逆正接関数・逆双曲線正接関数と多重対数関数の関係
逆正接関数・逆双曲線正接関数と多重対数関数の関係
(1)
\[ \Tan^{\bullet}z=\frac{i}{2}\left(-\Li_{1}\left(iz\right)+\Li_{1}\left(-iz\right)\right) \](2)
\[ \Tanh^{\bullet}z=\frac{1}{2}\left(\Li_{1}\left(z\right)-\Li_{1}\left(-z\right)\right) \](3)
\begin{align*} \int\frac{\Tan^{\bullet}z}{z}dz & =\frac{i}{2}\left(\Li_{2}\left(-iz\right)-\Li_{2}\left(iz\right)\right)+\C{} \end{align*}(4)
\begin{align*} \int\frac{\Tanh^{\bullet}z}{z}dz & =\frac{1}{2}\left(\Li_{2}\left(z\right)-\Li_{2}\left(-z\right)\right)+\C{} \end{align*}(1)
\begin{align*} \Tan^{\bullet}z & =\int_{0}^{z}\frac{1}{1+z^{2}}dz\\ & =\frac{1}{2}\int_{0}^{z}\left(\frac{1}{1-iz}+\frac{1}{1+iz}\right)dz\\ & =\frac{1}{2}\left(-i\Li_{1}\left(iz\right)+i\Li_{1}\left(-iz\right)\right)\\ & =\frac{i}{2}\left(-\Li_{1}\left(iz\right)+\Li_{1}\left(-iz\right)\right) \end{align*}(2)
\begin{align*} \Tanh^{\bullet}z & =-i\Tan^{\bullet}iz\\ & =-i\left(\frac{i}{2}\left(-\Li_{1}\left(-z\right)+\Li_{1}\left(z\right)\right)\right)\\ & =\frac{1}{2}\left(\Li_{1}\left(z\right)-\Li_{1}\left(-z\right)\right) \end{align*}(3)
\begin{align*} \int\frac{\Tan^{\bullet}z}{z}dz & =\frac{i}{2}\int\frac{\Li_{1}\left(-iz\right)-\Li_{1}\left(iz\right)}{z}dz\\ & =\frac{i}{2}\left(\Li_{2}\left(-iz\right)-\Li_{2}\left(iz\right)\right)+\C{} \end{align*}(4)
\begin{align*} \int\frac{\Tanh^{\bullet}z}{z}dz & =-i\int\frac{\Tan^{\bullet}iz}{z}dz\\ & =-i\int^{iz}\frac{\Tan^{\bullet}z}{z}dz\\ & =-i\left[\frac{i}{2}\left(\Li_{2}\left(-iz\right)-\Li_{2}\left(iz\right)\right)\right]^{z\rightarrow iz}+\C{}\\ & =\frac{1}{2}\left(\Li_{2}\left(z\right)-\Li_{2}\left(-z\right)\right)+\C{} \end{align*}
ページ情報
| タイトル | 逆正接関数・逆双曲線正接関数と多重対数関数の関係 |
| URL | https://www.nomuramath.com/vz64908m/ |
| SNSボタン |
三角関数と双曲線関数の微分
\[
\frac{d}{dx}\tan x=\cos^{-2}x
\]
1と3角関数・双曲線関数(半角の公式の拡張)
\[
1+\sin z=\left(\cos\frac{z}{2}+\sin\frac{z}{2}\right)^{2}
\]
正接関数・双曲線正接関数の多重対数関数表示
\[
\tan^{\pm1}z=i^{\pm1}\left(1+2\Li_{0}\left(\mp e^{2iz}\right)\right)
\]
3角関数・双曲線関数の総和
\[
\sum_{k=m_{1}}^{m_{2}}\sin\left(ak+b\right)=\sin^{-1}\left(\frac{a}{2}\right)\sin\left(\left(m_{1}+m_{2}\right)\frac{a}{2}+b\right)\sin\left(\left(1+m_{2}-m_{1}\right)\frac{a}{2}\right)
\]