三角関数と双曲線関数の対数の積分
三角関数の対数の積分
(1)
\[ \int\Log\sin^{\alpha}zdz=z\Log\sin^{\alpha}x+\frac{i\alpha}{2}z^{2}+\alpha z\Li_{1}\left(e^{2iz}\right)+\frac{i\alpha}{2}\Li_{2}\left(e^{2iz}\right)+\C{} \](2)
\[ \int\Log\cos^{\alpha}zdx=z\Log\cos^{\alpha}x+\frac{i\alpha}{2}z^{2}+\alpha z\Li_{1}\left(-e^{2iz}\right)+\frac{i\alpha}{2}\Li_{2}\left(-e^{2iz}\right)+\C{} \](3)
\[ \int\Log\tan^{\alpha}zdz=\frac{i}{2}\left\{ \Log\left(\tan^{\alpha}z\right)\left(-\Li_{1}\left(i\tan z\right)+\Li_{1}\left(-i\tan z\right)\right)+\alpha\Li_{2}\left(i\tan z\right)-\alpha\Li_{2}\left(-i\tan z\right)\right\} +\C{} \](1)
\begin{align*} \int\Log\sin^{\alpha}zdz & =z\Log\sin^{\alpha}z-\alpha\int z\tan^{-1}zdz\\ & =z\Log\sin^{\alpha}z-i^{-1}\alpha\left(\frac{z^{2}}{2}-iz\Li_{1}\left(e^{2iz}\right)+\frac{1}{2}\Li_{2}\left(e^{2iz}\right)\right)+\C{}\\ & =z\Log\sin^{\alpha}x+\frac{i\alpha}{2}z^{2}+\alpha z\Li_{1}\left(e^{2iz}\right)+\frac{i\alpha}{2}\Li_{2}\left(e^{2iz}\right)+\C{} \end{align*}(2)
\begin{align*} \int\Log\cos^{\alpha}zdx & =z\Log\cos^{\alpha}z+\alpha\int z\tan zdz\\ & =z\Log\cos^{\alpha}z+i\alpha\left(\frac{z^{2}}{2}-iz\Li_{1}\left(-e^{2iz}\right)+\frac{1}{2}\Li_{2}\left(-e^{2iz}\right)\right)+\C{}\\ & =z\Log\cos^{\alpha}z+\frac{i\alpha}{2}z^{2}+\alpha z\Li_{1}\left(-e^{2iz}\right)+\frac{i\alpha}{2}\Li_{2}\left(-e^{2iz}\right)+\C{} \end{align*}(3)
\begin{align*} \int\Log\tan^{\alpha}zdz & =\int\Log\left(\tan^{\alpha}z\right)\cos^{2}\left(z\right)d\tan z\\ & =\int\Log\left(\tan^{\alpha}z\right)\frac{1}{1+\tan^{2}z}d\tan z\\ & =\frac{1}{2}\int\Log\left(\tan^{\alpha}z\right)\left(\frac{1}{1-i\tan z}+\frac{1}{1+i\tan z}\right)d\tan z\\ & =\frac{1}{2}\left\{ \Log\left(\tan^{\alpha}z\right)\left(-i\Li_{1}\left(i\tan z\right)+i\Li_{1}\left(-i\tan z\right)\right)-\int\frac{\alpha\tan^{\alpha-1}x}{\tan^{\alpha}\left(x\right)}\left(-i\Li_{1}\left(i\tan z\right)+i\Li_{1}\left(-i\tan z\right)\right)d\tan z\right\} +\C{}\\ & =\frac{1}{2}\left\{ \Log\left(\tan^{\alpha}z\right)\left(-i\Li_{1}\left(i\tan z\right)+i\Li_{1}\left(-i\tan z\right)\right)+\alpha i\int\frac{1}{\tan\left(x\right)}\left(\Li_{1}\left(i\tan z\right)-\Li_{1}\left(-i\tan z\right)\right)d\tan z\right\} +\C{}\\ & =\frac{i}{2}\left\{ \Log\left(\tan^{\alpha}z\right)\left(-\Li_{1}\left(i\tan z\right)+\Li_{1}\left(-i\tan z\right)\right)+\alpha\left(\Li_{2}\left(i\tan z\right)-\Li_{2}\left(-i\tan z\right)\right)\right\} +\C{}\\ & =\frac{i}{2}\left\{ \Log\left(\tan^{\alpha}z\right)\left(-\Li_{1}\left(i\tan z\right)+\Li_{1}\left(-i\tan z\right)\right)+\alpha\Li_{2}\left(i\tan z\right)-\alpha\Li_{2}\left(-i\tan z\right)\right\} +\C{} \end{align*}双曲線関数の対数の積分
(1)
\[ \int\Log\sinh^{\alpha}zdz=z\Log\sinh^{\alpha}x-\frac{\alpha}{2}z^{2}+\alpha z\Li_{1}\left(e^{-2z}\right)+\frac{\alpha}{2}\Li_{2}\left(e^{-2z}\right)+\C{} \](2)
\[ \int\Log\cosh^{\alpha}zdz=z\Log\cosh^{\alpha}z-\frac{\alpha}{2}z^{2}+\alpha z\Li_{1}\left(-e^{-2z}\right)+\frac{\alpha}{2}\Li_{2}\left(-e^{-2z}\right)+\C{} \](3)
\[ \int\Log\tanh^{\alpha}zdz=\frac{1}{2}\left\{ \Log\left(\tanh^{\alpha}z\right)\left(\Li_{1}\left(\tanh z\right)-\Li_{1}\left(-\tanh z\right)\right)-\alpha\Li_{2}\left(\tanh z\right)+\alpha\Li_{2}\left(-\tanh z\right)\right\} +\C{} \](1)
\begin{align*} \int\Log\sinh^{\alpha}zdz & =z\Log\sinh^{\alpha}z-\alpha\int z\tanh^{-1}zdz\\ & =z\Log\sinh^{\alpha}z-\alpha\left(\frac{z^{2}}{2}-z\Li_{1}\left(e^{-2z}\right)-\frac{1}{2}\Li_{2}\left(e^{-2z}\right)\right)+\C{}\\ & =z\Log\sinh^{\alpha}x-\frac{\alpha}{2}z^{2}+\alpha z\Li_{1}\left(e^{-2z}\right)+\frac{\alpha}{2}\Li_{2}\left(e^{-2z}\right)+\C{} \end{align*}(2)
\begin{align*} \int\Log\cosh^{\alpha}zdz & =-i\int\Log\cos^{\alpha}\left(iz\right)d\left(iz\right)\\ & =-i\left(iz\Log\cos^{\alpha}\left(iz\right)-\frac{i\alpha}{2}z^{2}+i\alpha z\Li_{1}\left(-e^{-2z}\right)+\frac{i\alpha}{2}\Li_{2}\left(-e^{-2z}\right)\right)+\C{}\\ & =z\Log\cosh^{\alpha}z-\frac{\alpha}{2}z^{2}+\alpha z\Li_{1}\left(-e^{-2z}\right)+\frac{\alpha}{2}\Li_{2}\left(-e^{-2z}\right)+\C{} \end{align*}(3)
\begin{align*} \int\Log\tanh^{\alpha}zdz & =\int\Log\left(\tanh^{\alpha}z\right)\cosh^{2}\left(z\right)d\tanh z\\ & =\int\Log\left(\tanh^{\alpha}z\right)\frac{1}{1-\tanh^{2}z}d\tanh z\\ & =\frac{1}{2}\int\Log\left(\tanh^{\alpha}z\right)\left(\frac{1}{1-\tanh z}+\frac{1}{1+\tanh z}\right)d\tanh z\\ & =\frac{1}{2}\left\{ \Log\left(\tanh^{\alpha}z\right)\left(\Li_{1}\left(\tanh z\right)-\Li_{1}\left(-\tanh z\right)\right)-\int\frac{\alpha\tanh^{\alpha-1}x}{\tanh^{\alpha}\left(x\right)}\left(\Li_{1}\left(\tanh z\right)-\Li_{1}\left(-\tanh z\right)\right)d\tanh z\right\} +\C{}\\ & =\frac{1}{2}\left\{ \Log\left(\tanh^{\alpha}z\right)\left(\Li_{1}\left(\tanh z\right)-\Li_{1}\left(-\tanh z\right)\right)-\alpha\int\frac{1}{\tanh\left(x\right)}\left(\Li_{1}\left(\tanh z\right)-\Li_{1}\left(-\tanh z\right)\right)d\tanh z\right\} +\C{}\\ & =\frac{1}{2}\left\{ \Log\left(\tanh^{\alpha}z\right)\left(\Li_{1}\left(\tanh z\right)-\Li_{1}\left(-\tanh z\right)\right)-\alpha\left(\Li_{2}\left(\tanh z\right)-\Li_{2}\left(-\tanh z\right)\right)\right\} +\C{}\\ & =\frac{1}{2}\left\{ \Log\left(\tanh^{\alpha}z\right)\left(\Li_{1}\left(\tanh z\right)-\Li_{1}\left(-\tanh z\right)\right)-\alpha\Li_{2}\left(\tanh z\right)+\alpha\Li_{2}\left(-\tanh z\right)\right\} +\C{} \end{align*}ページ情報
| タイトル | 三角関数と双曲線関数の対数の積分 |
| URL | https://www.nomuramath.com/stzuupj6/ |
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ピタゴラスの基本三角関数公式
\[
\cos^{2}x+\sin^{2}x=1
\]
逆三角関数と逆双曲線関数の級数表示
\[
\sin^{\bullet}x=\sum_{k=0}^{\infty}\frac{C\left(2k,k\right)}{4^{k}(2k+1)}x^{2k+1}\qquad,(|x|\leq1)
\]
双曲線関数と三角関数の級数展開
\[
\tanh x=\sum_{k=1}^{\infty}\frac{2^{2k}\left(2^{2k}-1\right)B_{2k}}{(2k)!}x{}^{2k-1}
\]
逆三角関数と逆双曲線関数の負角
\[
\Sin^{\bullet}\left(-z\right)=-\Sin^{\bullet}z
\]