x tan(x)とx tanh(x)の積分
z tan(z)とz tanh(z)の積分
次の積分が成り立つ。
次の積分が成り立つ。
(1)
\[ \int z\tan^{\pm1}\left(z\right)dz=i^{\pm1}\left\{ \frac{1}{2}z^{2}-iz\Li_{1}\left(\mp e^{2iz}\right)+\frac{1}{2}\Li_{2}\left(\mp e^{2iz}\right)\right\} +C \](2)
\[ \int z\tanh^{\pm1}\left(z\right)dz=\frac{1}{2}z^{2}-z\Li_{1}\left(\mp e^{-2z}\right)-\frac{1}{2}\Li_{2}\left(\mp e^{-2z}\right)+C \]-
\(\Li_{n}\left(z\right)\)は多重対数関数(1)
\begin{align*} \int z\tan^{\pm1}\left(z\right)dz & =i^{\pm1}\int z\left(1+2\Li_{0}\left(\mp e^{2iz}\right)\right)dz\cmt{\Li_{0}\left(z\right)=\frac{z}{1-z}}\\ & =i^{\pm1}\int\left(z+2z\Li_{0}\left(\mp e^{2iz}\right)\right)dz\\ & =i^{\pm1}\left\{ \frac{1}{2}z^{2}+\frac{2}{2i}z\Li_{1}\left(\mp e^{2iz}\right)-\frac{2}{2i}\int\Li_{1}\left(\mp e^{2iz}\right)dz\right\} \cmt{\int\Li_{n}\left(\alpha e^{\beta z}\right)dz=\frac{1}{\beta}\Li_{n+1}\left(\alpha e^{\beta z}\right)+C}\\ & =i^{\pm1}\left\{ \frac{1}{2}z^{2}-iz\Li_{1}\left(\mp e^{2iz}\right)-\frac{2}{\left(2i\right)^{2}}\Li_{2}\left(\mp e^{2iz}\right)\right\} \\ & =i^{\pm1}\left\{ \frac{1}{2}z^{2}-iz\Li_{1}\left(\mp e^{2iz}\right)+\frac{1}{2}\Li_{2}\left(\mp e^{2iz}\right)\right\} +C \end{align*}(1)-2
\(x\in\mathbb{R}\)とする。\begin{align*} \int x\tan^{\pm1}\left(x\right)dx & =\int x\left(i\right)^{\mp1}\frac{e^{ix}\mp e^{-ix}}{e^{ix}\pm e^{-ix}}dx\\ & =i^{\mp1}\int x\frac{e^{2ix}\mp1}{e^{2ix}\pm1}dx\\ & =i^{\mp1}\int x\left(1\mp\frac{2}{e^{2ix}\pm1}\right)dx\\ & =i^{\mp1}\int x\left(1-\frac{2}{1\pm e^{2ix}}\right)dx\\ & =i^{\mp1}\left(\frac{x^{2}}{2}-2\int\frac{x}{1\pm e^{2ix}}dx\right)\\ & =i^{\mp1}\left(\frac{x^{2}}{2}-2\int\frac{1}{1\pm e^{2ix}}\frac{1}{2i}\log e^{2ix}\frac{e^{-2ix}}{2i}de^{2ix}\right)\\ & =i^{\mp1}\left(\frac{x^{2}}{2}+\frac{1}{2}\int\log e^{2ix}\frac{1}{e^{2ix}\left(1\pm e^{2ix}\right)}de^{2ix}\right)\\ & =i^{\mp1}\left(\frac{x^{2}}{2}+\frac{1}{2}\int\log e^{2ix}\left(\frac{1}{e^{2ix}}\mp\frac{1}{1\pm e^{2ix}}\right)de^{2ix}\right)\\ & =i^{\mp1}\left(\frac{x^{2}}{2}+\frac{1}{2}\int\log e^{2ix}d\left(\log e^{2ix}\right)\mp\frac{1}{2}\int\log e^{2ix}\frac{1}{1\pm e^{2ix}}de^{2ix}\right)\\ & =i^{\mp1}\left(\frac{x^{2}}{2}+\frac{1}{4}\left(\log e^{2ix}\right)^{2}\mp\frac{1}{2}\left(\pm\log e^{2ix}\log\left(1\pm e^{2ix}\right)\mp\int\frac{\log\left(1\pm e^{2ix}\right)}{e^{2ix}}de^{2ix}\right)\right)\\ & =i^{\mp1}\left(\frac{x^{2}}{2}-x^{2}\mp\frac{1}{2}\left(\mp2ix\Li_{1}\left(\mp e^{2ix}\right)\pm\int\frac{\Li_{1}\left(\mp e^{2ix}\right)}{e^{2ix}}de^{2ix}\right)\right)\\ & =i^{\mp1}\left(\frac{x^{2}}{2}-x^{2}\mp\frac{1}{2}\left(\mp2ix\Li_{1}\left(\mp e^{2ix}\right)\pm\Li_{2}\left(\mp e^{2ix}\right)\right)\right)\\ & =i^{\pm1}\left(\frac{x^{2}}{2}-ix\Li_{1}\left(\mp e^{2ix}\right)+\frac{1}{2}\Li_{2}\left(\mp e^{2ix}\right)\right)+C \end{align*}
(2)
\begin{align*} \int z\tanh^{\pm1}\left(z\right)dz & =\int z\left(-i\tan\left(iz\right)\right)^{\pm1}dz\\ & =\left(-i\right)^{\pm1}\int z\tan^{\pm1}\left(iz\right)dz\\ & =-\left(-i\right)^{\pm1}\int iz\tan^{\pm1}\left(iz\right)d\left(iz\right)\\ & =i^{\pm1}\left\{ i^{\pm1}\left\{ \frac{1}{2}\left(iz\right)^{2}-i\left(iz\right)\Li_{1}\left(\mp e^{-2z}\right)+\frac{1}{2}\Li_{2}\left(\mp e^{-2z}\right)\right\} \right\} \\ & =-\left\{ \frac{1}{2}\left(iz\right)^{2}-i\left(iz\right)\Li_{1}\left(\mp e^{-2z}\right)+\frac{1}{2}\Li_{2}\left(\mp e^{-2z}\right)\right\} \\ & =\frac{1}{2}z^{2}-z\Li_{1}\left(\mp e^{-2z}\right)-\frac{1}{2}\Li_{2}\left(\mp e^{-2z}\right)+C \end{align*}ページ情報
タイトル | x tan(x)とx tanh(x)の積分 |
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偏角の三角関数
\[
\sin\Arg z=\frac{\Im z}{\left|z\right|}
\]
逆正接関数・逆双曲線正接関数と多重対数関数の関係
\[
\Tan^{\bullet}z=\frac{i}{2}\left(-\Li_{1}\left(iz\right)+\Li_{1}\left(-iz\right)\right)
\]
逆三角関数の負角、余角、逆数
\[
\cos^{\bullet}x+\sin^{\bullet}x=\frac{\pi}{2}
\]
三角関数と双曲線関数の対数
\[
\log\sin x=-\log2+\frac{\pi}{2}i-ix-Li_{1}\left(e^{2ix}\right)
\]