x=0\;かつ\;y=0) \end{cases} \end{eqnarray}$$ $$\begin{eqnarray} \frac{\partial v(x,y)}{\partial y} &=&\frac{\partial}{\partial y}\mathrm{Arg}{\left(x+iy\right)}+2n\pi \\&=&\begin{cases} \frac{\partial}{\partial y}\left(\arctan{\left(\frac{y}{x}\right)}+2n\pi\right) & (x\gt0) \\\frac{\partial}{\partial y}\left(\arctan{\left(\frac{y}{x}\right)}+\pi+2n\pi\right) & (x\lt0\;かつ\;y\geq0) \\\frac{\partial}{\partial y}\left(\arctan{\left(\frac{y}{x}\right)}-\pi+2n\pi\right) & (x\lt0\;かつ\;y\lt0) \\\frac{\partial}{\partial y}\left(\frac{\pi}{2}+2n\pi\right) & (x=0\;かつ\;y\gt0) \\\frac{\partial}{\partial y}\left(-\frac{\pi}{2}+2n\pi\right) & (x=0\;かつ\;y\lt0) \\\mathrm{identerminate} & (x=0\;かつ\;y=0) \end{cases} \\&=&\begin{cases} \href{https://shikitenkai.blogspot.com/2021/07/blog-post_9.html}{\frac{x}{x^2+y^2}} & (x\gt0) \\\href{https://shikitenkai.blogspot.com/2021/07/blog-post_9.html}{\frac{x}{x^2+y^2}} & (x\lt0\;かつ\;y\geq0) \\\href{https://shikitenkai.blogspot.com/2021/07/blog-post_9.html}{\frac{x}{x^2+y^2}} & (x\lt0\;かつ\;y\lt0) \\\href{https://shikitenkai.blogspot.com/2021/07/blog-post_9.html}{0} & (x=0\;かつ\;y\gt0) \\\href{https://shikitenkai.blogspot.com/2021/07/blog-post_9.html}{0} & (x=0\;かつ\;y\lt0) \\\mathrm{identerminate}& (x=0\;かつ\;y=0) \end{cases} \end{eqnarray}$$

コーシー・リーマンの関係式を満たす

$$\href{https://shikitenkai.blogspot.com/2021/07/blog-post_19.html}{\left\{ \begin{eqnarray} \frac{\partial u}{\partial x}&=&\frac{\partial v}{\partial y} \\\frac{\partial v}{\partial x}&=&-\frac{\partial u}{\partial y} \end{eqnarray} \right.}$$

実軸方向での微分

$$\begin{eqnarray} \frac{\mathrm{d}}{\mathrm{d}z}\log{\left(z\right)} &=&\href{https://shikitenkai.blogspot.com/2021/07/blog-post_19.html}{\frac{\partial u(x,y)}{\partial x}+i\frac{\partial v(x,y)}{\partial x}} \\&=&\frac{x}{x^2+y^2}+i\frac{-y}{x^2+y^2} \\&=&\frac{x-iy}{x^2+y^2} \\&=&\frac{\cancel{x-iy}}{(x+iy)\cancel{(x-iy)}} \\&&\;\ldots\;(x+iy)(x-iy)=x^2\cancel{+ixy}\cancel{-ixy}-i^2y^2=x^2+y^2 \\&=&\frac{1}{x+iy} \\&=&\frac{1}{z}\;\ldots\;z=x+iy \end{eqnarray}$$

虚軸方向での微分

$$\begin{eqnarray} \frac{\mathrm{d}}{\mathrm{d}z}\log{\left(z\right)} &=&\href{https://shikitenkai.blogspot.com/2021/07/blog-post_19.html}{\frac{\partial v(x,y)}{\partial y}-i\frac{\partial u(x,y)}{\partial y}} \\&=&\frac{x}{x^2+y^2}-i\frac{y}{x^2+y^2} \\&=&\frac{x-iy}{x^2+y^2} \\&=&\frac{\cancel{x-iy}}{(x+iy)\cancel{(x-iy)}} \\&&\;\ldots\;(x+iy)(x-iy)=x^2\cancel{+ixy}\cancel{-ixy}-i^2y^2=x^2+y^2 \\&=&\frac{1}{x+iy} \\&=&\frac{1}{z}\;\ldots\;z=x+iy \end{eqnarray}$$

cot(az)の微分

\(cot\left(az\right)\)の微分

\begin{eqnarray} \frac{\mathrm{d}}{\mathrm{d}z}\cot{\left(az\right)} &=&\frac{\mathrm{d}}{\mathrm{d}w}\cot{\left(w\right)}\frac{\mathrm{d}w}{\mathrm{d}z} \\&&\;\ldots\;a\in\mathbb{R},\;z\in\mathbb{C} \\&&\;\ldots\;\href{https://shikitenkai.blogspot.com/2021/07/wz.html}{w=az,\;\frac{\mathrm{d}w}{\mathrm{d}z}=a} \\&=&\frac{-1}{\sin^2{\left(az\right)}}\cdot a \;\ldots\;\href{https://shikitenkai.blogspot.com/2021/07/cotz.html}{\frac{\mathrm{d}}{\mathrm{d}w}\cot{\left(w\right)}=\frac{-1}{\sin^2{\left(w\right)}}} \\&=&-\frac{a}{\sin^2{\left(az\right)}} \end{eqnarray}

cot(ax)の微分

\(\cot{\left(ax\right)}\)の微分

$$\begin{eqnarray} \frac{\mathrm{d}}{\mathrm{d}x}\cot{\left(ax\right)} &=&\frac{\mathrm{d}}{\mathrm{d}x}\left\{\tan{\left(ax\right)}\right\}^{-1}\;\ldots\;a,x\in\mathbb{R} \\&=&\frac{\mathrm{d}}{\mathrm{d}u}\left\{\tan{\left(u\right)}\right\}^{-1}\frac{\mathrm{d}u}{\mathrm{d}x} \;\ldots\;u=ax,\;\frac{\mathrm{d}u}{\mathrm{d}x}=a \\&=&\frac{\mathrm{d}}{\mathrm{d}v}\left\{ v\left(u\right)\right\}^{-1}\frac{\mathrm{d}v}{\mathrm{d}u}\frac{\mathrm{d}u}{\mathrm{d}x} \;\ldots\;v=\tan{\left(u\right)},\;\frac{\mathrm{d}v}{\mathrm{d}u}=1+\tan^2{\left(u\right)} \\&=&-\left\{ v\left(u\right)\right\}^{-2}\frac{\mathrm{d}v}{\mathrm{d}u}\frac{\mathrm{d}u}{\mathrm{d}x} \\&=&-\frac{1}{\tan^2{\left(u\right)}}\cdot \left(1+\tan^2{\left(u\right)}\right)\frac{\mathrm{d}u}{\mathrm{d}x} \\&=&-\frac{1}{\tan^2{\left(ax\right)}}\cdot \left(1+\tan^2{\left(ax\right)}\right)\cdot a \\&=&-\frac{a\left(1+\frac{\sin^2{\left(ax\right)}}{\cos^2{\left(ax\right)}}\right)}{\frac{\sin^2{\left(ax\right)}}{\cos^2{\left(ax\right)}}} \\&=&-\frac{\cos^2{\left(ax\right)}a\left(1+\frac{\sin^2{\left(ax\right)}}{\cos^2{\left(ax\right)}}\right)}{\sin^2{\left(ax\right)}} \\&=&-\frac{a\left(\cos^2{\left(ax\right)}+\cos^2{\left(ax\right)}\frac{\sin^2{\left(ax\right)}}{\cos^2{\left(ax\right)}}\right)}{\sin^2{\left(ax\right)}} \\&=&-\frac{a\left(\cos^2{\left(ax\right)}+\sin^2{\left(ax\right)}\right)}{\sin^2{\left(ax\right)}} \\&=&-\frac{a\cdot 1}{\sin^2{\left(ax\right)}} \\&=&-\frac{a}{\sin^2{\left(ax\right)}} \end{eqnarray}$$

cot(z)の微分

\(u(x,y)\)を\(x\)で偏微分する

$$\begin{eqnarray} \frac{\partial u}{\partial x} &=&\frac{\partial}{\partial x}\href{https://shikitenkai.blogspot.com/2021/07/cotzuxyivxy.html}{\frac{\cos{\left(x\right)}\sin{\left(x\right)}} {\sin^2{\left(x\right)}+\sinh^2{\left(y\right)}}} \\&=&\frac{\sinh^2\left(y\right)(\cos^2\left(x\right)-\sin^2\left(x\right))-\sin^2\left(x\right)}{\left(\sin^2{\left(x\right)}+\sinh^2{\left(y\right)}\right)^2} \\&=&\frac{\cos^2\left(x\right)\sinh^2\left(y\right)-\sinh^2\left(y\right)\sin^2\left(x\right)-\sin^2\left(x\right)}{\left(\sin^2{\left(x\right)}+\sinh^2{\left(y\right)}\right)^2} \\&=&\frac{\cos^2\left(x\right)\sinh^2\left(y\right)-\sin^2\left(x\right)(1+\sinh^2\left(y\right))}{\left(\sin^2{\left(x\right)}+\sinh^2{\left(y\right)}\right)^2} \\&=&\frac{\cos^2\left(x\right)\sinh^2\left(y\right)-\sin^2\left(x\right)\cosh^2\left(y\right)}{\left(\sin^2{\left(x\right)}+\sinh^2{\left(y\right)}\right)^2} \end{eqnarray}$$

\(u(x,y)\)を\(y\)で偏微分する

$$\begin{eqnarray} \frac{\partial u}{\partial y} &=&\frac{\partial}{\partial y}\href{https://shikitenkai.blogspot.com/2021/07/cotzuxyivxy.html}{\frac{\cos{\left(x\right)}\sin{\left(x\right)}} {\sin^2{\left(x\right)}+\sinh^2{\left(y\right)}}} \\&=&-\frac{2\sin\left(x\right)\cos\left(x\right)\sinh\left(y\right)\cosh\left(y\right)}{\left(\sin^2{\left(x\right)}+\sinh^2{\left(y\right)}\right)^2} \end{eqnarray}$$

\(v(x,y)\)を\(x\)で偏微分する

$$\begin{eqnarray} \frac{\partial v}{\partial x} &=&\frac{\partial}{\partial x}\href{https://shikitenkai.blogspot.com/2021/07/cotzuxyivxy.html}{\frac{-\cosh{\left(y\right)}\sinh{\left(y\right)}} {\sin^2{\left(x\right)}+\sinh^2{\left(y\right)}}} \\&=&\frac{2\sin\left(x\right)\cos\left(x\right)\sinh\left(y\right)\cosh\left(y\right)}{\left(\sin^2{\left(x\right)}+\sinh^2{\left(y\right)}\right)^2} \end{eqnarray}$$

\(v(x,y)\)を\(y\)で偏微分する

$$\begin{eqnarray} \frac{\partial v}{\partial y} &=&\frac{\partial}{\partial y}\href{https://shikitenkai.blogspot.com/2021/07/cotzuxyivxy.html}{\frac{-\cosh{\left(y\right)}\sinh{\left(y\right)}} {\sin^2{\left(x\right)}+\sinh^2{\left(y\right)}}} \\&=&\frac{\cosh^2\left(y\right)(\sinh^2\left(y\right)-\sin^2\left(x\right))-\sinh^2\left(y\right)(\sinh^2\left(y\right)+\sin^2\left(x\right))}{\left(\sin^2{\left(x\right)}+\sinh^2{\left(y\right)}\right)^2} \\&=&\frac{\cosh^2\left(y\right)\sinh^2\left(y\right)-\cosh^2\left(y\right)\sin^2\left(x\right)-\sinh^2\left(y\right)\sinh^2\left(y\right)-\sinh^2\left(y\right)\sin^2\left(x\right)}{\left(\sin^2{\left(x\right)}+\sinh^2{\left(y\right)}\right)^2} \\&=&\frac{\sinh^2\left(y\right)(\cosh^2\left(y\right)-\sinh^2\left(y\right))-\cosh^2\left(y\right)\sin^2\left(x\right)-\sinh^2\left(y\right)\sin^2\left(x\right)}{\left(\sin^2{\left(x\right)}+\sinh^2{\left(y\right)}\right)^2} \\&=&\frac{\sinh^2\left(y\right)-\cosh^2\left(y\right)\sin^2\left(x\right)-\sinh^2\left(y\right)\sin^2\left(x\right)}{\left(\sin^2{\left(x\right)}+\sinh^2{\left(y\right)}\right)^2} \\&=&\frac{\sinh^2\left(y\right)(1-\sin^2\left(x\right))-\cosh^2\left(y\right)\sin^2\left(x\right)}{\left(\sin^2{\left(x\right)}+\sinh^2{\left(y\right)}\right)^2} \\&=&\frac{\cos^2\left(x\right)\sinh^2\left(y\right)-\sin^2\left(x\right)\cosh^2\left(y\right)}{\left(\sin^2{\left(x\right)}+\sinh^2{\left(y\right)}\right)^2} \end{eqnarray}$$

コーシー・リーマンの関係式を満たす

$$\href{https://shikitenkai.blogspot.com/2021/07/blog-post_19.html}{ \left\{ \begin{eqnarray} \frac{\partial u}{\partial x}&=&\frac{\partial v}{\partial y} \\\frac{\partial u}{\partial y}&=&-\frac{\partial v}{\partial x} \end{eqnarray} \right. }$$

実軸(x)方向の微分

$$\begin{eqnarray} \frac{\mathrm{d}}{\mathrm{d} z}\cot{}&=&\href{https://shikitenkai.blogspot.com/2021/07/blog-post_19.html}{\frac{\partial u(x,y)}{\partial x}+i\frac{\partial v(x,y)}{\partial x}} \\&=&\frac{\cos^2\left(x\right)\sinh^2\left(y\right)-\sin^2\left(x\right)\cosh^2\left(y\right)}{\left(\sin^2{\left(x\right)}+\sinh^2{\left(y\right)}\right)^2} +i\frac{2\sin\left(x\right)\cos\left(x\right)\sinh\left(y\right)\cosh\left(y\right)}{\left(\sin^2{\left(x\right)}+\sinh^2{\left(y\right)}\right)^2} \\&=&\frac{\left(\cos\left(x\right)\sinh\left(y\right)+i\sin\left(x\right)\cosh\left(y\right)\right)^2}{\left(\sin^2{\left(x\right)}+\sinh^2{\left(y\right)}\right)^2} \\&=&\frac{\left(\cos\left(x\right)\sinh\left(y\right)+i\sin\left(x\right)\cosh\left(y\right)\right)^2}{\left(\sin^2{\left(x\right)}+\sinh^2{\left(y\right)}\right)^2} \\&=&\frac{\left(\cos\left(x\right)\sinh\left(y\right)+i\sin\left(x\right)\cosh\left(y\right)\right)^2}{\left(\sin^2{\left(x\right)}+\sinh^2{\left(y\right)}\right)^2} \\&=&\frac{\left(\cos\left(x\right)\frac{1}{i}\sin\left(iy\right)+i\sin\left(x\right)\cos\left(iy\right)\right)^2} {\left(\sin^2{\left(x\right)}+\left(\frac{1}{i}\sin\left(iy\right)\right)^2\right)^2} \;\ldots\;\cos\left(iy\right)=i\cosh\left(y\right),\;\sin\left(iy\right)=i\sinh\left(y\right),\;\sinh\left(y\right)=\frac{1}{i}\sin\left(iy\right) \\&=&\frac{i^2}{i^2}\frac{\left(\cos\left(x\right)\frac{1}{i}\sin\left(iy\right)+i\sin\left(x\right)\cos\left(iy\right)\right)^2} {\left(\sin^2{\left(x\right)}+\left(\frac{1}{i}\right)^2\left(\sin\left(iy\right)\right)^2\right)^2} \\&=&\frac{1}{i^2}\frac{\left(i\left(\cos\left(x\right)\frac{1}{i}\sin\left(iy\right)+i\sin\left(x\right)\cos\left(iy\right)\right)\right)^2} {\left(\sin^2{\left(x\right)}-\sin^2\left(iy\right)\right)^2} \\&=&\frac{1}{-1}\frac{\left(\cos\left(x\right)\sin\left(iy\right)-\sin\left(x\right)\cos\left(iy\right)\right)^2} {\left(\sin^2{\left(x\right)}-\sin^2\left(iy\right)\right)^2} \\&=&-\frac{\left(-\left(-\cos\left(x\right)\sin\left(iy\right)+\sin\left(x\right)\cos\left(iy\right)\right)\right)^2} {\left(\sin^2{\left(x\right)}-\sin^2\left(iy\right)\right)^2} \\&=&-\frac{\left(-\sin(x-iy)\right)^2} {\left(\sin{\left(x+iy\right)}\sin{\left(x-iy\right)}\right)^2} \\&&\;\ldots\;\sin^2{\left(x\right)}-\sin^2\left(iy\right) \\&&\;\ldots\;=\sin^2{\left(x\right)}-\sin^2\left(iy\right)\color{red}{-\sin^2{\left(x\right)}\sin^2\left(iy\right)+\sin^2{\left(x\right)}\sin^2\left(iy\right)} \\&&\;\ldots\;=\sin^2{\left(x\right)}\left(1-\sin^2\left(iy\right)\right)-\sin^2{\left(iy\right)}\left(1-\sin^2\left(x\right)\right) \\&&\;\ldots\;=\sin^2{\left(x\right)}\cos^2\left(iy\right)-\sin^2{\left(iy\right)}\cos^2{\left(x\right)} \\&&\;\ldots\;=\left\{\sin{\left(x\right)}\cos\left(iy\right)+\sin{\left(iy\right)}\cos{\left(x\right)}\right\} \left\{\sin{\left(x\right)}\cos\left(iy\right)-\sin{\left(iy\right)}\cos{\left(x\right)}\right\} \\&&\;\ldots\;=\sin{\left(x+iy\right)}\sin{\left(x-iy\right)} \\&=&-\frac{\cancel{\sin^2(x-iy)}} {\sin^2{\left(x+iy\right)}\cancel{\sin^2{\left(x-iy\right)}}} \\&=&\frac{-1}{\sin^2{\left(x+iy\right)}} \\&=&\frac{-1}{\sin^2{\left(z\right)}} \end{eqnarray}$$

cot(z)をu(x,y)+iv(x,y)で表す

$$\begin{eqnarray} \cot{\left(z\right)} &=&\frac{1}{\tan{\left(z\right)}}\;\ldots\;z\in\mathbb{C} \\&=&\frac{\cos{\left(z\right)}}{\sin{\left(z\right)}} \\&=&\frac{\cos{\left(x+iy\right)}}{\sin{\left(x+iy\right)}}\;\ldots\;x,y\in\mathbb{R} \\&=&\frac{ \cos{\left(x\right)}\cos{\left(iy\right)} - \sin{\left(x\right)}\sin{\left(iy\right)}} {\sin{\left(x\right)}\cos{\left(iy\right)}+\cos{\left(x\right)}\sin{\left(iy\right)}} \\&&\;\ldots\;\cos{\left(\alpha+\beta\right)}=\cos{\left(\alpha\right)}\cos{\left(\beta\right)}-\sin{\left(\alpha\right)}\sin{\left(\beta\right)} \\&&\;\ldots\;\sin{\left(\alpha+\beta\right)}=\cos{\left(\alpha\right)}\sin{\left(\beta\right)}+\sin{\left(\alpha\right)}\cos{\left(\beta\right)} \\&=&\frac{ \cos{\left(x\right)}\cosh{\left(y\right)} - \sin{\left(x\right)}i\sinh{\left(y\right)}} {\sin{\left(x\right)}\cosh{\left(y\right)}+\cos{\left(x\right)}i\sinh{\left(y\right)}} \\&&\;\ldots\;\cos{\left(iy\right)}=\frac{e^{i\left(iy\right)}+e^{-i\left(iy\right)}}{2 }=\frac{e^{-y}+e^{y}}{2 }=\cosh{\left(y\right)} \\&&\;\ldots\;\sin{\left(iy\right)}=\frac{e^{i\left(iy\right)}-e^{-i\left(iy\right)}}{2i}=\frac{e^{-y}-e^{y}}{2i}=\frac{1}{i}\frac{-\left(e^{y}-e^{-y}\right)}{2}=\frac{i}{i}\frac{-1}{i}\sinh{\left(y\right)}=i\sinh{\left(y\right)} \\&=&\frac{ \cos{\left(x\right)}\cosh{\left(y\right)} - i\sin{\left(x\right)}\sinh{\left(y\right)}} {\sin{\left(x\right)}\cosh{\left(y\right)}+i\cos{\left(x\right)}\sinh{\left(y\right)}} \frac{\sin{\left(x\right)}\cosh{\left(y\right)}-i\cos{\left(x\right)}\sinh{\left(y\right)}} {\sin{\left(x\right)}\cosh{\left(y\right)}-i\cos{\left(x\right)}\sinh{\left(y\right)}} \\&=&\frac{ \cos{\left(x\right)}\cosh{\left(y\right)} \sin{\left(x\right)}\cosh{\left(y\right)} +\cos{\left(x\right)}\cosh{\left(y\right)} (-i\cos{\left(x\right)}\sinh{\left(y\right)}) - i\sin{\left(x\right)}\sinh{\left(y\right)} \sin{\left(x\right)}\cosh{\left(y\right)} - i\sin{\left(x\right)}\sinh{\left(y\right)} (-i\cos{\left(x\right)}\sinh{\left(y\right)}) } {\sin^2{\left(x\right)}\cosh^2{\left(y\right)}+\cos^2{\left(x\right)}\sinh^2{\left(y\right)}} \\&=&\frac{ \cos{\left(x\right)}\sin{\left(x\right)}\cosh^2{\left(y\right)} -i\cos^2{\left(x\right)}\cosh{\left(y\right)}\sinh{\left(y\right)} - i\sin^2{\left(x\right)}\cosh{\left(y\right)}\sinh{\left(y\right)} - \cos {\left(x\right)}\sin {\left(x\right)}\sinh{\left(y\right)} } {\sin^2{\left(x\right)}\cosh^2{\left(y\right)}+\left(1-\sin^2{\left(x\right)}\right)\sinh^2{\left(y\right)}} \\&=&\frac{ \cos{\left(x\right)}\sin{\left(x\right)}\cosh^2{\left(y\right)} -i\cos^2{\left(x\right)}\cosh{\left(y\right)}\sinh{\left(y\right)} - i\sin^2{\left(x\right)}\cosh{\left(y\right)}\sinh{\left(y\right)} - \cos {\left(x\right)}\sin {\left(x\right)}\sinh{\left(y\right)} } {\sin^2{\left(x\right)}\cosh^2{\left(y\right)}+\sinh^2{\left(y\right)}-\sin^2{\left(x\right)}\sinh^2{\left(y\right)}} \\&=&\frac{ \cos{\left(x\right)}\sin{\left(x\right)}\left(\cosh^2{\left(y\right)}-\sinh{\left(y\right)}\right) -i\left\{ \left(\cos^2{\left(x\right)}+\sin^2{\left(x\right)}\right)\cosh{\left(y\right)}\sinh{\left(y\right)} \right\} } {\sin^2{\left(x\right)}\left(\cosh^2{\left(y\right)}-\sinh^2{\left(y\right)}\right)+\sinh^2{\left(y\right)}} \\&=&\frac{ \cos{\left(x\right)}\sin{\left(x\right)}-i\cosh{\left(y\right)}\sinh{\left(y\right)}} {\sin^2{\left(x\right)}+\sinh^2{\left(y\right)}} \\&=&\frac{\cos{\left(x\right)}\sin{\left(x\right)}} {\sin^2{\left(x\right)}+\sinh^2{\left(y\right)}} +i\frac{-\cosh{\left(y\right)}\sinh{\left(y\right)}} {\sin^2{\left(x\right)}+\sinh^2{\left(y\right)}} \\&=&u(x,y)+iv(x,y) \end{eqnarray}$$ $$\left\{\begin{eqnarray} u(x,y)&=&\frac{\cos{\left(x\right)}\sin{\left(x\right)}} {\sin^2{\left(x\right)}+\sinh^2{\left(y\right)}} \\v(x,y)&=&\frac{-\cosh{\left(y\right)}\sinh{\left(y\right)}} {\sin^2{\left(x\right)}+\sinh^2{\left(y\right)}} \end{eqnarray}\right.$$

wzの微分

\(wz\)の微分

\(u+iv\)で表す

$$\begin{eqnarray} wz&=&(a+ib)(x+iy)\;\ldots\;a,b,x,y\in\mathbb{R},\;w,z\in\mathbb{C} \\&=&ax-by+i(ay+bx) \\&=&u(x,y)+iv(x,y) \end{eqnarray}$$

\(u,v\)を\(x,y\)で偏微分する

$$\begin{eqnarray} \frac{\partial u(x,y)}{\partial x}&=&\frac{\partial }{\partial x}(ax-by) \\&=&a \\\frac{\partial u(x,y)}{\partial y}&=&\frac{\partial }{\partial y}(ax-by) \\&=&-b \\\frac{\partial v(x,y)}{\partial x}&=&\frac{\partial }{\partial x}(ay+bx) \\&=&b \\\frac{\partial v(x,y)}{\partial y}&=&\frac{\partial }{\partial y}(ay+bx) \\&=&a \end{eqnarray}$$

コーシー・リーマンの関係式を満たす

$$\href{https://shikitenkai.blogspot.com/2021/07/blog-post_19.html}{ \left\{ \begin{eqnarray} \frac{\partial u}{\partial x}&=&\frac{\partial v}{\partial y} \\\frac{\partial u}{\partial y}&=&-\frac{\partial v}{\partial x} \end{eqnarray} \right. }$$

実軸(x)方向の微分

$$\begin{eqnarray} \frac{\mathrm{d}}{\mathrm{d} z}wz&=&\href{https://shikitenkai.blogspot.com/2021/07/blog-post_19.html}{\frac{\partial u(x,y)}{\partial x}+i\frac{\partial v(x,y)}{\partial x}} \\&=&a+ib \\&=&w \end{eqnarray}$$

虚軸(y)方向の微分

$$\begin{eqnarray} \frac{\mathrm{d}}{\mathrm{d} z}wz&=&\href{https://shikitenkai.blogspot.com/2021/07/blog-post_19.html}{\frac{\partial v(x,y)}{\partial y}-i\frac{\partial u(x,y)}{\partial y}} \\&=&a-i(-b) \\&=&a+ib \\&=&w \end{eqnarray}$$

azの微分

\(az\)の微分

\(u+iv\)で表す

$$\begin{eqnarray} az&=&a(x+iy)\;\ldots\;a,x,y\in\mathbb{R},\;z\in\mathbb{C} \\&=&ax+iay \\&=&u(x,y)+iv(x,y) \end{eqnarray}$$ $$\left\{ \begin{eqnarray} u(x,y)&=&ax \\v(x,y)&=&ay \end{eqnarray} \right.$$

\(u,v\)を\(x,y\)で偏微分する

$$\begin{eqnarray} \frac{\partial u(x,y)}{\partial x}&=&\frac{\partial }{\partial x}ax \\&=&a \\\frac{\partial u(x,y)}{\partial y}&=&\frac{\partial }{\partial y}ax \\&=&0 \\\frac{\partial v(x,y)}{\partial x}&=&\frac{\partial }{\partial x}ay \\&=&0 \\\frac{\partial v(x,y)}{\partial y}&=&\frac{\partial }{\partial y}ay \\&=&a \end{eqnarray}$$

コーシー・リーマンの関係式を満たす

$$\href{https://shikitenkai.blogspot.com/2021/07/blog-post_19.html}{ \left\{ \begin{eqnarray} \frac{\partial u}{\partial x}&=&\frac{\partial v}{\partial y} \\\frac{\partial u}{\partial y}&=&-\frac{\partial v}{\partial x} \end{eqnarray} \right. }$$

実軸方向での微分

\begin{eqnarray} \frac{\mathrm{d}}{\mathrm{d} z}az&=&\href{https://shikitenkai.blogspot.com/2021/07/blog-post_19.html}{\frac{\partial u(x,y)}{\partial x}+i\frac{\partial v(x,y)}{\partial x}} \\&=&a+i0 \\&=&a \end{eqnarray}

虚軸方向での微分

\begin{eqnarray} \frac{\mathrm{d}}{\mathrm{d} z}az&=&\href{https://shikitenkai.blogspot.com/2021/07/blog-post_19.html}{\frac{\partial v(x,y)}{\partial y}-i\frac{\partial u(x,y)}{\partial y}} \\&=&a-i0 \\&=&a \end{eqnarray}

コーシー・リーマンの関係式

コーシー・リーマンの関係式

複素平面の実軸方向の微分(偏微分)

$$\begin{eqnarray} \lim_{\Delta z\rightarrow0}\frac{f(z_0+\Delta z)-f(z_0)}{\Delta z} &=& \lim_{\Delta x\rightarrow0}\frac{\left\{u\left(x_0+\Delta x, y_0\right)+iv\left(x_0+\Delta x, y_0\right)\right\} -\left\{u\left(x_0, y_0\right)+iv\left(x_0, y_0\right)\right\}}{\Delta x} \;\ldots\;z_0,\Delta zin\mathbb{C},\;x_0,y_0,\Delta x\in\mathbb{R} \\&=& \lim_{\Delta x\rightarrow0}\left\{ \frac{ u\left(x_0+\Delta x, y_0\right) -u\left(x_0, y_0\right) }{\Delta x} +i\frac{ v\left(x_0+\Delta x, y_0\right) -v\left(x_0, y_0\right) }{\Delta x} \right\} \\&=&\frac{\partial u\left(x_0, y_0\right)}{\partial x}+i\frac{\partial v\left(x_0, y_0\right)}{\partial x} \end{eqnarray}$$

複素平面の虚軸方向の微分(偏微分)

$$\begin{eqnarray} \lim_{\Delta z\rightarrow0}\frac{f(z_0+\Delta z)-f(z_0)}{\Delta z} &=& \lim_{\Delta y\rightarrow0}\frac{\left\{u\left(x_0, y_0+\Delta y\right)+iv\left(x_0, y_0+\Delta y\right)\right\} -\left\{u\left(x_0, y_0\right)+iv\left(x_0, y_0\right)\right\}}{i\Delta y} \;\ldots\;z_0,\Delta zin\mathbb{C},\;x_0,y_0,\Delta y\in\mathbb{R} \\&=& \lim_{\Delta y\rightarrow0}\left\{ \frac{ u\left(x_0, y_0+\Delta y\right) -u\left(x_0, y_0\right) }{i\Delta y} +i\frac{ v\left(x_0, y_0+\Delta y\right) -v\left(x_0, y_0\right) }{i\Delta y} \right\} \\&=&\frac{1}{i}\frac{\partial u\left(x_0, y_0\right)}{\partial y}+\frac{i}{i}\frac{\partial v\left(x_0, y_0\right)}{\partial y} \\&=&\frac{i}{i}\frac{1}{i}\frac{\partial u\left(x_0, y_0\right)}{\partial y}+\frac{\partial v\left(x_0, y_0\right)}{\partial y} \\&=&\frac{i}{-1}\frac{\partial u\left(x_0, y_0\right)}{\partial y}+\frac{\partial v\left(x_0, y_0\right)}{\partial y} \\&=&-i\frac{\partial u\left(x_0, y_0\right)}{\partial y}+\frac{\partial v\left(x_0, y_0\right)}{\partial y} \\&=&\frac{\partial v\left(x_0, y_0\right)}{\partial y}+i\left\{-\frac{\partial u\left(x_0, y_0\right)}{\partial y}\right\} \end{eqnarray}$$

複素平面の各軸微分結果の実部同士，虚部同士が等しくなる場合という関係

$$\left\{ \begin{eqnarray} \frac{\partial u}{\partial x}&=&\frac{\partial v}{\partial y} \\\frac{\partial v}{\partial x}&=&-\frac{\partial u}{\partial y} \end{eqnarray} \right.$$

ディガンマ凾数同士の差

ディガンマ凾数同士の差

$$\begin{eqnarray} \psi\left(y\right)-\psi\left(x\right) &=&\lim_{n\rightarrow\infty} \left\{ \log{\left(n \right)} -\sum_{k=0}^n \frac{1}{y+k} \right\} - \lim_{n\rightarrow\infty} \left\{ \log{\left(n \right)} -\sum_{k=0}^n \frac{1}{x+k} \right\} \;\ldots\;\href{https://shikitenkai.blogspot.com/2021/07/blog-post_47.html}{\psi\left(x\right)=\lim_{n\rightarrow\infty} \left\{ \log{\left(n \right)} -\sum_{k=0}^n \frac{1}{x+k} \right\}} \\&=&\lim_{n\rightarrow\infty}\left\{ \log{\left(n \right)} -\sum_{k=0}^n \frac{1}{y+k} -\log{\left(n \right)} +\sum_{k=0}^n \frac{1}{x+k} \right\} \\&=&\lim_{n\rightarrow\infty}\left\{ \sum_{k=0}^n \frac{1}{x+k} -\sum_{k=0}^n \frac{1}{y+k} \right\} \\&=& \sum_{k=0}^\infty \frac{1}{x+k} -\sum_{k=0}^\infty \frac{1}{y+k} \\&=& \int_{0}^1 \frac{u^{x-1}}{1-u} \mathrm{d}u -\int_{0}^1 \frac{u^{y-1}}{1-u} \mathrm{d}u \\&&\;\ldots\;\href{https://shikitenkai.blogspot.com/2021/07/uz-11-u011zk.html}{\sum_{k=0}^\infty \frac{1}{z+k}=\int_{0}^1 \frac{u^{z-1}}{1-u} \mathrm{d}u} \\&=& \int_{0}^1 \left(\frac{u^{x-1}}{1-u} - \frac{u^{y-1}}{1-u} \right)\mathrm{d}u \\&=& \int_{0}^1 \frac{u^{x-1}-u^{y-1}}{1-u}\mathrm{d}u \end{eqnarray}$$

u^(z-1)/(1-u)における定積分[0,1],もしくは1/(z+k)の無限級数

\(\frac{u^{z-1}}{1-u}\)における定積分[0,1],もしくは\(\frac{1}{z+k}\)の無限級数

$$\begin{eqnarray} \int_{0}^1 \frac{u^{z-1}}{1-u} \mathrm{d}u &=&\int_{0}^1 u^{z-1}\frac{1}{1-u} \mathrm{d}u \\&=&\int_{0}^1 u^{z-1}\left(1+u+u^2+\cdots\right) \mathrm{d}u \\&&\;\ldots\href{https://shikitenkai.blogspot.com/2021/07/11-x-1.html}{\frac{1}{1-u}=1+u+u^2+\cdots\;\;\;\left(|u|\lt1\right)} \\&=&\int_{0}^1 \left(u^{z-1}+u^{z-1}u+u^{z-1}u^2+\cdots\right) \mathrm{d}u \\&=&\int_{0}^1 \left(u^{z-1}+u^{z-1+1}+u^{z-1+2}+\cdots\right) \mathrm{d}u \\&=&\int_{0}^1 \left(u^{z-1}+u^{z}+u^{z+1}+\cdots\right) \mathrm{d}u \\&=&\left[\frac{1}{z-1+1}u^{z-1+1}+\frac{1}{z+1}u^{z+1}+\frac{1}{z+1+1}u^{z+1+1}+\cdots\right]_0^1 \\&=&\left[\frac{1}{z}u^z+\frac{1}{z+1}u^{z+1}+\frac{1}{z+2}u^{z+2}+\cdots\right]_0^1 \\&=&\left[\frac{1}{z}1^z-\frac{1}{z}0^z+\frac{1}{z+1}1^{z+1}-\frac{1}{z+1}0^{z+1}+\frac{1}{z+2}1^{z+2}-\frac{1}{z+2}0^{z+2}+\cdots\right] \\&=&\frac{1}{z}+\frac{1}{z+1}+\frac{1}{z+2}+\cdots \\&=&\sum_{k=0}^\infty \frac{1}{z+k} \end{eqnarray}$$

1/(1-x)のマクローリン展開(ただし-1<x<1)

\(\frac{1}{1-x}\)のマクローリン展開(\(ただし-1<x<1\))

a点まわりのテイラー展開

\begin{eqnarray} f(x) &=& \sum_{k=0}^{\infty}\frac{f^{(k)}(a)}{x!}(x-a)^k \;\ldots\;a点まわりのテイラー展開 \\&=& \frac{1}{0!}f^{(0)}(a)(x-a)^0+\frac{1}{1!}f^{(1)}(a)(x-a)^1+\frac{1}{2!}f^{(2)}(a)(x-a)^2+\dotsb \\&&\;\ldots\;f^{(n)}(x): f(x)のn階微分 \end{eqnarray}

マクローリン展開(0点まわりのテイラー展開)

\begin{eqnarray} f(x) &=& \left.\sum_{k=0}^{\infty}\frac{f^{(k)}(a)}{x!}(x-a)^k\right|_{a=0} \\&=& \sum_{k=0}^{\infty}\frac{f^{(k)}(0)}{x!}(x-0)^k \\&=& \sum_{k=0}^{\infty}\frac{f^{(k)}(0)}{x!}(x)^k \\&=& \frac{1}{0!}f^{(0)}(0)x^0+\frac{1}{1!}f^{(1)}(0)x+\frac{1}{2!}f^{(2)}(0)x^2+\cdots \end{eqnarray}

マクローリン展開(0点まわりのテイラー展開)

$$\begin{eqnarray} \frac{1}{1-x}&=&\sum_{k=0}^\infty \frac{1}{k!}f^{\left(k\right)}\left(0\right)x^k\;\ldots\;(ただし\;-1\lt x \lt1) \\&=&\frac{1}{0!}\left[\left(1-x\right)^{-1}\right]_{x=0}x^0 +\frac{1}{1!}\left[-\left(1-x\right)^{-1-1}(-1)\right]_{x=0}x^1 +\frac{1}{2!}\left[-2\left(1-x\right)^{-2-1}(-1)\right]_{x=0}x^2 +\frac{1}{3!}\left[-6\left(1-x\right)^{-3-1}(-1)\right]_{x=0}x^3 +\cdots \\&=&1+x+x^2+x^3+\cdots \end{eqnarray}$$

ディガンマ凾数の極限表示

ディガンマ凾数

$$\begin{eqnarray} \psi\left(z\right)&=&\frac{\mathrm{d}}{\mathrm{d}z} \log{\left(\Gamma\left(z\right)\right)} \\&=&\frac{\Gamma^\prime\left(z\right)}{\Gamma\left(z\right)} \end{eqnarray}$$

ディガンマ凾数の極限表示

$$\begin{eqnarray} \Gamma\left(z\right) &=&\href{https://shikitenkai.blogspot.com/2021/07/blog-post_16.html}{\lim_{n\rightarrow\infty}\frac{n^z n!}{\prod_{k=0}^n\left(z+k\right)}} \\\log{\left(\Gamma\left(z\right)\right)} &=&\log{\left(\lim_{n\rightarrow\infty}\frac{n^z n!}{\prod_{k=0}^n\left(z+k\right)}\right)} \\&=&\lim_{n\rightarrow\infty}\log{\left(\frac{n^z n!}{\prod_{k=0}^n\left(z+k\right)}\right)} \\&=&\lim_{n\rightarrow\infty} \left\{ \log{\left(n^z \right)} +\log{\left(n! \right)} -\log{\left( \prod_{k=0}^n\left(z+k\right) \right)} \right\} \\&=&\lim_{n\rightarrow\infty} \left\{ \log{\left(n^z \right)} +\log{\left(n! \right)} -\log{\left(z+0\right)}-\log{\left(z+1\right)}-\log{\left(z+2\right)}-\cdots-\log{\left(z+n\right)} \right\} \\\psi\left(z\right)=\frac{\mathrm{d}}{\mathrm{d}z} \log{\left(\Gamma\left(z\right)\right)} &=&\frac{\mathrm{d}}{\mathrm{d}z}\left[ \lim_{n\rightarrow\infty} \left\{ \log{\left(n^z \right)} +\log{\left(n! \right)} -\log{\left(z+0\right)}-\log{\left(z+1\right)}-\log{\left(z+2\right)}-\cdots-\log{\left(z+n\right)} \right\}\right] \\&=&\lim_{n\rightarrow\infty} \left[ \frac{\mathrm{d}}{\mathrm{d}z}\left\{ \log{\left(n^z \right)} +\log{\left(n! \right)} -\log{\left(z+0\right)}-\log{\left(z+1\right)}-\log{\left(z+2\right)}-\cdots-\log{\left(z+n\right)} \right\}\right] \\&=&\lim_{n\rightarrow\infty} \left\{ \log{\left(n \right)} -\frac{1}{z}-\frac{1}{z+1}-\frac{1}{z+2}-\cdots-\frac{1}{z+n} \right\} \\&=&\lim_{n\rightarrow\infty} \left[ \log{\left(n \right)} -\left\{\frac{1}{z}+\frac{1}{z+1}+\frac{1}{z+2}+\cdots+\frac{1}{z+n}\right\} \right] \\&=&\lim_{n\rightarrow\infty} \left\{ \log{\left(n \right)} -\sum_{k=0}^n \frac{1}{z+k} \right\} \end{eqnarray}$$

ガンマ凾数の極限表示

ガンマ凾数の極限表示

ガンマ凾数の定義と極限表示

$$\begin{eqnarray} \Gamma\left(z\right) &=&\int_0^\infty t^{z-1}e^{-t}\mathrm{d}t\;\;\;\left(\mathrm{Re}\left(z\right)\gt0\right)\;\ldots\;定義 \\&=&\lim_{n\rightarrow\infty}\frac{n^z n!}{\prod_{k=0}^n\left(z+k\right)}\;\ldots\;極限表示 \end{eqnarray}$$

\(G_n\)の導入

$$\begin{eqnarray} G_n\left(z\right)&=&\int_0^n t^{z-1}\left(1-\frac{t}{n}\right)^{n}\mathrm{d}t \end{eqnarray}$$

\(G_n\)の総乗表示

$$\begin{eqnarray} G_n\left(z\right)&=&\int_0^n t^{z-1}\left(1-\frac{t}{n}\right)^{n}\mathrm{d}t \\&=&\int_0^1 \left(nu\right)^{z-1}\left(1-\frac{nu}{n}\right)^{n}n\mathrm{d}u \\&&\;\ldots\;t=nu,\;u=\frac{t}{n},\;\frac{\mathrm{d}t}{\mathrm{d}u}=n,\;\mathrm{d}t=n\mathrm{d}u \\&&\;\ldots\;t:0\rightarrow n,\;u:0\rightarrow1 \\&=&n^z\int_0^1 u^{z-1}\left(1-u\right)^{n}\mathrm{d}u \\&=&n^z\cdot g_n\left(z\right)\;\ldots\;g_n\left(z\right)=\int_0^1 u^{z-1}\left(1-u\right)^{n}\mathrm{d}u \\g_0\left(z\right)&=&\int_0^1 u^{z-1}\left(1-u\right)^{0}\mathrm{d}u \\&=&\int_0^1 u^{z-1}\mathrm{d}u \\&=&\left[\frac{u^z}{z}\right]_{u=0}^1 \\&=&\frac{1}{z} \\g_n\left(z\right)&=&\int_0^1 u^{z-1}\left(1-u\right)^{n}\mathrm{d}u \\&=&\int_0^1 \left(\frac{u^{z}}{z}\right)^\prime\left(1-u\right)^{n}\mathrm{d}u \;\ldots\;\frac{\mathrm{d}}{\mathrm{d}u}\frac{u^{z}}{z}=\frac{1}{z}zu^{z-1}=u^{z-1} \\&=&\left[\frac{u^z}{z}\left(1-u\right)^{n}\right]_{u=0}^1 -\int_0^1 \frac{u^{z}}{z}\left(\left(1-u\right)^{n}\right)^\prime\mathrm{d}u \\&=&\left[\frac{1^z}{z}\left(1-1\right)^{n}-\frac{0^z}{z}\left(1-0\right)^{n}\right] -\int_0^1 \frac{u^{z}}{z}\left(n\left(1-u\right)^{n-1}(-1)\right)\mathrm{d}u \\&=&0+\frac{n}{z}\int_0^1 u^{z}\left(1-u\right)^{n-1}\mathrm{d}u \\&=&\frac{n}{z}g_{n-1}\left(z+1\right) \\&=&\frac{n}{z}\frac{n-1}{z+1}g_{n-2}\left(z+2\right) \\&=&\frac{n}{z}\frac{n-1}{z+1}\frac{n-2}{z+2}g_{n-3}\left(z+3\right) \\&=&\frac{n}{z}\frac{n-1}{z+1}\frac{n-2}{z+2}\ldots\frac{n-(n-1)}{z+(n-1)}g_{n-n}\left(z+n\right) \\&=&\frac{n}{z}\frac{n-1}{z+1}\frac{n-2}{z+2}\ldots\frac{1}{z+(n-1)}g_{0}\left(z+n\right) \\&=&\frac{n}{z}\frac{n-1}{z+1}\frac{n-2}{z+2}\ldots\frac{1}{z+(n-1)}\frac{1}{z+n} \\&=&\frac{n!}{\prod_{k=0}^n (z+k)} \\G_n\left(z\right)&=&n^z\cdot g_n\left(z\right) \\&=&n^z\frac{n!}{\prod_{k=0}^n (z+k)} \\&=&\frac{n^zn!}{\prod_{k=0}^n (z+k)} \end{eqnarray}$$

\(G_n\)の極限

$$\begin{eqnarray} \lim_{n\rightarrow\infty}G_n\left(z\right) &=&\lim_{n\rightarrow\infty}\int_0^n t^{z-1}\left(1-\frac{t}{n}\right)^{n}\mathrm{d}t \\&=&\int_0^\infty t^{z-1} e^{-t}\mathrm{d}t \\&&\;\ldots\;\lim_{n\rightarrow\infty}\left(1-\frac{t}{n}\right)^{n} =\lim_{n\rightarrow\infty}\left(1+\frac{-t}{n}\right)^{n} =e^{-t} \\&=&\Gamma\left(z\right) \end{eqnarray}$$

よって\(G_n\)の総乗表示での極限も\(\Gamma\left(z\right)\)

$$\begin{eqnarray} \lim_{n\rightarrow\infty}G_n\left(z\right) &=&\lim_{n\rightarrow\infty}\frac{n^zn!}{\prod_{k=0}^n (z+k)} \\&=&\Gamma\left(z\right) \end{eqnarray}$$

偏角の微分

$$\begin{eqnarray} z&=&x+iy\;\ldots\;z\in\mathbb{Z},\;x,y\in\mathbb{R} \\&=&re^{i\theta}\;\ldots\;r,\theta\in\mathbb{R},\;r\geq0,\;-\pi\lt\theta\leq\pi \\r&=&|z|=\sqrt{x^2+y^2} \\\arg{\left(z\right)}&=&\theta+2n\pi\;\left(n\in\mathbb{Z}\right) \\&=&\mathrm{Arg}{\left(z\right)}+2n\pi\;\left(n\in\mathbb{Z}\right) \end{eqnarray}$$ $$\begin{eqnarray} \mathrm{Arg}{\left(z\right)}&=&\mathrm{Arg}{\left(x+iy\right)} \\&=&\begin{cases} \tan^{-1}{\left(\frac{y}{x}\right)} & (x\gt0) \\\tan^{-1}{\left(\frac{y}{x}\right)+\pi} & (x\lt0\;かつ\;y\geq0) \\\tan^{-1}{\left(\frac{y}{x}\right)-\pi} & (x\lt0\;かつ\;y\lt0) \\\frac{\pi}{2} & (x=0\;かつ\;y\gt0) \\-\frac{\pi}{2} & (x=0\;かつ\;y\lt0) \\不定 & (x=0\;かつ\;y=0) \end{cases} \end{eqnarray}$$

xでの偏微分

$$\begin{eqnarray} \frac{\partial}{\partial x}\arg{\left(z\right)}&=&\frac{\partial}{\partial x}\left\{\theta+2n\pi\right\} \\&=&\frac{\partial}{\partial x}\left\{\mathrm{Arg}{\left(z\right)}+2n\pi\right\} \\&=&\frac{\partial}{\partial x}\mathrm{Arg}{\left(z\right)}+\frac{\partial}{\partial x}2n\pi \\&=&\frac{\partial}{\partial x}\mathrm{Arg}{\left(z\right)}\;\ldots\;\frac{\partial}{\partial x} C=0\;(Cは定数．xの凾数でない) \end{eqnarray}$$ $$\begin{eqnarray} \frac{\partial}{\partial x}\mathrm{Arg}{\left(z\right)}&=&\frac{\partial}{\partial x}\mathrm{Arg}{\left(x+iy\right)} \\&=&\begin{cases} \frac{\partial}{\partial x}\tan^{-1}{\left(\frac{y}{x}\right)} &=&\frac{-y}{x^2+y^2}& (x\gt0) \\\frac{\partial}{\partial x}\tan^{-1}{\left(\frac{y}{x}\right)+\pi} &=&\frac{-y}{x^2+y^2}& (x\lt0\;かつ\;y\geq0) \\\frac{\partial}{\partial x}\tan^{-1}{\left(\frac{y}{x}\right)-\pi} &=&\frac{-y}{x^2+y^2}& (x\lt0\;かつ\;y\lt0) \\\frac{\partial}{\partial x}\frac{\pi}{2} &=&0& (x=0\;かつ\;y\gt0) \\\frac{\partial}{\partial x}-\frac{\pi}{2} &=&0& (x=0\;かつ\;y\lt0) \\不定 &&& (x=0\;かつ\;y=0) \end{cases} \ldots\href{https://shikitenkai.blogspot.com/2021/07/tan^{-1}yx.html}{\frac{\partial}{\partial x}\tan^{-1}{\left(\frac{y}{x}\right)}=\frac{-y}{x^2+y^2}} \end{eqnarray}$$

yでの偏微分

$$\begin{eqnarray} \frac{\partial}{\partial y}\arg{\left(z\right)}&=&\frac{\partial}{\partial y}\left\{\theta+2n\pi\right\} \\&=&\frac{\partial}{\partial y}\left\{\mathrm{Arg}{\left(z\right)}+2n\pi\right\} \\&=&\frac{\partial}{\partial y}\mathrm{Arg}{\left(z\right)}+\frac{\partial}{\partial y}2n\pi \\&=&\frac{\partial}{\partial y}\mathrm{Arg}{\left(z\right)}\;\ldots\;\frac{\partial}{\partial y} C=0\;(Cは定数．xの凾数でない) \end{eqnarray}$$ $$\begin{eqnarray} \frac{\partial}{\partial y}\mathrm{Arg}{\left(z\right)}&=&\frac{\partial}{\partial y}\mathrm{Arg}{\left(x+iy\right)} \\&=&\begin{cases} \frac{\partial}{\partial y}\tan^{-1}{\left(\frac{y}{x}\right)} &=&\frac{x}{x^2+y^2}& (x\gt0) \\\frac{\partial}{\partial y}\tan^{-1}{\left(\frac{y}{x}\right)+\pi} &=&\frac{x}{x^2+y^2}& (x\lt0\;かつ\;y\geq0) \\\frac{\partial}{\partial y}\tan^{-1}{\left(\frac{y}{x}\right)-\pi} &=&\frac{x}{x^2+y^2}& (x\lt0\;かつ\;y\lt0) \\\frac{\partial}{\partial y}\frac{\pi}{2} &=&0& (x=0\;かつ\;y\gt0) \\\frac{\partial}{\partial y}-\frac{\pi}{2} &=&0& (x=0\;かつ\;y\lt0) \\不定 &&& (x=0\;かつ\;y=0) \end{cases} \ldots\href{https://shikitenkai.blogspot.com/2021/07/tan^{-1}yx.html}{\frac{\partial}{\partial y}\tan^{-1}{\left(\frac{y}{x}\right)}=\frac{x}{x^2+y^2}} \end{eqnarray}$$

arctan(y/x)の偏微分

\(\tan^{-1}{\left(\frac{y}{x}\right)}\)の偏微分

$$\begin{eqnarray} \theta&=&\tan^{-1}{\left(\frac{y}{x}\right)} \end{eqnarray}$$

\(\tan^{-1}{\left(\frac{y}{x}\right)}\)の\(x\)での偏微分

$$\begin{eqnarray} \frac{\partial }{\partial x}\tan^{-1}{\left(\frac{y}{x}\right)} &=&\left\{\frac{\partial }{\partial u}\tan^{-1}{\left(u\right)}\right\}\frac{\partial u}{\partial x} \;\ldots\;u=\frac{y}{x} \\&=&\frac{1}{1+u^2}\frac{\partial }{\partial x}\frac{y}{x} \;\ldots\;\href{https://shikitenkai.blogspot.com/2021/07/arctanx.html}{\frac{\partial }{\partial u}\tan^{-1}{\left(u\right)}=\frac{1}{1+u^2}} \\&=&\frac{1}{1+\frac{y^2}{x^2}}y\frac{\partial }{\partial x}x^{-1} \\&=&\frac{1}{\frac{x^2+y^2}{x^2}}y\left(-x^{-2}\right) \\&=&\frac{x^2}{x^2+y^2}y\left(-x^{-2}\right) \\&=&\frac{-y}{x^2+y^2} \end{eqnarray}$$

\(\tan^{-1}{\left(\frac{y}{x}\right)}\)の\(y\)での偏微分

$$\begin{eqnarray} \\\frac{\partial }{\partial y}\tan^{-1}{\left(\frac{y}{x}\right)} &=&\left\{\frac{\partial }{\partial u}\tan^{-1}{\left(u\right)}\right\}\frac{\partial u}{\partial y} \;\ldots\;u=\frac{y}{x} \\&=&\frac{1}{1+u^2}\frac{\partial }{\partial y}\frac{y}{x} \;\ldots\;\href{https://shikitenkai.blogspot.com/2021/07/arctanx.html}{\frac{\partial }{\partial u}\tan^{-1}{\left(u\right)}=\frac{1}{1+u^2}} \\&=&\frac{1}{1+\frac{y^2}{x^2}}x^{-1}\frac{\partial }{\partial y}y \\&=&\frac{1}{\frac{x^2+y^2}{x^2}}x^{-1}\cdot1 \\&=&\frac{x^2}{x^2+y^2}x^{-1} \\&=&\frac{x}{x^2+y^2} \end{eqnarray}$$

arctan(x)の微分

\(\tan^{-1}{\left(x\right)}\)の微分

$$\begin{eqnarray} y&=&\tan^{-1}{\left(x\right)} \\x&=&\tan{\left(y\right)} \\\frac{\mathrm{d}x}{\mathrm{d}y}&=&\frac{\mathrm{d}}{\mathrm{d}y}\tan{\left(y\right)} \\&=&1+\tan^2{\left(y\right)} \\\frac{\mathrm{d}y}{\mathrm{d}x}&=&\frac{1}{1+\tan^2{\left(y\right)}} \;\ldots\;\href{https://shikitenkai.blogspot.com/2021/07/blog-post.html}{\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{1}{\frac{\mathrm{d}x}{\mathrm{d}y}}\;\left(逆凾数の微分\right)} \\&=&\frac{1}{1+x^2} \;\ldots\;\href{https://shikitenkai.blogspot.com/2021/07/tanx-2.html}{x=\tan{\left(y\right)}} \end{eqnarray}$$

逆凾数の微分 (微分可能な凾数の逆凾数の微分)

逆凾数の微分 (微分可能な凾数の逆凾数の微分)

凾数，逆凾数における独立変数の差と従属変数の差の関係

$$\begin{eqnarray} y&=&f\left(x\right)\;\ldots\;微分可能な凾数 \\x&=&f^{-1}\left(y\right)\;\ldots\;f(x)の逆凾数 \end{eqnarray}$$ $$\begin{eqnarray} k&=&f\left(x+h\right)-f\left(x\right) \\&&\;\ldots\;凾数において独立変数の差hがある際の従属変数の差をkとする \\f\left(x\right)+k&=&y+k \\&=&f\left(x+h\right) \\x+h&=&f^{-1}\left(y+k\right) \end{eqnarray}$$ $$\begin{eqnarray} x+h&=&f^{-1}\left(y+k\right) \\h&=&f^{-1}\left(y+k\right)-x \\&=&f^{-1}\left(y+k\right)-f^{-1}\left(y\right) \\&&\;\ldots\;逆凾数において独立変数の差kがある際の従属変数の差がhである \end{eqnarray}$$ $$\begin{eqnarray} \lim_{k\rightarrow0}f^{-1}\left(y+k\right)&=&f^{-1}\left(y\right) \\h&=&\lim_{k\rightarrow0}\left(f^{-1}\left(y+k\right)-f^{-1}\left(y\right)\right) \\&=&\left(\lim_{k\rightarrow0}f^{-1}\left(y+k\right)\right)-f^{-1}\left(y\right) \\&=&f^{-1}\left(y\right)-f^{-1}\left(y\right) \\&=&0 \\&&\;\ldots\;kを0に近づけることはhを0に近づけることと等しい \end{eqnarray}$$ $$\begin{eqnarray} k&=&f\left(x+h\right)-f\left(x\right) \\\lim_{h\rightarrow0}f\left(x+h\right)&=&f\left(x\right) \\k&=&\lim_{h\rightarrow0}\left(f\left(x+h\right)-f\left(x\right)\right) \\&=&\left(\lim_{h\rightarrow0}f\left(x+h\right)\right)-f\left(x\right) \\&=&f\left(x\right)-f\left(x\right) \\&=&0 \\&&\;\ldots\;hを0に近づけることはkを0に近づけることと等しい \end{eqnarray}$$ $$\begin{eqnarray} h\rightarrow0 \iff k\rightarrow0 \\\lim_{k\rightarrow0}\frac{h}{k}=\lim_{h\rightarrow0}\frac{h}{k} \end{eqnarray}$$

逆凾数の微分

$$\begin{eqnarray} \frac{\mathrm{d}x}{\mathrm{d}y}&=&\lim_{k\rightarrow0}\frac{f^{-1}\left(y+k\right)-f^{-1}\left(y\right)}{k} \\&=&\lim_{k\rightarrow0}\frac{x+h-x}{k} \\&=&\lim_{k\rightarrow0}\frac{h}{k} \\&=&\lim_{h\rightarrow0}\frac{h}{k} \\&=&\lim_{h\rightarrow0}\frac{1}{\frac{k}{h}} \\&=&\frac{1}{\lim_{h\rightarrow0}\frac{k}{h}} \\&=&\frac{1}{\lim_{h\rightarrow0}\frac{f\left(x+h\right)-f\left(x\right)}{h}} \\&=&\frac{1}{\frac{\mathrm{d}y}{\mathrm{d}x}} \end{eqnarray}$$

tan(x)の微分 2(微分の定義からの場合)

\(\tan{\left(x\right)}\)の微分 2

(sin, cosの微分を既知とした場合) $$\begin{eqnarray} \\y&=&\tan{\left(x\right)} \\\frac{\mathrm{d}}{\mathrm{d}x}\tan{\left(x\right)} &=&\lim_{h\rightarrow0}\frac{\tan{\left(x+h\right)}-\tan{\left(x\right)}}{h} \\&=&\lim_{h\rightarrow0}\frac{1}{h}\left\{ \frac{\tan{\left(x\right)}+\tan{\left(h\right)}}{1-\tan{\left(x\right)}\tan{\left(h\right)}} -\tan{\left(x\right)} \right\} \\&=&\lim_{h\rightarrow0}\frac{1}{h} \frac{ \tan{\left(x\right)}+\tan{\left(h\right)} -\tan{\left(x\right)}\left(1-\tan{\left(x\right)}\tan{\left(h\right)}\right) }{1-\tan{\left(x\right)}\tan{\left(h\right)}} \\&=&\lim_{h\rightarrow0}\frac{1}{h} \frac{ \tan{\left(x\right)}+\tan{\left(h\right)} -\tan{\left(x\right)}+\tan^2{\left(x\right)}\tan{\left(h\right)} }{1-\tan{\left(x\right)}\tan{\left(h\right)}} \\&=&\lim_{h\rightarrow0}\frac{1}{h} \frac{ \tan{\left(h\right)}+\tan^2{\left(x\right)}\tan{\left(h\right)} }{1-\tan{\left(x\right)}\tan{\left(h\right)}} \\&=&\lim_{h\rightarrow0}\frac{1}{h} \frac{ \tan{\left(h\right)}\left(1+\tan^2{\left(x\right)}\right) }{1-\tan{\left(x\right)}\tan{\left(h\right)}} \\&=&\lim_{h\rightarrow0}\frac{\tan{\left(h\right)}}{h} \frac{1+\tan^2{\left(x\right)}}{1-\tan{\left(x\right)}\tan{\left(h\right)}} \\&=&1\cdot\frac{1+\tan^2{\left(x\right)}}{1-\tan{\left(x\right)}\cdot0} \\&&\;\ldots\;\lim_{h\rightarrow0}\frac{\tan{\left(h\right)}}{h} =\lim_{h\rightarrow0}\frac{1}{h}\frac{\sin{\left(h\right)}}{\cos{\left(h\right)}} =\lim_{h\rightarrow0}\frac{\sin{\left(h\right)}}{h}\frac{1}{\cos{\left(h\right)}} =1\cdot1 =1 \\&&\;\ldots\;\lim_{h\rightarrow0}\frac{\sin{\left(h\right)}}{h}=1 \\&&\;\ldots\;\lim_{h\rightarrow0}\cos{\left(h\right)}=1 \\&=&1+\tan^2{\left(x\right)} \end{eqnarray}$$ $$\begin{eqnarray} 1+\tan^2{\left(x\right)}&=&1+\left\{\frac{\sin{\left(x\right)}}{\cos{\left(x\right)}}\right\}^2 \\&=&\frac{\cos^2{\left(x\right)}+\sin^2{\left(x\right)}}{\cos^2{\left(x\right)}} \\&=&\frac{1}{ \cos^2{\left(x\right)} } \;\cdots\;\cos^2{\left(x\right)}+\sin^2{\left(x\right)}=1 \\&=&\sec^{2}{\left(x\right)} \end{eqnarray}$$

tan(x)の微分 1 (sin, cosの微分を既知とした場合)

\(\tan{\left(x\right)}\)の微分 1

(微分の定義からの場合) $$\begin{eqnarray} y&=&\tan{\left(x\right)} \\\frac{\mathrm{d}}{\mathrm{d}x}\tan{\left(x\right)} &=&\frac{\mathrm{d}}{\mathrm{d}x}\frac{\sin{\left(x\right)}}{\cos{\left(x\right)}} \\&=&\frac{\mathrm{d}}{\mathrm{d}x}\sin{\left(x\right)}\left\{\cos{\left(x\right)}\right\}^{-1} \\&=&\left\{\cos{\left(x\right)}\right\}^{-1}\frac{\mathrm{d}}{\mathrm{d}x}\sin{\left(x\right)} +\sin{\left(x\right)}\frac{\mathrm{d}}{\mathrm{d}x}\left\{\cos{\left(x\right)}\right\}^{-1} \\&=&\left\{\cos{\left(x\right)}\right\}^{-1}\left\{\cos\left(x\right)\right\} +\sin{\left(x\right)} \left[-\left\{\cos{\left(x\right)}\right\}^{-2}\right] \frac{\mathrm{d}}{\mathrm{d}x}\left\{\cos{\left(x\right)}\right\} \\&=&1+\frac{-\sin{\left(x\right)}}{\cos^2{\left(x\right)}} \left\{-\sin{\left(x\right)}\right\} \\&=&1+\frac{\sin^2{\left(x\right)}}{\cos^2{\left(x\right)}} \\&=&1+\left\{\frac{\sin{\left(x\right)}}{\cos{\left(x\right)}}\right\}^2 \\&=&1+\tan^2{\left(x\right)} \end{eqnarray}$$

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$$\begin{eqnarray} 1+\tan^2{\left(x\right)}&=&1+\left\{\frac{\sin{\left(x\right)}}{\cos{\left(x\right)}}\right\}^2 \\&=&\frac{\cos^2{\left(x\right)}+\sin^2{\left(x\right)}}{\cos^2{\left(x\right)}} \\&=&\frac{1}{ \cos^2{\left(x\right)} } \;\cdots\;\cos^2{\left(x\right)}+\sin^2{\left(x\right)}=1 \\&=&\sec^{2}{\left(x\right)} \end{eqnarray}$$