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)

$$\begin{array}{rcl}\sum _{k=1}^{\mathrm{\infty }}\frac{1}{k^2}& =& \frac{\pi }{6}\end{array}$$

ディガンマ凾数の相反公式

$$\begin{array}{rcl}\psi (1-z)-\psi \left(z\right)& =& \pi \cot \left(\pi z\right)\end{array}$$
その微分 $$\begin{array}{rcl}\frac{\partial }{\partial z}\pi \cot \left(\pi z\right)& =& \pi \frac{\partial }{\partial z}\cot \left(\pi z\right)\\ & =& \pi \{-\frac{\pi }{{\sin }^2\left(\pi z\right)}\}\dots \frac{\partial }{\partial w}\cot \left(aw\right)=-\frac{a}{{\sin }^2\left(aw\right)}\\ & =& -\frac{{\pi }^2}{{\sin }^2\left(\pi z\right)}\\ & =& f(z)\end{array}$$

$f(z)$の$z=\frac{1}{3}$の値

$$\begin{array}{rcl}-\frac{{\pi }^2}{{\sin }^2\left(\pi \frac{1}{3}\right)}& =& -\frac{{\pi }^2}{{\left(\frac{\sqrt{3}}{2}\right)}^2}\dots \sin \left(\frac{\pi }{3}\right)=\frac{\sqrt{3}}{2}\\ & =& -\frac{{\pi }^2}{\frac{3}{4}}\\ & =& \frac{-4{\pi }^2}{3}\end{array}$$

ディガンマ凾数の相反公式の積分表示

$$\begin{array}{rcl}\psi (1-z)-\psi \left(z\right)& =& {\int }_0^1\frac{u^{(z)-1}-u^{(1-z)-1}}{1-u}\mathrm{d}u\dots \psi \left(y\right)-\psi \left(x\right)={\int }_0^1\frac{u^{x-1}-u^{y-1}}{1-u}\mathrm{d}u\\ & =& {\int }_0^1\frac{u^{z-1}-u^{-z}}{1-u}\mathrm{d}u\\ & =& {\int }_{\mathrm{\infty }}^0\frac{(e^{-x}{)}^{z-1}-(e^{-x}{)}^{-z}}{1-e^{-x}}(-e^{-x})\mathrm{d}x\\ & & \dots u=e^{-x},\frac{\mathrm{d}u}{\mathrm{d}x}=-e^{-x}\\ & & \dots u=e^{-x},u:0\to 1,x:\mathrm{\infty }\to 0\\ & =& {\int }_0^{\mathrm{\infty }}\frac{e^{-x}\{e^{-x(z-1)}-e^{-x(-z)}\}}{1-e^{-x}}\mathrm{d}x\\ & =& {\int }_0^{\mathrm{\infty }}\frac{e^{-x}{e}^{-xz+x}-e^{-x}{e}^{xz}}{1-e^{-x}}\mathrm{d}x\\ & =& {\int }_0^{\mathrm{\infty }}\frac{e^{-x-xz+x}-e^{-x+xz}}{1-e^{-x}}\mathrm{d}x\\ & =& {\int }_0^{\mathrm{\infty }}\frac{e^{-xz}-e^{-x(1-z)}}{1-e^{-x}}\mathrm{d}x\end{array}$$
その微分 $$\begin{array}{rcl}\frac{\partial }{\partial z}{\int }_0^{\mathrm{\infty }}\frac{e^{-xz}-e^{-x(1-z)}}{1-e^{-x}}\mathrm{d}x& =& {\int }_0^{\mathrm{\infty }}\frac{1}{1-e^{-x}}(\frac{\partial }{\partial z}{e}^{-xz}-\frac{\partial }{\partial z}{e}^{-x(1-z)})\mathrm{d}x\\ & =& {\int }_0^{\mathrm{\infty }}\frac{1}{1-e^{-x}}(-x{e}^{-xz}-(-x)(-1)e^{-x(1-z)})\mathrm{d}x\\ & =& {\int }_0^{\mathrm{\infty }}\frac{1}{1-e^{-x}}(-x{e}^{-xz}-x{e}^{-x(1-z)})\mathrm{d}x\\ & =& {\int }_0^{\mathrm{\infty }}\frac{1}{1-e^{-x}}\{-x(e^{-xz}+e^{-x(1-z)})\}\mathrm{d}x\\ & =& -{\int }_0^{\mathrm{\infty }}\frac{1}{1-e^{-x}}\left\{x(e^{-xz}+e^{-x(1-z)})\right\}\mathrm{d}x\\ & =& g(z)\end{array}$$

$g(z)$の$z=\frac{1}{3}$の値

$$\begin{array}{rcl}g\left(\frac{1}{3}\right)& =& -{\int }_0^{\mathrm{\infty }}\frac{1}{1-e^{-x}}\left\{x(e^{-x\frac{1}{3}}+e^{-x(1-\frac{1}{3})})\right\}\mathrm{d}x\\ & =& -{\int }_0^{\mathrm{\infty }}\frac{1}{1-e^{-x}}\left\{x(e^{-\frac{x}{3}}+e^{\frac{-2x}{3}})\right\}\mathrm{d}x\\ & =& -{\int }_0^{\mathrm{\infty }}\frac{e^x}{e^x}\frac{1}{1-e^{-x}}\left\{x(e^{-\frac{x}{3}}+e^{\frac{-2x}{3}})\right\}\mathrm{d}x\\ & =& -{\int }_0^{\mathrm{\infty }}\frac{e^x}{e^x(1-e^{-x})}\left\{x(e^{-\frac{x}{3}}+e^{\frac{-2x}{3}})\right\}\mathrm{d}x\\ & =& -{\int }_0^{\mathrm{\infty }}\frac{e^x}{e^x-e^x{e}^{-x}}\left\{x(e^{-\frac{x}{3}}+e^{\frac{-2x}{3}})\right\}\mathrm{d}x\\ & =& -{\int }_0^{\mathrm{\infty }}\frac{e^x}{e^x-1}\left\{x(e^{-\frac{x}{3}}+e^{\frac{-2x}{3}})\right\}\mathrm{d}x\\ & =& -{\int }_0^{\mathrm{\infty }}\left(\sum _{k=0}^{\mathrm{\infty }}{e}^{-kx}\right)\left\{x(e^{-\frac{x}{3}}+e^{\frac{-2x}{3}})\right\}\mathrm{d}x\dots \sum _{k=0}^{\mathrm{\infty }}{e}^{-kx}=\frac{e^x}{e^x-1},\sum _{k=0}^{\mathrm{\infty }}{a}^{-kx}=\frac{a^x}{a^x-1}\\ & =& -\sum _{k=0}^{\mathrm{\infty }}{\int }_0^{\mathrm{\infty }}{e}^{-kx}\left\{x(e^{-\frac{x}{3}}+e^{\frac{-2x}{3}})\right\}\mathrm{d}x\\ & =& -\sum _{k=0}^{\mathrm{\infty }}{\int }_0^{\mathrm{\infty }}x(e^{-kx}{e}^{-\frac{x}{3}}+e^{-kx}{e}^{\frac{-2x}{3}})\mathrm{d}x\\ & =& -\sum _{k=0}^{\mathrm{\infty }}{\int }_0^{\mathrm{\infty }}x(e^{-x(k+\frac{1}{3})}+e^{-x(k+\frac{2}{3})})\mathrm{d}x\\ & =& -\sum _{k=0}^{\mathrm{\infty }}({\int }_0^{\mathrm{\infty }}x{e}^{-x(k+\frac{1}{3})}\mathrm{d}x+{\int }_0^{\mathrm{\infty }}x{e}^{-x(k+\frac{2}{3})}\mathrm{d}x)\\ & =& -\sum _{k=0}^{\mathrm{\infty }}(\frac{1}{{(k+\frac{1}{3})}^2}+\frac{1}{{(k+\frac{2}{3})}^2})\dots {\int }_0^{\mathrm{\infty }}x{e}^{-ax}\mathrm{d}x=\frac{1}{a^2}\\ & =& -\sum _{k=0}^{\mathrm{\infty }}(\frac{1}{{\left(\frac{3k+1}{3}\right)}^2}+\frac{1}{{\left(\frac{3k+2}{3}\right)}^2})\\ & =& -\sum _{k=0}^{\mathrm{\infty }}(\frac{3^2}{{(3k+1)}^2}+\frac{3^2}{{(3k+2)}^2})\\ & =& -\sum _{k=0}^{\mathrm{\infty }}(\frac{3^2}{{(3k+1)}^2}+\frac{3^2}{{(3k+2)}^2}+\frac{3^2}{{(3k+3)}^2}-\frac{3^2}{{(3k+3)}^2})\\ & =& -\sum _{k=0}^{\mathrm{\infty }}(\frac{3^2}{{(3k+1)}^2}+\frac{3^2}{{(3k+2)}^2}+\frac{3^2}{{(3k+3)}^2})-\sum _{k=0}^{\mathrm{\infty }}(-\frac{3^2}{{(3k+3)}^2})\\ & =& -\sum _{k=0}^{\mathrm{\infty }}(\frac{3^2}{{(3k+1)}^2}+\frac{3^2}{{(3k+2)}^2}+\frac{3^2}{{(3k+3)}^2})+\sum _{k=0}^{\mathrm{\infty }}\frac{3^2}{{(3k+3)}^2}\\ & =& -\sum _{k=0}^{\mathrm{\infty }}(\frac{3^2}{{(3k+1)}^2}+\frac{3^2}{{(3k+2)}^2}+\frac{3^2}{{(3k+3)}^2})+\sum _{k=0}^{\mathrm{\infty }}\frac{3^2}{3^2{(k+1)}^2}\\ & =& -\sum _{k=0}^{\mathrm{\infty }}(\frac{3^2}{{(3k+1)}^2}+\frac{3^2}{{(3k+2)}^2}+\frac{3^2}{{(3k+3)}^2})+\sum _{k=0}^{\mathrm{\infty }}\frac{1}{{(k+1)}^2}\\ & =& -\sum _{k=0}^{\mathrm{\infty }}(\frac{3^2}{{(3k+1)}^2}+\frac{3^2}{{(3k+2)}^2}+\frac{3^2}{{(3k+3)}^2})+\sum _{k=1}^{\mathrm{\infty }}\frac{1}{k^2}\\ & & \dots \sum _{k=0}^{\mathrm{\infty }}\frac{1}{{(k+1)}^2}=\frac{1}{{(0+1)}^2}+\frac{1}{{(1+1)}^2}+\frac{1}{{(2+1)}^2}+\cdots =\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\cdots =\sum _{k=1}^{\mathrm{\infty }}\frac{1}{k^2}\\ & =& -3^2\sum _{k=0}^{\mathrm{\infty }}(\frac{1}{{(3k+1)}^2}+\frac{1}{{(3k+2)}^2}+\frac{1}{{(3k+3)}^2})+\sum _{k=1}^{\mathrm{\infty }}\frac{1}{k^2}\\ & =& -9\sum _{k=1}^{\mathrm{\infty }}\frac{1}{k^2}+\sum _{k=1}^{\mathrm{\infty }}\frac{1}{k^2}\\ & & \dots \sum _{k=0}^{\mathrm{\infty }}(\frac{1}{{(3k+1)}^2}+\frac{1}{{(3k+2)}^2}+\frac{1}{{(3k+3)}^2})=(\frac{1}{{(3\cdot 0+1)}^2}+\frac{1}{{(3\cdot 0+2)}^2}+\frac{1}{{(3\cdot 0+3)}^2})+(\frac{1}{{(3\cdot 1+1)}^2}+\frac{1}{{(3\cdot 1+2)}^2}+\frac{1}{{(3\cdot 1+3)}^2})+\cdots =(\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2})+(\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2})+\cdots =\sum _{k=1}^{\mathrm{\infty }}\frac{1}{k^2}\\ & =& -8\sum _{k=1}^{\mathrm{\infty }}\frac{1}{k^2}\end{array}$$

fとgが等しいことを利用

$$\begin{array}{rcl}g\left(\frac{1}{3}\right)& =& f\left(\frac{1}{3}\right)\\ -8\sum _{k=1}^{\mathrm{\infty }}\frac{1}{k^2}& =& \frac{-4{\pi }^2}{3}\\ \sum _{k=1}^{\mathrm{\infty }}\frac{1}{k^2}& =& \frac{-1}{8}\frac{-4{\pi }^2}{3}\\ & =& \frac{{\pi }^2}{6}\end{array}$$

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

$$\begin{array}{rcl}S=\sum _{k=0}^{\mathrm{\infty }}{a}^{-k}& =& \sum _{k=0}^{\mathrm{\infty }}{\left(a^{-1}\right)}^k\dots a\in \mathbb{R},{\left|a\right|}^{-1}=\frac{1}{\left|a\right|}<1\\ & =& \sum _{k=0}^{\mathrm{\infty }}{\left(\frac{1}{a}\right)}^k\\ & =& \sum _{k=0}^{\mathrm{\infty }}\frac{1}{a^k}\\ & =& \frac{1}{a^0}+\frac{1}{a}+\frac{1}{a^2}+\frac{1}{a^3}+\cdots \\ & =& 1+\frac{1}{a}+\frac{1}{a\cdot a}+\frac{1}{a\cdot a^2}+\cdots \\ & =& 1+\frac{1}{a}+\frac{1}{a}\frac{1}{a}+\frac{1}{a}\frac{1}{a^2}+\cdots \\ & =& 1+\frac{1}{a}(1+\frac{1}{a}+\frac{1}{a^2}+\cdots )\\ & =& 1+\frac{1}{a}S\\ S-\frac{1}{a}S& =& 1\\ S(1-\frac{1}{a})& =& 1\\ S& =& \frac{1}{1-\frac{1}{a}}\\ & =& \frac{1}{\frac{a-1}{a}}\\ & =& \frac{a}{a-1}\end{array}$$

$a^{-kx}$の無限級数

$$\begin{array}{rcl}S=\sum _{k=0}^{\mathrm{\infty }}{a}^{-kx}& =& \sum _{k=0}^{\mathrm{\infty }}{\left(a^{-x}\right)}^k\dots a,x\in \mathbb{R},|a{|}^{-x}=\frac{1}{|a{|}^x}<1\\ & =& \sum _{k=0}^{\mathrm{\infty }}{\left(\frac{1}{a^x}\right)}^k\\ & =& \sum _{k=0}^{\mathrm{\infty }}\frac{1}{a^{kx}}\\ & =& \frac{1}{a^{0x}}+\frac{1}{a^x}+\frac{1}{a^{2x}}+\frac{1}{a^{3x}}+\cdots \\ & =& 1+\frac{1}{a^x}+\frac{1}{a^x{a}^x}+\frac{1}{a^x{a}^{2x}}+\cdots \\ & =& 1+\frac{1}{a^x}+\frac{1}{a^x}\frac{1}{a^x}+\frac{1}{a^x}\frac{1}{a^{2x}}+\cdots \\ & =& 1+\frac{1}{a^x}(1+\frac{1}{a^x}+\frac{1}{a^{2x}}+\cdots )\\ & =& 1+\frac{1}{a^x}S\\ S-\frac{1}{a^x}S& =& 1\\ S(1-\frac{1}{a^x})& =& 1\\ S& =& \frac{1}{1-\frac{1}{a^x}}\\ & =& \frac{1}{\frac{a^x-1}{a^x}}\\ & =& \frac{a^x}{a^x-1}\end{array}$$

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

$$\begin{array}{rcl}{B(p,q)|}_{p=1-q}& =& {\frac{\mathrm{\Gamma }\left(p\right)\mathrm{\Gamma }\left(q\right)}{\mathrm{\Gamma }(p+q)}|}_{p=1-q}\dots p,q\in \mathbb{C},\mathrm{Re}\left(p\right),\mathrm{Re}\left(q\right)>0\\ & =& \frac{\mathrm{\Gamma }(1-q)\mathrm{\Gamma }\left(q\right)}{\mathrm{\Gamma }(1-q+q)}\\ & =& \frac{\mathrm{\Gamma }(1-q)\mathrm{\Gamma }\left(q\right)}{\mathrm{\Gamma }\left(1\right)}\\ & =& \frac{\mathrm{\Gamma }(1-q)\mathrm{\Gamma }\left(q\right)}{1}\\ & & \dots \mathrm{\Gamma }\left(1\right)={\int }_0^{\mathrm{\infty }}{t}^{1-1}{e}^{-t}\mathrm{d}t={\int }_0^{\mathrm{\infty }}{t}^0{e}^{-t}\mathrm{d}t={\int }_0^{\mathrm{\infty }}1\cdot e^{-t}\mathrm{d}t={\int }_0^{\mathrm{\infty }}{e}^{-t}\mathrm{d}t\\ & & \dots {\int }_0^{\mathrm{\infty }}{e}^{-t}\mathrm{d}t=\frac{1}{1}=1\\ & =& \mathrm{\Gamma }(1-q)\mathrm{\Gamma }\left(q\right)\end{array}$$

ベータ凾数の定義での表示

$$\begin{array}{rcl}{B(p,q)|}_{p=1-q}& =& {\int }_0^1{t}^{p-1}{(1-t)}^{q-1}\mathrm{d}t\\ & =& {\int }_{\mathrm{\infty }}^0{\left(\frac{1}{x+1}\right)}^{p-1}{(1-\frac{1}{x+1})}^{q-1}\cdot \frac{-1}{{(x+1)}^2}\mathrm{d}x\\ & & \dots t=\frac{1}{x+1},x=\frac{1}{t}-1,\frac{\mathrm{d}t}{\mathrm{d}x}=\frac{-1}{{(x+1)}^2}\\ & & \dots t:0\to 1,x:\mathrm{\infty }\to 0\\ & =& {\int }_0^{\mathrm{\infty }}{\left(\frac{1}{x+1}\right)}^{p-1}{(1-\frac{1}{x+1})}^{q-1}\cdot \frac{1}{{(x+1)}^2}\mathrm{d}x\\ & =& {\int }_0^{\mathrm{\infty }}{\left(\frac{1}{x+1}\right)}^{p-1}{\left(\frac{x}{x+1}\right)}^{q-1}\cdot \frac{1}{{(x+1)}^2}\mathrm{d}x\\ & =& {\int }_0^{\mathrm{\infty }}\frac{1^{p-1}\cdot x^{q-1}\cdot 1}{{(x+1)}^{p-1}{(x+1)}^{q-1}{(x+1)}^2}\mathrm{d}x\\ & =& {\int }_0^{\mathrm{\infty }}\frac{x^{q-1}}{{(x+1)}^{(p-1)+(q-1)+2}}\mathrm{d}x\\ & =& {\int }_0^{\mathrm{\infty }}\frac{x^{q-1}}{{(x+1)}^{p+q}}\mathrm{d}x\\ & =& {\int }_0^{\mathrm{\infty }}\frac{x^{q-1}}{{(x+1)}^{1-q+q}}\mathrm{d}x\\ & =& {\int }_0^{\mathrm{\infty }}\frac{x^{q-1}}{x+1}\mathrm{d}x\end{array}$$

ガンマ凾数の相反公式の積分表示

$$\begin{array}{rcl}\mathrm{\Gamma }(1-q)\mathrm{\Gamma }\left(q\right)& =& {\int }_0^{\mathrm{\infty }}\frac{x^{q-1}}{x+1}\mathrm{d}x\end{array}$$

ガンマ凾数の相反公式

ガンマ凾数の相反公式

$$\begin{array}{rcl}\mathrm{\Gamma }\left(z\right)\mathrm{\Gamma }(1-z)& =& \mathrm{\Gamma }\left(z\right)\mathrm{\Gamma }(-z+1)\dots \mathrm{Re}\left(z\right)>0\mathrm{か}\mathrm{つ}\mathrm{Re}(1-z)>0\mathrm{な}\mathrm{の}\mathrm{で}0<\mathrm{Re}\left(z\right)<1\\ & =& \mathrm{\Gamma }\left(z\right)(-z\mathrm{\Gamma }(-z))\dots \mathrm{\Gamma }(z+1)=z\mathrm{\Gamma }\left(z\right)\\ & =& -z\mathrm{\Gamma }\left(z\right)\mathrm{\Gamma }(-z)\\ & =& -z\left(\underset{n\to \mathrm{\infty }}{lim}\frac{n^{z}n!}{\prod _{k=0}^n(z+k)}\right)\left(\underset{n\to \mathrm{\infty }}{lim}\frac{n^{-z}n!}{\prod _{k=0}^n(-z+k)}\right)\dots \mathrm{\Gamma }\left(z\right)=\underset{n\to \mathrm{\infty }}{lim}\frac{n^{z}n!}{\prod _{k=0}^n(z+k)}\\ & =& -z\left(\underset{n\to \mathrm{\infty }}{lim}\frac{n^{z}n!\cdot n^{-z}n!}{\prod _{k=0}^n(z+k)(-z+k)}\right)\\ & =& -z\left(\underset{n\to \mathrm{\infty }}{lim}\frac{n!n!}{\prod _{k=0}^n(k^2-z^2)}\right)\dots A^B{A}^{-B}=A^{B-B}=A^0=1\\ & =& -z\left(\underset{n\to \mathrm{\infty }}{lim}\frac{\left(\prod _{k=1}^{n}k\right)\left(\prod _{k=1}^{n}k\right)}{\prod _{k=0}^n(k^2-z^2)}\right)\dots n!=\prod _{k=1}^{n}k\\ & =& -z\left(\underset{n\to \mathrm{\infty }}{lim}\frac{\prod _{k=1}^n{k}^2}{\prod _{k=0}^n(k^2-z^2)}\right)\dots \left(\prod _{k=m}^n{A}_k\right)\left(\prod _{k=m}^n{B}_k\right)=\prod _{k=m}^n{A}_k{B}_k\\ & =& -z\left(\underset{n\to \mathrm{\infty }}{lim}\frac{\prod _{k=1}^n{k}^2}{(0^2-z^2)\prod _{k=1}^n(k^2-z^2)}\right)\\ & =& -z\frac{1}{-z^2}\left(\underset{n\to \mathrm{\infty }}{lim}\prod _{k=1}^n\frac{k^2}{k^2-z^2}\right)\\ & =& \frac{1}{z}\prod _{k=1}^{\mathrm{\infty }}\frac{k^2}{k^2-z^2}\\ & =& \frac{1}{\frac{1}{\frac{1}{z}\prod _{k=0}^{\mathrm{\infty }}\frac{k^2}{k^2-z^2}}}\dots A=\frac{1}{\frac{1}{A}}\\ & =& \frac{1}{z\prod _{k=1}^{\mathrm{\infty }}\frac{k^2-z^2}{k^2}}\\ & =& \frac{\pi }{\pi }\frac{1}{z\prod _{k=0}^{\mathrm{\infty }}\frac{k^2-z^2}{k^2}}\\ & =& \frac{\pi }{\pi z\prod _{k=0}^{\mathrm{\infty }}(1-\frac{z^2}{k^2})}\\ & =& \frac{\pi }{\sin \left(\pi z\right)}\\ & & \dots \sin \left(\pi z\right)=\pi z\prod _{k=1}^{\mathrm{\infty }}(1-{\left(\frac{z}{k}\right)}^2)(\mathrm{三}\mathrm{角}\mathrm{凾}\mathrm{数}\mathrm{の}\mathrm{無}\mathrm{限}\mathrm{乗}\mathrm{積}\mathrm{展}\mathrm{開}wikipedia)\end{array}$$

xe^(-ax)の広義積分[0,∞]

$x{e}^{-ax}$の広義積分$[0,\mathrm{\infty }]$

$$\begin{array}{rcl}{\int }_0^{\mathrm{\infty }}x{e}^{-ax}\mathrm{d}x& =& \underset{n\to \mathrm{\infty }}{lim}{\int }_0^{n}x{e}^{-ax}\mathrm{d}x\\ & =& \underset{n\to \mathrm{\infty }}{lim}[{[x\cdot \frac{-1}{a}{e}^{-ax}]}_0^n-{\int }_0^n\frac{-1}{a}\cdot e^{-ax}\mathrm{d}x]\\ & =& \underset{n\to \mathrm{\infty }}{lim}[[n\frac{-1}{a}{e}^{-an}-0\cdot \frac{-1}{a}{e}^{-a0}]+\frac{1}{a}{\int }_0^n{e}^{-ax}\mathrm{d}x]\\ & =& \underset{n\to \mathrm{\infty }}{lim}[n\frac{-1}{a}{e}^{-an}+\frac{1}{a}{\int }_0^n\cdot e^{-ax}\mathrm{d}x]\\ & =& \frac{-1}{a}\underset{n\to \mathrm{\infty }}{lim}n{e}^{-an}+\frac{1}{a}\underset{n\to \mathrm{\infty }}{lim}{\int }_0^n\cdot e^{-ax}\mathrm{d}x\\ & =& 0+\frac{1}{a}\underset{n\to \mathrm{\infty }}{lim}{\int }_0^n{e}^{-ax}\mathrm{d}x\\ & & \mathrm{指}\mathrm{数}\mathrm{凾}\mathrm{数}\left(e^{-x}\right)\mathrm{の}\mathrm{方}\mathrm{が}\mathrm{線}\mathrm{形}\mathrm{凾}\mathrm{数}\left(x\right)\mathrm{よ}\mathrm{り}\mathrm{早}\mathrm{く}\mathrm{漸}\mathrm{近}\mathrm{化}\mathrm{す}\mathrm{る}\mathrm{の}\mathrm{で}\underset{n\to \mathrm{\infty }}{lim}{e}^{-ax}=0\mathrm{が}\mathrm{収}\mathrm{束}\mathrm{先}\mathrm{と}\mathrm{な}\mathrm{る}．\\ & =& \frac{1}{a}\underset{n\to \mathrm{\infty }}{lim}{\int }_0^n{e}^{-ax}\mathrm{d}x\\ & =& \frac{1}{a}\cdot \frac{1}{a}\dots {\int }_0^{\mathrm{\infty }}{e}^{-ax}\mathrm{d}x=\frac{1}{a}\\ & =& \frac{1}{a^2}\end{array}$$

e^(-ax)の広義積分[0,∞]

$e^{-ax}$の広義積分$[0,\mathrm{\infty }]$

$$\begin{array}{rcl}{\int }_0^{\mathrm{\infty }}{e}^{-ax}\mathrm{d}x& =& \underset{n\to \mathrm{\infty }}{lim}{\int }_0^n{e}^{-ax}\mathrm{d}x\\ & =& \underset{n\to \mathrm{\infty }}{lim}{\int }_0^{-\mathrm{\infty }}{e}^u\left(\frac{-1}{a}\right)\mathrm{d}u\\ & & \dots u=-ax,\frac{\mathrm{d}u}{\mathrm{d}x}=-a,\mathrm{d}x=\frac{-1}{a}\mathrm{d}u\\ & & \dots x:0\to n,u:0\to -an\\ & =& \underset{n\to \mathrm{\infty }}{lim}\frac{-1}{a}{\int }_0^{-an}{e}^u\mathrm{d}u\\ & =& \underset{n\to \mathrm{\infty }}{lim}\frac{-1}{a}{\left[e^u\right]}_0^{-an}\\ & =& \underset{n\to \mathrm{\infty }}{lim}\frac{-1}{a}[e^{-an}-e^0]\\ & =& \frac{-1}{a}[0-1]\\ & =& \frac{-1}{a}\cdot (-1)\\ & =& \frac{1}{a}\end{array}$$

cos(z)の微分

$\cos \left(z\right)$の微分

$u+iv$で表す

$$\begin{array}{rcl}\cos \left(z\right)& =& \cos \left(x\right)\cos \left(iy\right)-\sin \left(x\right)\sin \left(iy\right)\\ & =& \cos \left(x\right)\cosh \left(y\right)-\sin \left(x\right)i\sinh \left(y\right)\\ & & \dots \cos \left(iy\right)=i\cosh \left(y\right),\sin \left(iy\right)=i\sinh \left(y\right)\\ & =& \cos \left(x\right)\cosh \left(y\right)-i\sin \left(x\right)\sinh \left(y\right)\\ & =& u(x,y)+iv(x,y)\end{array}$$ $$\{\begin{array}{rcl}u(x,y)& =& \cos \left(x\right)\cosh \left(y\right)\\ v(x,y)& =& -\sin \left(x\right)\sinh \left(y\right)\end{array}\dots x,y\in \mathbb{R}$$

$u,v$を$x,y$で偏微分する

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

コーシー・リーマンの関係式を満たす

$$\{\begin{array}{rcl}\frac{\partial u}{\partial x}& =& \frac{\partial v}{\partial y}\\ \frac{\partial v}{\partial x}& =& -\frac{\partial u}{\partial y}\end{array}$$

実軸方向での微分

$$\begin{array}{rcl}\frac{\mathrm{d}}{\mathrm{d}z}\cos \left(z\right)& =& \frac{\partial u(x,y)}{\partial x}+i\frac{\partial v(x,y)}{\partial x}\\ & =& -\sin \left(x\right)\cosh \left(y\right)+i(-\cos \left(x\right)\sinh \left(y\right))\\ & =& -\sin \left(x\right)\cosh \left(y\right)-i\cos \left(x\right)\sinh \left(y\right)\\ & =& -(\sin \left(x\right)\cos \left(iy\right)+\cos \left(x\right)\sin \left(iy\right))\\ & & \dots \cos \left(iy\right)=i\cosh \left(y\right),\sin \left(iy\right)=i\sinh \left(y\right)\\ & =& -\sin (x+iy)\\ & =& -\sin \left(z\right)\end{array}$$

cos(i x), sin(i x) (純虚数に対するcos, sin)

$\cos (ix),\sin (ix)$ (純虚数に対する$\cos ,\sin$)

$\cos \left(ix\right)$

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

$\sin \left(ix\right)$

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

$cosh(ix),sinh(ix)$ (純虚数に対する$\cosh ,\sinh$)

$cosh(ix),sinh(ix)$ (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

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

$1-z$のディガンマ凾数

$$\begin{array}{rcl}\psi (1-z)& =& \frac{\mathrm{d}}{\mathrm{d}z}\log \left(\mathrm{\Gamma }(1-z)\right)\\ & =& \frac{\mathrm{d}}{\mathrm{d}u}\log \left(\mathrm{\Gamma }\left(u\right)\right)\frac{\mathrm{d}u}{\mathrm{d}z}\dots u=1-z,\frac{\mathrm{d}u}{\mathrm{d}z}=-1\\ & =& -\frac{\mathrm{d}}{\mathrm{d}z}\log \left(\mathrm{\Gamma }(1-z)\right)\\ & =& -\frac{\mathrm{d}}{\mathrm{d}z}\log \left(\frac{\mathrm{\Gamma }\left(z\right)}{\mathrm{\Gamma }\left(z\right)}\mathrm{\Gamma }(1-z)\right)\\ & =& -\frac{\mathrm{d}}{\mathrm{d}z}\log \left(\frac{1}{\mathrm{\Gamma }\left(z\right)}\frac{\pi }{\sin \left(\pi z\right)}\right)\dots \mathrm{\Gamma }\left(z\right)\mathrm{\Gamma }(1-z)=\frac{\pi }{\sin \left(\pi z\right)}\\ & =& -\{\frac{\mathrm{d}}{\mathrm{d}z}\log \left(\pi \right)-\frac{\mathrm{d}}{\mathrm{d}z}\log (\sin \left(\pi z\right))-\frac{\mathrm{d}}{\mathrm{d}z}\log \left(\mathrm{\Gamma }\left(z\right)\right)\}\\ & =& -\{0-\pi \cot \left(\pi z\right)-\psi \left(z\right)\}\\ & & \dots \frac{\mathrm{d}}{\mathrm{d}z}\log \sin \left(\pi z\right)=\pi \cot \left(\pi z\right)\\ & & \dots \frac{\mathrm{d}}{\mathrm{d}z}\log \mathrm{\Gamma }\left(z\right)=\psi \left(z\right),(\mathrm{デ}\mathrm{ィ}\mathrm{ガ}\mathrm{ン}\mathrm{マ}\mathrm{凾}\mathrm{数}\mathrm{の}\mathrm{定}\mathrm{義})\\ & =& \pi \cot \left(\pi z\right)+\psi \left(z\right)\end{array}$$

ディガンマ凾数の相反公式

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

log(sin(πz))の微分

$\log (\sin \left(\pi z\right))$の微分

$$\begin{array}{rcl}\frac{\mathrm{d}}{\mathrm{d}z}\log (\sin \left(\pi z\right))& =& \frac{\mathrm{d}}{\mathrm{d}f}\log (\sin \left(f\right))\frac{\mathrm{d}f}{\mathrm{d}z}\dots f=\pi z,f\in \mathbb{C}\\ & =& \frac{\mathrm{d}}{\mathrm{d}g}\log \left(g\right)\frac{\mathrm{d}g}{\mathrm{d}f}\frac{\mathrm{d}f}{\mathrm{d}z}\dots g=\sin \left(f\right),g\in \mathbb{C}\\ & =& \frac{1}{g}\cos \left(f\right)\pi \\ & & \dots \frac{\mathrm{d}}{\mathrm{d}g}\log \left(g\right)=\frac{1}{g}\\ & & \dots \frac{\mathrm{d}}{\mathrm{d}f}\sin \left(f\right)=\cos \left(f\right)\\ & & \dots \frac{\mathrm{d}}{\mathrm{d}z}\pi z=\pi \\ & =& \frac{1}{\sin \left(f\right)}\cos \left(f\right)\pi \\ & =& \frac{1}{\sin \left(\pi z\right)}\cos \left(\pi z\right)\pi \\ & =& \pi \frac{\cos \left(\pi z\right)}{\sin \left(\pi z\right)}\\ & =& \pi \frac{1}{\tan \left(\pi z\right)}\dots \tan \left(\pi z\right)=\frac{\sin \left(\pi z\right)}{\cos \left(\pi z\right)}\\ & =& \pi \cot \left(\pi z\right)\end{array}$$

sin(z)の微分

$\sin \left(z\right)$の微分

$u+iv$で表す

$$\begin{array}{rcl}\sin \left(z\right)& =& \sin \left(x\right)\cos \left(iy\right)+\cos \left(x\right)\sin \left(iy\right)\\ & =& \sin \left(x\right)\cosh \left(y\right)+\cos \left(x\right)i\sinh \left(y\right)\\ & & \dots \cos \left(iy\right)=i\cosh \left(y\right),\sin \left(iy\right)=i\sinh \left(y\right)\\ & =& \sin \left(x\right)\cosh \left(y\right)+i\cos \left(x\right)\sinh \left(y\right)\\ & =& u(x,y)+iv(x,y)\end{array}$$ $$\{\begin{array}{rcl}u(x,y)& =& \sin \left(x\right)\cosh \left(y\right)\\ v(x,y)& =& \cos \left(x\right)\sinh \left(y\right)\end{array}\dots x,y\in \mathbb{R}$$

$u,v$を$x,y$で偏微分する

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

コーシー・リーマンの関係式を満たす

$$\{\begin{array}{rcl}\frac{\partial u}{\partial x}& =& \frac{\partial v}{\partial y}\\ \frac{\partial v}{\partial x}& =& -\frac{\partial u}{\partial y}\end{array}$$

実軸方向での微分

$$\begin{array}{rcl}\frac{\mathrm{d}}{\mathrm{d}z}\sin \left(z\right)& =& \frac{\partial u(x,y)}{\partial x}+i\frac{\partial v(x,y)}{\partial x}\\ & =& \cos \left(x\right)\cosh \left(y\right)+i(-\sin \left(x\right)\sinh \left(y\right))\\ & =& \cos \left(x\right)\cosh \left(y\right)-i\sin \left(x\right)\sinh \left(y\right)\\ & =& \cos \left(x\right)\cos \left(iy\right)-\sin \left(x\right)\sin \left(iy\right)\\ & & \dots \cos \left(iy\right)=\cosh \left(y\right),\sin \left(iy\right)=i\sinh \left(y\right)\\ & =& \cos (x+iy)\\ & =& \cos \left(z\right)\end{array}$$

log(z)の微分

$\log \left(z\right)$の微分

$u+iv$で表す

$$\begin{array}{rcl}\log \left(z\right)& =& \log (x+iy)\dots z=x+iy,z\in \mathbb{C},x,y\in \mathbb{R}\\ & =& \log (|z|e^{i\arg \left(z\right)})\\ & & \dots z=|z|e^{i\arg \left(z\right)},|z|=|x+iy|=\sqrt{x^2+y^2}\mathrm{は}\mathrm{実}\mathrm{数},\arg \left(z\right)\mathrm{は}\mathrm{実}\mathrm{数}\mathrm{で}\mathrm{多}\mathrm{価}(\mathrm{集}\mathrm{合})\\ & & \dots \arg \left(z\right)=\mathrm{Arg}\left(z\right)+2n\pi ,n\in \mathbb{Z}\\ & & \dots -\pi <\mathrm{Arg}\left(z\right)\le \pi ,\mathrm{Arg}\left(z\right)\mathrm{は}\mathrm{実}\mathrm{数}\mathrm{で}\mathrm{一}\mathrm{価}\\ & =& \log (|z|)+\log \left(e^{i\arg \left(z\right)}\right)\\ & =& \log (|z|)+\log \left(e^{i(\mathrm{Arg}\left(z\right)+2n\pi )}\right)\\ & =& \log (|z|)+i(\mathrm{Arg}\left(z\right)+2n\pi )\\ & =& u(x,y)+iv(x,y)\dots v(x,y)\mathrm{は}\mathrm{実}\mathrm{数}\mathrm{で}\mathrm{多}\mathrm{価}(\mathrm{集}\mathrm{合})\end{array}$$ $$\{\begin{array}{rcl}u(x,y)& =& \log \left(\sqrt{x^2+y^2}\right)\\ v(x,y)& =& \mathrm{Arg}\left(z\right)+2n\pi =\{\begin{array}{lll}\arctan \left(\frac{y}{x}\right)& +2n\pi & (x>0)\\ \arctan \left(\frac{y}{x}\right)+\pi & +2n\pi & (x<0\mathrm{か}\mathrm{つ}y\ge 0)\\ \arctan \left(\frac{y}{x}\right)-\pi & +2n\pi & (x<0\mathrm{か}\mathrm{つ}y<0)\\ \frac{\pi }{2}& +2n\pi & (x=0\mathrm{か}\mathrm{つ}y>0)\\ -\frac{\pi }{2}& +2n\pi & (x=0\mathrm{か}\mathrm{つ}y<0)\\ \mathrm{identerminate}& & (x=0\mathrm{か}\mathrm{つ}y=0)\end{array}\end{array}\dots x,y\in \mathbb{R},n\in \mathbb{Z}$$

$u,v$を$x,y$で偏微分する

$$\begin{array}{rcl}\frac{\partial u(x,y)}{\partial x}& =& \frac{\partial }{\partial x}\log \left(\sqrt{x^2+y^2}\right)\\ & =& \frac{\partial }{\partial f}\log \left(f\right)\frac{\partial f}{\partial x}\dots f=\sqrt{x^2+y^2}={(x^2+y^2)}^{\frac{1}{2}},f\in \mathbb{R}\\ & =& \frac{\partial }{\partial f}\log \left(f\right)\frac{\partial f}{\partial g}\frac{\partial g}{\partial x}\dots g=x^2+y^2,g\in \mathbb{R}\\ & =& \frac{1}{f}\frac{1}{2}{\left(g\right)}^{-\frac{1}{2}}2x\\ & =& \frac{1}{\sqrt{x^2+y^2}}\frac{1}{2\sqrt{x^2+y^2}}2x\\ & =& \frac{x}{x^2+y^2}\end{array}$$ $$\begin{array}{rcl}\frac{\partial u(x,y)}{\partial y}& =& \frac{\partial }{\partial y}\log \left(\sqrt{x^2+y^2}\right)\\ & =& \frac{\partial }{\partial f}\log \left(f\right)\frac{\partial f}{\partial y}\dots f=\sqrt{x^2+y^2}={(x^2+y^2)}^{\frac{1}{2}},f\in \mathbb{R}\\ & =& \frac{\partial }{\partial f}\log \left(f\right)\frac{\partial f}{\partial g}\frac{\partial g}{\partial y}\dots g=x^2+y^2,g\in \mathbb{R}\\ & =& \frac{1}{f}\frac{1}{2}{\left(g\right)}^{-\frac{1}{2}}2y\\ & =& \frac{1}{\sqrt{x^2+y^2}}\frac{1}{2\sqrt{x^2+y^2}}2y\\ & =& \frac{y}{x^2+y^2}\end{array}$$ $$\begin{array}{rcl}\frac{\partial v(x,y)}{\partial x}& =& \frac{\partial }{\partial x}\mathrm{Arg}(x+iy)+2n\pi \\ & =& \{\begin{array}{ll}\frac{\partial }{\partial x}(\arctan \left(\frac{y}{x}\right)+2n\pi )& (x>0)\\ \frac{\partial }{\partial x}(\arctan \left(\frac{y}{x}\right)+\pi +2n\pi )& (x<0\mathrm{か}\mathrm{つ}y\ge 0)\\ \frac{\partial }{\partial x}(\arctan \left(\frac{y}{x}\right)-\pi +2n\pi )& (x<0\mathrm{か}\mathrm{つ}y<0)\\ \frac{\partial }{\partial x}(\frac{\pi }{2}+2n\pi )& (x=0\mathrm{か}\mathrm{つ}y>0)\\ \frac{\partial }{\partial x}(-\frac{\pi }{2}+2n\pi )& (x=0\mathrm{か}\mathrm{つ}y<0)\\ \mathrm{identerminate}& (x=0\mathrm{か}\mathrm{つ}y=0)\end{array}\\ & =& \{\begin{array}{ll}\frac{-y}{x^2+y^2}& (x>0)\\ \frac{-y}{x^2+y^2}& (x<0\mathrm{か}\mathrm{つ}y\ge 0)\\ \frac{-y}{x^2+y^2}& (x<0\mathrm{か}\mathrm{つ}y<0)\\ 0& (x=0\mathrm{か}\mathrm{つ}y>0)\\ 0& (x=0\mathrm{か}\mathrm{つ}y<0)\\ \mathrm{identerminate}& (x=0\mathrm{か}\mathrm{つ}y=0)\end{array}\end{array}$$ $$\begin{array}{rcl}\frac{\partial v(x,y)}{\partial y}& =& \frac{\partial }{\partial y}\mathrm{Arg}(x+iy)+2n\pi \\ & =& \{\begin{array}{ll}\frac{\partial }{\partial y}(\arctan \left(\frac{y}{x}\right)+2n\pi )& (x>0)\\ \frac{\partial }{\partial y}(\arctan \left(\frac{y}{x}\right)+\pi +2n\pi )& (x<0\mathrm{か}\mathrm{つ}y\ge 0)\\ \frac{\partial }{\partial y}(\arctan \left(\frac{y}{x}\right)-\pi +2n\pi )& (x<0\mathrm{か}\mathrm{つ}y<0)\\ \frac{\partial }{\partial y}(\frac{\pi }{2}+2n\pi )& (x=0\mathrm{か}\mathrm{つ}y>0)\\ \frac{\partial }{\partial y}(-\frac{\pi }{2}+2n\pi )& (x=0\mathrm{か}\mathrm{つ}y<0)\\ \mathrm{identerminate}& (x=0\mathrm{か}\mathrm{つ}y=0)\end{array}\\ & =& \{\begin{array}{ll}\frac{x}{x^2+y^2}& (x>0)\\ \frac{x}{x^2+y^2}& (x<0\mathrm{か}\mathrm{つ}y\ge 0)\\ \frac{x}{x^2+y^2}& (x<0\mathrm{か}\mathrm{つ}y<0)\\ 0& (x=0\mathrm{か}\mathrm{つ}y>0)\\ 0& (x=0\mathrm{か}\mathrm{つ}y<0)\\ \mathrm{identerminate}& (x=0\mathrm{か}\mathrm{つ}y=0)\end{array}\end{array}$$

コーシー・リーマンの関係式を満たす

$$\{\begin{array}{rcl}\frac{\partial u}{\partial x}& =& \frac{\partial v}{\partial y}\\ \frac{\partial v}{\partial x}& =& -\frac{\partial u}{\partial y}\end{array}$$

実軸方向での微分

$$\begin{array}{rcl}\frac{\mathrm{d}}{\mathrm{d}z}\log \left(z\right)& =& \frac{\partial u(x,y)}{\partial x}+i\frac{\partial v(x,y)}{\partial x}\\ & =& \frac{x}{x^2+y^2}+i\frac{-y}{x^2+y^2}\\ & =& \frac{x-iy}{x^2+y^2}\\ & =& \frac{\cancel{x-iy}}{(x+iy)\cancel{(x-iy)}}\\ & & \dots (x+iy)(x-iy)=x^2\cancel{+ixy}\cancel{-ixy}-i^2{y}^2=x^2+y^2\\ & =& \frac{1}{x+iy}\\ & =& \frac{1}{z}\dots z=x+iy\end{array}$$

虚軸方向での微分

$$\begin{array}{rcl}\frac{\mathrm{d}}{\mathrm{d}z}\log \left(z\right)& =& \frac{\partial v(x,y)}{\partial y}-i\frac{\partial u(x,y)}{\partial y}\\ & =& \frac{x}{x^2+y^2}-i\frac{y}{x^2+y^2}\\ & =& \frac{x-iy}{x^2+y^2}\\ & =& \frac{\cancel{x-iy}}{(x+iy)\cancel{(x-iy)}}\\ & & \dots (x+iy)(x-iy)=x^2\cancel{+ixy}\cancel{-ixy}-i^2{y}^2=x^2+y^2\\ & =& \frac{1}{x+iy}\\ & =& \frac{1}{z}\dots z=x+iy\end{array}$$

cot(az)の微分

$cot\left(az\right)$の微分

$$\begin{array}{rcl}\frac{\mathrm{d}}{\mathrm{d}z}\cot \left(az\right)& =& \frac{\mathrm{d}}{\mathrm{d}w}\cot \left(w\right)\frac{\mathrm{d}w}{\mathrm{d}z}\\ & & \dots a\in \mathbb{R},z\in \mathbb{C}\\ & & \dots w=az,\frac{\mathrm{d}w}{\mathrm{d}z}=a\\ & =& \frac{-1}{{\sin }^2\left(az\right)}\cdot a\dots \frac{\mathrm{d}}{\mathrm{d}w}\cot \left(w\right)=\frac{-1}{{\sin }^2\left(w\right)}\\ & =& -\frac{a}{{\sin }^2\left(az\right)}\end{array}$$

cot(ax)の微分

$\cot \left(ax\right)$の微分

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

cot(z)の微分

$u(x,y)$を$x$で偏微分する

$$\begin{array}{rcl}\frac{\partial u}{\partial x}& =& \frac{\partial }{\partial x}\frac{\cos \left(x\right)\sin \left(x\right)}{{\sin }^2\left(x\right)+{\sinh }^2\left(y\right)}\\ & =& \frac{{\sinh }^2\left(y\right)({\cos }^2\left(x\right)-{\sin }^2\left(x\right))-{\sin }^2\left(x\right)}{{({\sin }^2\left(x\right)+{\sinh }^2\left(y\right))}^2}\\ & =& \frac{{\cos }^2\left(x\right){\sinh }^2\left(y\right)-{\sinh }^2\left(y\right){\sin }^2\left(x\right)-{\sin }^2\left(x\right)}{{({\sin }^2\left(x\right)+{\sinh }^2\left(y\right))}^2}\\ & =& \frac{{\cos }^2\left(x\right){\sinh }^2\left(y\right)-{\sin }^2\left(x\right)(1+{\sinh }^2\left(y\right))}{{({\sin }^2\left(x\right)+{\sinh }^2\left(y\right))}^2}\\ & =& \frac{{\cos }^2\left(x\right){\sinh }^2\left(y\right)-{\sin }^2\left(x\right){\cosh }^2\left(y\right)}{{({\sin }^2\left(x\right)+{\sinh }^2\left(y\right))}^2}\end{array}$$

$u(x,y)$を$y$で偏微分する

$$\begin{array}{rcl}\frac{\partial u}{\partial y}& =& \frac{\partial }{\partial y}\frac{\cos \left(x\right)\sin \left(x\right)}{{\sin }^2\left(x\right)+{\sinh }^2\left(y\right)}\\ & =& -\frac{2\sin \left(x\right)\cos \left(x\right)\sinh \left(y\right)\cosh \left(y\right)}{{({\sin }^2\left(x\right)+{\sinh }^2\left(y\right))}^2}\end{array}$$

$v(x,y)$を$x$で偏微分する

$$\begin{array}{rcl}\frac{\partial v}{\partial x}& =& \frac{\partial }{\partial x}\frac{-\cosh \left(y\right)\sinh \left(y\right)}{{\sin }^2\left(x\right)+{\sinh }^2\left(y\right)}\\ & =& \frac{2\sin \left(x\right)\cos \left(x\right)\sinh \left(y\right)\cosh \left(y\right)}{{({\sin }^2\left(x\right)+{\sinh }^2\left(y\right))}^2}\end{array}$$

$v(x,y)$を$y$で偏微分する

$$\begin{array}{rcl}\frac{\partial v}{\partial y}& =& \frac{\partial }{\partial y}\frac{-\cosh \left(y\right)\sinh \left(y\right)}{{\sin }^2\left(x\right)+{\sinh }^2\left(y\right)}\\ & =& \frac{{\cosh }^2\left(y\right)({\sinh }^2\left(y\right)-{\sin }^2\left(x\right))-{\sinh }^2\left(y\right)({\sinh }^2\left(y\right)+{\sin }^2\left(x\right))}{{({\sin }^2\left(x\right)+{\sinh }^2\left(y\right))}^2}\\ & =& \frac{{\cosh }^2\left(y\right){\sinh }^2\left(y\right)-{\cosh }^2\left(y\right){\sin }^2\left(x\right)-{\sinh }^2\left(y\right){\sinh }^2\left(y\right)-{\sinh }^2\left(y\right){\sin }^2\left(x\right)}{{({\sin }^2\left(x\right)+{\sinh }^2\left(y\right))}^2}\\ & =& \frac{{\sinh }^2\left(y\right)({\cosh }^2\left(y\right)-{\sinh }^2\left(y\right))-{\cosh }^2\left(y\right){\sin }^2\left(x\right)-{\sinh }^2\left(y\right){\sin }^2\left(x\right)}{{({\sin }^2\left(x\right)+{\sinh }^2\left(y\right))}^2}\\ & =& \frac{{\sinh }^2\left(y\right)-{\cosh }^2\left(y\right){\sin }^2\left(x\right)-{\sinh }^2\left(y\right){\sin }^2\left(x\right)}{{({\sin }^2\left(x\right)+{\sinh }^2\left(y\right))}^2}\\ & =& \frac{{\sinh }^2\left(y\right)(1-{\sin }^2\left(x\right))-{\cosh }^2\left(y\right){\sin }^2\left(x\right)}{{({\sin }^2\left(x\right)+{\sinh }^2\left(y\right))}^2}\\ & =& \frac{{\cos }^2\left(x\right){\sinh }^2\left(y\right)-{\sin }^2\left(x\right){\cosh }^2\left(y\right)}{{({\sin }^2\left(x\right)+{\sinh }^2\left(y\right))}^2}\end{array}$$

コーシー・リーマンの関係式を満たす

$$\{\begin{array}{rcl}\frac{\partial u}{\partial x}& =& \frac{\partial v}{\partial y}\\ \frac{\partial u}{\partial y}& =& -\frac{\partial v}{\partial x}\end{array}$$

実軸(x)方向の微分

$$\begin{array}{rcl}\frac{\mathrm{d}}{\mathrm{d}z}\cot & =& \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)}{{({\sin }^2\left(x\right)+{\sinh }^2\left(y\right))}^2}+i\frac{2\sin \left(x\right)\cos \left(x\right)\sinh \left(y\right)\cosh \left(y\right)}{{({\sin }^2\left(x\right)+{\sinh }^2\left(y\right))}^2}\\ & =& \frac{{(\cos \left(x\right)\sinh \left(y\right)+i\sin \left(x\right)\cosh \left(y\right))}^2}{{({\sin }^2\left(x\right)+{\sinh }^2\left(y\right))}^2}\\ & =& \frac{{(\cos \left(x\right)\sinh \left(y\right)+i\sin \left(x\right)\cosh \left(y\right))}^2}{{({\sin }^2\left(x\right)+{\sinh }^2\left(y\right))}^2}\\ & =& \frac{{(\cos \left(x\right)\sinh \left(y\right)+i\sin \left(x\right)\cosh \left(y\right))}^2}{{({\sin }^2\left(x\right)+{\sinh }^2\left(y\right))}^2}\\ & =& \frac{{(\cos \left(x\right)\frac{1}{i}\sin \left(iy\right)+i\sin \left(x\right)\cos \left(iy\right))}^2}{{({\sin }^2\left(x\right)+{(\frac{1}{i}\sin \left(iy\right))}^2)}^2}\dots \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{{(\cos \left(x\right)\frac{1}{i}\sin \left(iy\right)+i\sin \left(x\right)\cos \left(iy\right))}^2}{{({\sin }^2\left(x\right)+{\left(\frac{1}{i}\right)}^2{(\sin \left(iy\right))}^2)}^2}\\ & =& \frac{1}{i^2}\frac{{\left(i(\cos \left(x\right)\frac{1}{i}\sin \left(iy\right)+i\sin \left(x\right)\cos \left(iy\right))\right)}^2}{{({\sin }^2\left(x\right)-{\sin }^2\left(iy\right))}^2}\\ & =& \frac{1}{-1}\frac{{(\cos \left(x\right)\sin \left(iy\right)-\sin \left(x\right)\cos \left(iy\right))}^2}{{({\sin }^2\left(x\right)-{\sin }^2\left(iy\right))}^2}\\ & =& -\frac{{(-(-\cos \left(x\right)\sin \left(iy\right)+\sin \left(x\right)\cos \left(iy\right)))}^2}{{({\sin }^2\left(x\right)-{\sin }^2\left(iy\right))}^2}\\ & =& -\frac{{(-\sin (x-iy))}^2}{{(\sin (x+iy)\sin (x-iy))}^2}\\ & & \dots {\sin }^2\left(x\right)-{\sin }^2\left(iy\right)\\ & & \dots ={\sin }^2\left(x\right)-{\sin }^2\left(iy\right)-{\sin }^2\left(x\right){\sin }^2\left(iy\right)+{\sin }^2\left(x\right){\sin }^2\left(iy\right)\\ & & \dots ={\sin }^2\left(x\right)(1-{\sin }^2\left(iy\right))-{\sin }^2\left(iy\right)(1-{\sin }^2\left(x\right))\\ & & \dots ={\sin }^2\left(x\right){\cos }^2\left(iy\right)-{\sin }^2\left(iy\right){\cos }^2\left(x\right)\\ & & \dots =\{\sin \left(x\right)\cos \left(iy\right)+\sin \left(iy\right)\cos \left(x\right)\}\{\sin \left(x\right)\cos \left(iy\right)-\sin \left(iy\right)\cos \left(x\right)\}\\ & & \dots =\sin (x+iy)\sin (x-iy)\\ & =& -\frac{\cancel{{\sin }^2(x-iy)}}{{\sin }^2(x+iy)\cancel{{\sin }^2(x-iy)}}\\ & =& \frac{-1}{{\sin }^2(x+iy)}\\ & =& \frac{-1}{{\sin }^2\left(z\right)}\end{array}$$

cot(z)をu(x,y)+iv(x,y)で表す

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

wzの微分

$wz$の微分

$u+iv$で表す

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

$u,v$を$x,y$で偏微分する

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

コーシー・リーマンの関係式を満たす

$$\{\begin{array}{rcl}\frac{\partial u}{\partial x}& =& \frac{\partial v}{\partial y}\\ \frac{\partial u}{\partial y}& =& -\frac{\partial v}{\partial x}\end{array}$$

実軸(x)方向の微分

$$\begin{array}{rcl}\frac{\mathrm{d}}{\mathrm{d}z}wz& =& \frac{\partial u(x,y)}{\partial x}+i\frac{\partial v(x,y)}{\partial x}\\ & =& a+ib\\ & =& w\end{array}$$

虚軸(y)方向の微分

$$\begin{array}{rcl}\frac{\mathrm{d}}{\mathrm{d}z}wz& =& \frac{\partial v(x,y)}{\partial y}-i\frac{\partial u(x,y)}{\partial y}\\ & =& a-i(-b)\\ & =& a+ib\\ & =& w\end{array}$$

azの微分

$az$の微分

$u+iv$で表す

$$\begin{array}{rcl}az& =& a(x+iy)\dots a,x,y\in \mathbb{R},z\in \mathbb{C}\\ & =& ax+iay\\ & =& u(x,y)+iv(x,y)\end{array}$$ $$\{\begin{array}{rcl}u(x,y)& =& ax\\ v(x,y)& =& ay\end{array}$$

$u,v$を$x,y$で偏微分する

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

コーシー・リーマンの関係式を満たす

$$\{\begin{array}{rcl}\frac{\partial u}{\partial x}& =& \frac{\partial v}{\partial y}\\ \frac{\partial u}{\partial y}& =& -\frac{\partial v}{\partial x}\end{array}$$

実軸方向での微分

$$\begin{array}{rcl}\frac{\mathrm{d}}{\mathrm{d}z}az& =& \frac{\partial u(x,y)}{\partial x}+i\frac{\partial v(x,y)}{\partial x}\\ & =& a+i0\\ & =& a\end{array}$$

虚軸方向での微分

$$\begin{array}{rcl}\frac{\mathrm{d}}{\mathrm{d}z}az& =& \frac{\partial v(x,y)}{\partial y}-i\frac{\partial u(x,y)}{\partial y}\\ & =& a-i0\\ & =& a\end{array}$$

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)

$$\begin{array}{rcl}\underset{\mathrm{\Delta }z\to 0}{lim}\frac{f(z_0+\mathrm{\Delta }z)-f(z_0)}{\mathrm{\Delta }z}& =& \underset{\mathrm{\Delta }x\to 0}{lim}\frac{\{u(x_0+\mathrm{\Delta }x,y_0)+iv(x_0+\mathrm{\Delta }x,y_0)\}-\{u(x_0,y_0)+iv(x_0,y_0)\}}{\mathrm{\Delta }x}\dots z_0,\mathrm{\Delta }zin\mathbb{C},x_0,y_0,\mathrm{\Delta }x\in \mathbb{R}\\ & =& \underset{\mathrm{\Delta }x\to 0}{lim}\{\frac{u(x_0+\mathrm{\Delta }x,y_0)-u(x_0,y_0)}{\mathrm{\Delta }x}+i\frac{v(x_0+\mathrm{\Delta }x,y_0)-v(x_0,y_0)}{\mathrm{\Delta }x}\}\\ & =& \frac{\partial u(x_0,y_0)}{\partial x}+i\frac{\partial v(x_0,y_0)}{\partial x}\end{array}$$

複素平面の虚軸方向の微分(偏微分)

$$\begin{array}{rcl}\underset{\mathrm{\Delta }z\to 0}{lim}\frac{f(z_0+\mathrm{\Delta }z)-f(z_0)}{\mathrm{\Delta }z}& =& \underset{\mathrm{\Delta }y\to 0}{lim}\frac{\{u(x_0,y_0+\mathrm{\Delta }y)+iv(x_0,y_0+\mathrm{\Delta }y)\}-\{u(x_0,y_0)+iv(x_0,y_0)\}}{i\mathrm{\Delta }y}\dots z_0,\mathrm{\Delta }zin\mathbb{C},x_0,y_0,\mathrm{\Delta }y\in \mathbb{R}\\ & =& \underset{\mathrm{\Delta }y\to 0}{lim}\{\frac{u(x_0,y_0+\mathrm{\Delta }y)-u(x_0,y_0)}{i\mathrm{\Delta }y}+i\frac{v(x_0,y_0+\mathrm{\Delta }y)-v(x_0,y_0)}{i\mathrm{\Delta }y}\}\\ & =& \frac{1}{i}\frac{\partial u(x_0,y_0)}{\partial y}+\frac{i}{i}\frac{\partial v(x_0,y_0)}{\partial y}\\ & =& \frac{i}{i}\frac{1}{i}\frac{\partial u(x_0,y_0)}{\partial y}+\frac{\partial v(x_0,y_0)}{\partial y}\\ & =& \frac{i}{-1}\frac{\partial u(x_0,y_0)}{\partial y}+\frac{\partial v(x_0,y_0)}{\partial y}\\ & =& -i\frac{\partial u(x_0,y_0)}{\partial y}+\frac{\partial v(x_0,y_0)}{\partial y}\\ & =& \frac{\partial v(x_0,y_0)}{\partial y}+i\{-\frac{\partial u(x_0,y_0)}{\partial y}\}\end{array}$$

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

$$\{\begin{array}{rcl}\frac{\partial u}{\partial x}& =& \frac{\partial v}{\partial y}\\ \frac{\partial v}{\partial x}& =& -\frac{\partial u}{\partial y}\end{array}$$

ディガンマ凾数同士の差

ディガンマ凾数同士の差

$$\begin{array}{rcl}\psi \left(y\right)-\psi \left(x\right)& =& \underset{n\to \mathrm{\infty }}{lim}\{\log \left(n\right)-\sum _{k=0}^n\frac{1}{y+k}\}-\underset{n\to \mathrm{\infty }}{lim}\{\log \left(n\right)-\sum _{k=0}^n\frac{1}{x+k}\}\dots \psi \left(x\right)=\underset{n\to \mathrm{\infty }}{lim}\{\log \left(n\right)-\sum _{k=0}^n\frac{1}{x+k}\}\\ & =& \underset{n\to \mathrm{\infty }}{lim}\{\log \left(n\right)-\sum _{k=0}^n\frac{1}{y+k}-\log \left(n\right)+\sum _{k=0}^n\frac{1}{x+k}\}\\ & =& \underset{n\to \mathrm{\infty }}{lim}\{\sum _{k=0}^n\frac{1}{x+k}-\sum _{k=0}^n\frac{1}{y+k}\}\\ & =& \sum _{k=0}^{\mathrm{\infty }}\frac{1}{x+k}-\sum _{k=0}^{\mathrm{\infty }}\frac{1}{y+k}\\ & =& {\int }_0^1\frac{u^{x-1}}{1-u}\mathrm{d}u-{\int }_0^1\frac{u^{y-1}}{1-u}\mathrm{d}u\\ & & \dots \sum _{k=0}^{\mathrm{\infty }}\frac{1}{z+k}={\int }_0^1\frac{u^{z-1}}{1-u}\mathrm{d}u\\ & =& {\int }_0^1(\frac{u^{x-1}}{1-u}-\frac{u^{y-1}}{1-u})\mathrm{d}u\\ & =& {\int }_0^1\frac{u^{x-1}-u^{y-1}}{1-u}\mathrm{d}u\end{array}$$

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

$$\begin{array}{rcl}{\int }_0^1\frac{u^{z-1}}{1-u}\mathrm{d}u& =& {\int }_0^1{u}^{z-1}\frac{1}{1-u}\mathrm{d}u\\ & =& {\int }_0^1{u}^{z-1}(1+u+u^2+\cdots )\mathrm{d}u\\ & & \dots \frac{1}{1-u}=1+u+u^2+\cdots (|u|<1)\\ & =& {\int }_0^1(u^{z-1}+u^{z-1}u+u^{z-1}{u}^2+\cdots )\mathrm{d}u\\ & =& {\int }_0^1(u^{z-1}+u^{z-1+1}+u^{z-1+2}+\cdots )\mathrm{d}u\\ & =& {\int }_0^1(u^{z-1}+u^z+u^{z+1}+\cdots )\mathrm{d}u\\ & =& {[\frac{1}{z-1+1}{u}^{z-1+1}+\frac{1}{z+1}{u}^{z+1}+\frac{1}{z+1+1}{u}^{z+1+1}+\cdots ]}_0^1\\ & =& {[\frac{1}{z}{u}^z+\frac{1}{z+1}{u}^{z+1}+\frac{1}{z+2}{u}^{z+2}+\cdots ]}_0^1\\ & =& [\frac{1}{z}{1}^z-\frac{1}{z}{0}^z+\frac{1}{z+1}{1}^{z+1}-\frac{1}{z+1}{0}^{z+1}+\frac{1}{z+2}{1}^{z+2}-\frac{1}{z+2}{0}^{z+2}+\cdots ]\\ & =& \frac{1}{z}+\frac{1}{z+1}+\frac{1}{z+2}+\cdots \\ & =& \sum _{k=0}^{\mathrm{\infty }}\frac{1}{z+k}\end{array}$$

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($\mathrm{た}\mathrm{だ}\mathrm{し}-1<x<1$)

a点まわりのテイラー展開

$$\begin{array}{rcl}f(x)& =& \sum _{k=0}^{\mathrm{\infty }}\frac{f^{(k)}(a)}{x!}(x-a{)}^k\dots a\mathrm{点}\mathrm{ま}\mathrm{わ}\mathrm{り}\mathrm{の}\mathrm{テ}\mathrm{イ}\mathrm{ラ}\mathrm{ー}\mathrm{展}\mathrm{開}\\ & =& \frac{1}{0!}{f}^{(0)}(a)(x-a{)}^0+\frac{1}{1!}{f}^{(1)}(a)(x-a{)}^1+\frac{1}{2!}{f}^{(2)}(a)(x-a{)}^2+\cdots \\ & & \dots f^{(n)}(x):f(x)\mathrm{の}n\mathrm{階}\mathrm{微}\mathrm{分}\end{array}$$

マクローリン展開(0点まわりのテイラー展開)

$$\begin{array}{rcl}f(x)& =& {\sum _{k=0}^{\mathrm{\infty }}\frac{f^{(k)}(a)}{x!}(x-a{)}^k|}_{a=0}\\ & =& \sum _{k=0}^{\mathrm{\infty }}\frac{f^{(k)}(0)}{x!}(x-0{)}^k\\ & =& \sum _{k=0}^{\mathrm{\infty }}\frac{f^{(k)}(0)}{x!}(x{)}^k\\ & =& \frac{1}{0!}{f}^{(0)}(0)x^0+\frac{1}{1!}{f}^{(1)}(0)x+\frac{1}{2!}{f}^{(2)}(0)x^2+\cdots \end{array}$$

マクローリン展開(0点まわりのテイラー展開)

$$\begin{array}{rcl}\frac{1}{1-x}& =& \sum _{k=0}^{\mathrm{\infty }}\frac{1}{k!}{f}^{\left(k\right)}\left(0\right)x^k\dots (\mathrm{た}\mathrm{だ}\mathrm{し}-1<x<1)\\ & =& \frac{1}{0!}{\left[{(1-x)}^{-1}\right]}_{x=0}{x}^0+\frac{1}{1!}{[-{(1-x)}^{-1-1}(-1)]}_{x=0}{x}^1+\frac{1}{2!}{[-2{(1-x)}^{-2-1}(-1)]}_{x=0}{x}^2+\frac{1}{3!}{[-6{(1-x)}^{-3-1}(-1)]}_{x=0}{x}^3+\cdots \\ & =& 1+x+x^2+x^3+\cdots \end{array}$$

ディガンマ凾数の極限表示

ディガンマ凾数

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

ディガンマ凾数の極限表示

$$\begin{array}{rcl}\mathrm{\Gamma }\left(z\right)& =& \underset{n\to \mathrm{\infty }}{lim}\frac{n^{z}n!}{\prod _{k=0}^n(z+k)}\\ \log \left(\mathrm{\Gamma }\left(z\right)\right)& =& \log \left(\underset{n\to \mathrm{\infty }}{lim}\frac{n^{z}n!}{\prod _{k=0}^n(z+k)}\right)\\ & =& \underset{n\to \mathrm{\infty }}{lim}\log \left(\frac{n^{z}n!}{\prod _{k=0}^n(z+k)}\right)\\ & =& \underset{n\to \mathrm{\infty }}{lim}\{\log \left(n^z\right)+\log (n!)-\log \left(\prod _{k=0}^n(z+k)\right)\}\\ & =& \underset{n\to \mathrm{\infty }}{lim}\{\log \left(n^z\right)+\log (n!)-\log (z+0)-\log (z+1)-\log (z+2)-\cdots -\log (z+n)\}\\ \psi \left(z\right)=\frac{\mathrm{d}}{\mathrm{d}z}\log \left(\mathrm{\Gamma }\left(z\right)\right)& =& \frac{\mathrm{d}}{\mathrm{d}z}\left[\underset{n\to \mathrm{\infty }}{lim}\{\log \left(n^z\right)+\log (n!)-\log (z+0)-\log (z+1)-\log (z+2)-\cdots -\log (z+n)\}\right]\\ & =& \underset{n\to \mathrm{\infty }}{lim}\left[\frac{\mathrm{d}}{\mathrm{d}z}\{\log \left(n^z\right)+\log (n!)-\log (z+0)-\log (z+1)-\log (z+2)-\cdots -\log (z+n)\}\right]\\ & =& \underset{n\to \mathrm{\infty }}{lim}\{\log \left(n\right)-\frac{1}{z}-\frac{1}{z+1}-\frac{1}{z+2}-\cdots -\frac{1}{z+n}\}\\ & =& \underset{n\to \mathrm{\infty }}{lim}[\log \left(n\right)-\{\frac{1}{z}+\frac{1}{z+1}+\frac{1}{z+2}+\cdots +\frac{1}{z+n}\}]\\ & =& \underset{n\to \mathrm{\infty }}{lim}\{\log \left(n\right)-\sum _{k=0}^n\frac{1}{z+k}\}\end{array}$$

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

$$\begin{array}{rcl}\mathrm{\Gamma }\left(z\right)& =& {\int }_0^{\mathrm{\infty }}{t}^{z-1}{e}^{-t}\mathrm{d}t(\mathrm{Re}\left(z\right)>0)\dots \mathrm{定}\mathrm{義}\\ & =& \underset{n\to \mathrm{\infty }}{lim}\frac{n^{z}n!}{\prod _{k=0}^n(z+k)}\dots \mathrm{極}\mathrm{限}\mathrm{表}\mathrm{示}\end{array}$$

$G_n$の導入

$$\begin{array}{rcl}{G}_n\left(z\right)& =& {\int }_0^n{t}^{z-1}{(1-\frac{t}{n})}^n\mathrm{d}t\end{array}$$

$G_n$の総乗表示

$$\begin{array}{rcl}{G}_n\left(z\right)& =& {\int }_0^n{t}^{z-1}{(1-\frac{t}{n})}^n\mathrm{d}t\\ & =& {\int }_0^1{\left(nu\right)}^{z-1}{(1-\frac{nu}{n})}^{n}n\mathrm{d}u\\ & & \dots t=nu,u=\frac{t}{n},\frac{\mathrm{d}t}{\mathrm{d}u}=n,\mathrm{d}t=n\mathrm{d}u\\ & & \dots t:0\to n,u:0\to 1\\ & =& n^z{\int }_0^1{u}^{z-1}{(1-u)}^n\mathrm{d}u\\ & =& n^z\cdot g_n\left(z\right)\dots g_n\left(z\right)={\int }_0^1{u}^{z-1}{(1-u)}^n\mathrm{d}u\\ g_0\left(z\right)& =& {\int }_0^1{u}^{z-1}{(1-u)}^0\mathrm{d}u\\ & =& {\int }_0^1{u}^{z-1}\mathrm{d}u\\ & =& {\left[\frac{u^z}{z}\right]}_{u=0}^1\\ & =& \frac{1}{z}\\ g_n\left(z\right)& =& {\int }_0^1{u}^{z-1}{(1-u)}^n\mathrm{d}u\\ & =& {\int }_0^1{\left(\frac{u^z}{z}\right)}^{\prime }{(1-u)}^n\mathrm{d}u\dots \frac{\mathrm{d}}{\mathrm{d}u}\frac{u^z}{z}=\frac{1}{z}z{u}^{z-1}=u^{z-1}\\ & =& {\left[\frac{u^z}{z}{(1-u)}^n\right]}_{u=0}^1-{\int }_0^1\frac{u^z}{z}{\left({(1-u)}^n\right)}^{\prime }\mathrm{d}u\\ & =& [\frac{1^z}{z}{(1-1)}^n-\frac{0^z}{z}{(1-0)}^n]-{\int }_0^1\frac{u^z}{z}(n{(1-u)}^{n-1}(-1))\mathrm{d}u\\ & =& 0+\frac{n}{z}{\int }_0^1{u}^z{(1-u)}^{n-1}\mathrm{d}u\\ & =& \frac{n}{z}{g}_{n-1}(z+1)\\ & =& \frac{n}{z}\frac{n-1}{z+1}{g}_{n-2}(z+2)\\ & =& \frac{n}{z}\frac{n-1}{z+1}\frac{n-2}{z+2}{g}_{n-3}(z+3)\\ & =& \frac{n}{z}\frac{n-1}{z+1}\frac{n-2}{z+2}\dots \frac{n-(n-1)}{z+(n-1)}{g}_{n-n}(z+n)\\ & =& \frac{n}{z}\frac{n-1}{z+1}\frac{n-2}{z+2}\dots \frac{1}{z+(n-1)}{g}_0(z+n)\\ & =& \frac{n}{z}\frac{n-1}{z+1}\frac{n-2}{z+2}\dots \frac{1}{z+(n-1)}\frac{1}{z+n}\\ & =& \frac{n!}{\prod _{k=0}^n(z+k)}\\ G_n\left(z\right)& =& n^z\cdot g_n\left(z\right)\\ & =& n^z\frac{n!}{\prod _{k=0}^n(z+k)}\\ & =& \frac{n^{z}n!}{\prod _{k=0}^n(z+k)}\end{array}$$

$G_n$の極限

$$\begin{array}{rcl}\underset{n\to \mathrm{\infty }}{lim}{G}_n\left(z\right)& =& \underset{n\to \mathrm{\infty }}{lim}{\int }_0^n{t}^{z-1}{(1-\frac{t}{n})}^n\mathrm{d}t\\ & =& {\int }_0^{\mathrm{\infty }}{t}^{z-1}{e}^{-t}\mathrm{d}t\\ & & \dots \underset{n\to \mathrm{\infty }}{lim}{(1-\frac{t}{n})}^n=\underset{n\to \mathrm{\infty }}{lim}{(1+\frac{-t}{n})}^n=e^{-t}\\ & =& \mathrm{\Gamma }\left(z\right)\end{array}$$

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

$$\begin{array}{rcl}\underset{n\to \mathrm{\infty }}{lim}{G}_n\left(z\right)& =& \underset{n\to \mathrm{\infty }}{lim}\frac{n^{z}n!}{\prod _{k=0}^n(z+k)}\\ & =& \mathrm{\Gamma }\left(z\right)\end{array}$$

偏角の微分

$$\begin{array}{rcl}z& =& x+iy\dots z\in \mathbb{Z},x,y\in \mathbb{R}\\ & =& r{e}^{i\theta }\dots r,\theta \in \mathbb{R},r\ge 0,-\pi <\theta \le \pi \\ r& =& |z|=\sqrt{x^2+y^2}\\ \arg \left(z\right)& =& \theta +2n\pi (n\in \mathbb{Z})\\ & =& \mathrm{Arg}\left(z\right)+2n\pi (n\in \mathbb{Z})\end{array}$$ $$\begin{array}{rcl}\mathrm{Arg}\left(z\right)& =& \mathrm{Arg}(x+iy)\\ & =& \{\begin{array}{ll}{\tan }^{-1}\left(\frac{y}{x}\right)& (x>0)\\ {\tan }^{-1}\left(\frac{y}{x}\right)+\pi & (x<0\mathrm{か}\mathrm{つ}y\ge 0)\\ {\tan }^{-1}\left(\frac{y}{x}\right)-\pi & (x<0\mathrm{か}\mathrm{つ}y<0)\\ \frac{\pi }{2}& (x=0\mathrm{か}\mathrm{つ}y>0)\\ -\frac{\pi }{2}& (x=0\mathrm{か}\mathrm{つ}y<0)\\ \mathrm{不}\mathrm{定}& (x=0\mathrm{か}\mathrm{つ}y=0)\end{array}\end{array}$$

xでの偏微分

$$\begin{array}{rcl}\frac{\partial }{\partial x}\arg \left(z\right)& =& \frac{\partial }{\partial x}\{\theta +2n\pi \}\\ & =& \frac{\partial }{\partial x}\{\mathrm{Arg}\left(z\right)+2n\pi \}\\ & =& \frac{\partial }{\partial x}\mathrm{Arg}\left(z\right)+\frac{\partial }{\partial x}2n\pi \\ & =& \frac{\partial }{\partial x}\mathrm{Arg}\left(z\right)\dots \frac{\partial }{\partial x}C=0(C\mathrm{は}\mathrm{定}\mathrm{数}．x\mathrm{の}\mathrm{凾}\mathrm{数}\mathrm{で}\mathrm{な}\mathrm{い})\end{array}$$ $$\begin{array}{rcl}\frac{\partial }{\partial x}\mathrm{Arg}\left(z\right)& =& \frac{\partial }{\partial x}\mathrm{Arg}(x+iy)\\ & =& \{\begin{array}{llll}\frac{\partial }{\partial x}{\tan }^{-1}\left(\frac{y}{x}\right)& =& \frac{-y}{x^2+y^2}& (x>0)\\ \frac{\partial }{\partial x}{\tan }^{-1}\left(\frac{y}{x}\right)+\pi & =& \frac{-y}{x^2+y^2}& (x<0\mathrm{か}\mathrm{つ}y\ge 0)\\ \frac{\partial }{\partial x}{\tan }^{-1}\left(\frac{y}{x}\right)-\pi & =& \frac{-y}{x^2+y^2}& (x<0\mathrm{か}\mathrm{つ}y<0)\\ \frac{\partial }{\partial x}\frac{\pi }{2}& =& 0& (x=0\mathrm{か}\mathrm{つ}y>0)\\ \frac{\partial }{\partial x}-\frac{\pi }{2}& =& 0& (x=0\mathrm{か}\mathrm{つ}y<0)\\ \mathrm{不}\mathrm{定}& & & (x=0\mathrm{か}\mathrm{つ}y=0)\end{array}\dots \frac{\partial }{\partial x}{\tan }^{-1}\left(\frac{y}{x}\right)=\frac{-y}{x^2+y^2}\end{array}$$

yでの偏微分

$$\begin{array}{rcl}\frac{\partial }{\partial y}\arg \left(z\right)& =& \frac{\partial }{\partial y}\{\theta +2n\pi \}\\ & =& \frac{\partial }{\partial y}\{\mathrm{Arg}\left(z\right)+2n\pi \}\\ & =& \frac{\partial }{\partial y}\mathrm{Arg}\left(z\right)+\frac{\partial }{\partial y}2n\pi \\ & =& \frac{\partial }{\partial y}\mathrm{Arg}\left(z\right)\dots \frac{\partial }{\partial y}C=0(C\mathrm{は}\mathrm{定}\mathrm{数}．x\mathrm{の}\mathrm{凾}\mathrm{数}\mathrm{で}\mathrm{な}\mathrm{い})\end{array}$$ $$\begin{array}{rcl}\frac{\partial }{\partial y}\mathrm{Arg}\left(z\right)& =& \frac{\partial }{\partial y}\mathrm{Arg}(x+iy)\\ & =& \{\begin{array}{llll}\frac{\partial }{\partial y}{\tan }^{-1}\left(\frac{y}{x}\right)& =& \frac{x}{x^2+y^2}& (x>0)\\ \frac{\partial }{\partial y}{\tan }^{-1}\left(\frac{y}{x}\right)+\pi & =& \frac{x}{x^2+y^2}& (x<0\mathrm{か}\mathrm{つ}y\ge 0)\\ \frac{\partial }{\partial y}{\tan }^{-1}\left(\frac{y}{x}\right)-\pi & =& \frac{x}{x^2+y^2}& (x<0\mathrm{か}\mathrm{つ}y<0)\\ \frac{\partial }{\partial y}\frac{\pi }{2}& =& 0& (x=0\mathrm{か}\mathrm{つ}y>0)\\ \frac{\partial }{\partial y}-\frac{\pi }{2}& =& 0& (x=0\mathrm{か}\mathrm{つ}y<0)\\ \mathrm{不}\mathrm{定}& & & (x=0\mathrm{か}\mathrm{つ}y=0)\end{array}\dots \frac{\partial }{\partial y}{\tan }^{-1}\left(\frac{y}{x}\right)=\frac{x}{x^2+y^2}\end{array}$$

arctan(y/x)の偏微分

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

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

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

$$\begin{array}{rcl}\frac{\partial }{\partial x}{\tan }^{-1}\left(\frac{y}{x}\right)& =& \{\frac{\partial }{\partial u}{\tan }^{-1}\left(u\right)\}\frac{\partial u}{\partial x}\dots u=\frac{y}{x}\\ & =& \frac{1}{1+u^2}\frac{\partial }{\partial x}\frac{y}{x}\dots \frac{\partial }{\partial u}{\tan }^{-1}\left(u\right)=\frac{1}{1+u^2}\\ & =& \frac{1}{1+\frac{y^2}{x^2}}y\frac{\partial }{\partial x}{x}^{-1}\\ & =& \frac{1}{\frac{x^2+y^2}{x^2}}y(-x^{-2})\\ & =& \frac{x^2}{x^2+y^2}y(-x^{-2})\\ & =& \frac{-y}{x^2+y^2}\end{array}$$

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

$$\begin{array}{r}\\ \frac{\partial }{\partial y}{\tan }^{-1}\left(\frac{y}{x}\right)& =& \{\frac{\partial }{\partial u}{\tan }^{-1}\left(u\right)\}\frac{\partial u}{\partial y}\dots u=\frac{y}{x}\\ & =& \frac{1}{1+u^2}\frac{\partial }{\partial y}\frac{y}{x}\dots \frac{\partial }{\partial u}{\tan }^{-1}\left(u\right)=\frac{1}{1+u^2}\\ & =& \frac{1}{1+\frac{y^2}{x^2}}{x}^{-1}\frac{\partial }{\partial y}y\\ & =& \frac{1}{\frac{x^2+y^2}{x^2}}{x}^{-1}\cdot 1\\ & =& \frac{x^2}{x^2+y^2}{x}^{-1}\\ & =& \frac{x}{x^2+y^2}\end{array}$$

arctan(x)の微分

${\tan }^{-1}\left(x\right)$の微分

$$\begin{array}{rcl}y& =& {\tan }^{-1}\left(x\right)\\ x& =& \tan \left(y\right)\\ \frac{\mathrm{d}x}{\mathrm{d}y}& =& \frac{\mathrm{d}}{\mathrm{d}y}\tan \left(y\right)\\ & =& 1+{\tan }^2\left(y\right)\\ \frac{\mathrm{d}y}{\mathrm{d}x}& =& \frac{1}{1+{\tan }^2\left(y\right)}\dots \frac{\mathrm{d}y}{\mathrm{d}x}=\frac{1}{\frac{\mathrm{d}x}{\mathrm{d}y}}\left(\mathrm{逆}\mathrm{凾}\mathrm{数}\mathrm{の}\mathrm{微}\mathrm{分}\right)\\ & =& \frac{1}{1+x^2}\dots x=\tan \left(y\right)\end{array}$$

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

$$\begin{array}{rcl}y& =& f\left(x\right)\dots \mathrm{微}\mathrm{分}\mathrm{可}\mathrm{能}\mathrm{な}\mathrm{凾}\mathrm{数}\\ x& =& f^{-1}\left(y\right)\dots f(x)\mathrm{の}\mathrm{逆}\mathrm{凾}\mathrm{数}\end{array}$$ $$\begin{array}{rcl}k& =& f(x+h)-f\left(x\right)\\ & & \dots \mathrm{凾}\mathrm{数}\mathrm{に}\mathrm{お}\mathrm{い}\mathrm{て}\mathrm{独}\mathrm{立}\mathrm{変}\mathrm{数}\mathrm{の}\mathrm{差}h\mathrm{が}\mathrm{あ}\mathrm{る}\mathrm{際}\mathrm{の}\mathrm{従}\mathrm{属}\mathrm{変}\mathrm{数}\mathrm{の}\mathrm{差}\mathrm{を}k\mathrm{と}\mathrm{す}\mathrm{る}\\ f\left(x\right)+k& =& y+k\\ & =& f(x+h)\\ x+h& =& f^{-1}(y+k)\end{array}$$ $$\begin{array}{rcl}x+h& =& f^{-1}(y+k)\\ h& =& f^{-1}(y+k)-x\\ & =& f^{-1}(y+k)-f^{-1}\left(y\right)\\ & & \dots \mathrm{逆}\mathrm{凾}\mathrm{数}\mathrm{に}\mathrm{お}\mathrm{い}\mathrm{て}\mathrm{独}\mathrm{立}\mathrm{変}\mathrm{数}\mathrm{の}\mathrm{差}k\mathrm{が}\mathrm{あ}\mathrm{る}\mathrm{際}\mathrm{の}\mathrm{従}\mathrm{属}\mathrm{変}\mathrm{数}\mathrm{の}\mathrm{差}\mathrm{が}h\mathrm{で}\mathrm{あ}\mathrm{る}\end{array}$$ $$\begin{array}{rcl}\underset{k\to 0}{lim}{f}^{-1}(y+k)& =& f^{-1}\left(y\right)\\ h& =& \underset{k\to 0}{lim}(f^{-1}(y+k)-f^{-1}\left(y\right))\\ & =& \left(\underset{k\to 0}{lim}{f}^{-1}(y+k)\right)-f^{-1}\left(y\right)\\ & =& f^{-1}\left(y\right)-f^{-1}\left(y\right)\\ & =& 0\\ & & \dots k\mathrm{を}0\mathrm{に}\mathrm{近}\mathrm{づ}\mathrm{け}\mathrm{る}\mathrm{こ}\mathrm{と}\mathrm{は}h\mathrm{を}0\mathrm{に}\mathrm{近}\mathrm{づ}\mathrm{け}\mathrm{る}\mathrm{こ}\mathrm{と}\mathrm{と}\mathrm{等}\mathrm{し}\mathrm{い}\end{array}$$ $$\begin{array}{rcl}k& =& f(x+h)-f\left(x\right)\\ \underset{h\to 0}{lim}f(x+h)& =& f\left(x\right)\\ k& =& \underset{h\to 0}{lim}(f(x+h)-f\left(x\right))\\ & =& \left(\underset{h\to 0}{lim}f(x+h)\right)-f\left(x\right)\\ & =& f\left(x\right)-f\left(x\right)\\ & =& 0\\ & & \dots h\mathrm{を}0\mathrm{に}\mathrm{近}\mathrm{づ}\mathrm{け}\mathrm{る}\mathrm{こ}\mathrm{と}\mathrm{は}k\mathrm{を}0\mathrm{に}\mathrm{近}\mathrm{づ}\mathrm{け}\mathrm{る}\mathrm{こ}\mathrm{と}\mathrm{と}\mathrm{等}\mathrm{し}\mathrm{い}\end{array}$$ $$\begin{array}{r}h\to 0⟺k\to 0\\ \underset{k\to 0}{lim}\frac{h}{k}=\underset{h\to 0}{lim}\frac{h}{k}\end{array}$$

逆凾数の微分

$$\begin{array}{rcl}\frac{\mathrm{d}x}{\mathrm{d}y}& =& \underset{k\to 0}{lim}\frac{f^{-1}(y+k)-f^{-1}\left(y\right)}{k}\\ & =& \underset{k\to 0}{lim}\frac{x+h-x}{k}\\ & =& \underset{k\to 0}{lim}\frac{h}{k}\\ & =& \underset{h\to 0}{lim}\frac{h}{k}\\ & =& \underset{h\to 0}{lim}\frac{1}{\frac{k}{h}}\\ & =& \frac{1}{\underset{h\to 0}{lim}\frac{k}{h}}\\ & =& \frac{1}{\underset{h\to 0}{lim}\frac{f(x+h)-f\left(x\right)}{h}}\\ & =& \frac{1}{\frac{\mathrm{d}y}{\mathrm{d}x}}\end{array}$$

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) $$\begin{array}{r}\\ y& =& \tan \left(x\right)\\ \frac{\mathrm{d}}{\mathrm{d}x}\tan \left(x\right)& =& \underset{h\to 0}{lim}\frac{\tan (x+h)-\tan \left(x\right)}{h}\\ & =& \underset{h\to 0}{lim}\frac{1}{h}\{\frac{\tan \left(x\right)+\tan \left(h\right)}{1-\tan \left(x\right)\tan \left(h\right)}-\tan \left(x\right)\}\\ & =& \underset{h\to 0}{lim}\frac{1}{h}\frac{\tan \left(x\right)+\tan \left(h\right)-\tan \left(x\right)(1-\tan \left(x\right)\tan \left(h\right))}{1-\tan \left(x\right)\tan \left(h\right)}\\ & =& \underset{h\to 0}{lim}\frac{1}{h}\frac{\tan \left(x\right)+\tan \left(h\right)-\tan \left(x\right)+{\tan }^2\left(x\right)\tan \left(h\right)}{1-\tan \left(x\right)\tan \left(h\right)}\\ & =& \underset{h\to 0}{lim}\frac{1}{h}\frac{\tan \left(h\right)+{\tan }^2\left(x\right)\tan \left(h\right)}{1-\tan \left(x\right)\tan \left(h\right)}\\ & =& \underset{h\to 0}{lim}\frac{1}{h}\frac{\tan \left(h\right)(1+{\tan }^2\left(x\right))}{1-\tan \left(x\right)\tan \left(h\right)}\\ & =& \underset{h\to 0}{lim}\frac{\tan \left(h\right)}{h}\frac{1+{\tan }^2\left(x\right)}{1-\tan \left(x\right)\tan \left(h\right)}\\ & =& 1\cdot \frac{1+{\tan }^2\left(x\right)}{1-\tan \left(x\right)\cdot 0}\\ & & \dots \underset{h\to 0}{lim}\frac{\tan \left(h\right)}{h}=\underset{h\to 0}{lim}\frac{1}{h}\frac{\sin \left(h\right)}{\cos \left(h\right)}=\underset{h\to 0}{lim}\frac{\sin \left(h\right)}{h}\frac{1}{\cos \left(h\right)}=1\cdot 1=1\\ & & \dots \underset{h\to 0}{lim}\frac{\sin \left(h\right)}{h}=1\\ & & \dots \underset{h\to 0}{lim}\cos \left(h\right)=1\\ & =& 1+{\tan }^2\left(x\right)\end{array}$$ $$\begin{array}{rcl}1+{\tan }^2\left(x\right)& =& 1+{\left\{\frac{\sin \left(x\right)}{\cos \left(x\right)}\right\}}^2\\ & =& \frac{{\cos }^2\left(x\right)+{\sin }^2\left(x\right)}{{\cos }^2\left(x\right)}\\ & =& \frac{1}{{\cos }^2\left(x\right)}\cdots {\cos }^2\left(x\right)+{\sin }^2\left(x\right)=1\\ & =& {\sec }^2\left(x\right)\end{array}$$

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) $$\begin{array}{rcl}y& =& \tan \left(x\right)\\ \frac{\mathrm{d}}{\mathrm{d}x}\tan \left(x\right)& =& \frac{\mathrm{d}}{\mathrm{d}x}\frac{\sin \left(x\right)}{\cos \left(x\right)}\\ & =& \frac{\mathrm{d}}{\mathrm{d}x}\sin \left(x\right){\{\cos \left(x\right)\}}^{-1}\\ & =& {\{\cos \left(x\right)\}}^{-1}\frac{\mathrm{d}}{\mathrm{d}x}\sin \left(x\right)+\sin \left(x\right)\frac{\mathrm{d}}{\mathrm{d}x}{\{\cos \left(x\right)\}}^{-1}\\ & =& {\{\cos \left(x\right)\}}^{-1}\{\cos \left(x\right)\}+\sin \left(x\right)[-{\{\cos \left(x\right)\}}^{-2}]\frac{\mathrm{d}}{\mathrm{d}x}\{\cos \left(x\right)\}\\ & =& 1+\frac{-\sin \left(x\right)}{{\cos }^2\left(x\right)}\{-\sin \left(x\right)\}\\ & =& 1+\frac{{\sin }^2\left(x\right)}{{\cos }^2\left(x\right)}\\ & =& 1+{\left\{\frac{\sin \left(x\right)}{\cos \left(x\right)}\right\}}^2\\ & =& 1+{\tan }^2\left(x\right)\end{array}$$

---

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