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/ |
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三角関数・双曲線関数の一次結合の逆数の積分
\[
\int\frac{1}{\alpha\sin z+\beta\cos z+\gamma}dz=-\frac{2}{\sqrt{\alpha^{2}+\beta^{2}-\gamma^{2}}}\tanh^{\bullet}\frac{\left(\gamma-\beta\right)\tan\frac{z}{2}+\alpha}{\sqrt{\alpha^{2}+\beta^{2}-\gamma^{2}}}+C
\]
逆三角関数と逆双曲線関数の積分表示
\[
\sin^{\bullet}x=\int_{0}^{x}\frac{1}{\sqrt{1-t^{2}}}dt
\]
三角関数の積
\[
\prod_{k=1}^{n-1}\sin\frac{k\pi}{n}=\frac{n}{2^{n-1}}
\]
逆三角関数と逆双曲線関数の積分
\[
\int\sin^{\bullet}xdx=x\sin^{\bullet}x+\sqrt{1-x^{2}}
\]