巾関数と逆三角関数・逆双曲線関数の積の積分
巾関数と逆三角関数の積の積分
(1)
\[ \int z^{\alpha}\Sin^{\bullet}zdz=\frac{1}{\alpha+1}\left(z^{\alpha+1}\Sin^{\bullet}z-\frac{z^{\alpha+2}}{\alpha+2}F\left(\frac{1}{2},\frac{\alpha}{2}+1;\frac{\alpha}{2}+2;z^{2}\right)\right)+C \](2)
\[ \int z^{\alpha}\Cos^{\bullet}zdz=\frac{1}{\alpha+1}\left(z^{\alpha+1}\Cos^{\bullet}z+\frac{z^{\alpha+2}}{\alpha+2}F\left(\frac{1}{2},\frac{\alpha}{2}+1;\frac{\alpha}{2}+2;z^{2}\right)\right)+C \](3)
\[ \int z^{\alpha}\Tan^{\bullet}zdz=\frac{1}{\alpha+1}\left(z^{\alpha+1}\Tan^{\bullet}z-\frac{z^{\alpha+2}}{\alpha+2}F\left(1,\frac{\alpha}{2}+1;\frac{\alpha}{2}+2;-z^{2}\right)\right)+C \]-
\(F\)は一般化超幾何関数(1)
\begin{align*} \int z^{\alpha}\Sin^{\bullet}zdz & =\frac{1}{\alpha+1}z^{\alpha+1}\Sin^{\bullet}z-\frac{1}{\alpha+1}\int z^{\alpha+1}\frac{1}{\sqrt{1-z^{2}}}dz\\ & =\frac{1}{\alpha+1}z^{\alpha+1}\Sin^{\bullet}z-\frac{1}{\alpha+1}\int z^{\alpha+1}F\left(\frac{1}{2};;z^{2}\right)dz\\ & =\frac{1}{\alpha+1}z^{\alpha+1}\Sin^{\bullet}z-\frac{1}{\alpha+1}\frac{z^{\alpha+2}}{\alpha+2}F\left(\frac{1}{2},\frac{\alpha+2}{2};\frac{\alpha+2}{2}+1;z^{2}\right)+C\\ & =\frac{1}{\alpha+1}\left(z^{\alpha+1}\Sin^{\bullet}z-\frac{z^{\alpha+2}}{\alpha+2}F\left(\frac{1}{2},\frac{\alpha}{2}+1;\frac{\alpha}{2}+2;z^{2}\right)\right)+C \end{align*}(2)
\begin{align*} \int z^{\alpha}\Cos^{\bullet}zdz & =\frac{1}{\alpha+1}z^{\alpha+1}\Cos^{\bullet}z+\frac{1}{\alpha+1}\int z^{\alpha+1}\frac{1}{\sqrt{1-z^{2}}}dz\\ & =\frac{1}{\alpha+1}z^{\alpha+1}\Cos^{\bullet}z+\frac{1}{\alpha+1}\int z^{\alpha+1}F\left(\frac{1}{2};;z^{2}\right)dz\\ & =\frac{1}{\alpha+1}z^{\alpha+1}\Cos^{\bullet}z+\frac{1}{\alpha+1}\frac{z^{\alpha+2}}{\alpha+2}F\left(\frac{1}{2},\frac{\alpha+2}{2};\frac{\alpha+2}{2}+1;z^{2}\right)+C\\ & =\frac{1}{\alpha+1}\left(z^{\alpha+1}\Cos^{\bullet}z+\frac{z^{\alpha+2}}{\alpha+2}F\left(\frac{1}{2},\frac{\alpha}{2}+1;\frac{\alpha}{2}+2;z^{2}\right)\right)+C \end{align*}(3)
\begin{align*} \int z^{\alpha}\Tan^{\bullet}zdz & =\frac{z^{\alpha+1}}{\alpha+1}\Tan^{\bullet}z-\frac{1}{\alpha+1}\int\frac{z^{\alpha+1}}{1+z^{2}}dz\\ & =\frac{z^{\alpha+1}}{\alpha+1}\Tan^{\bullet}z-\frac{1}{\alpha+1}\int z^{\alpha+1}F\left(1;;-z^{2}\right)dz\\ & =\frac{z^{\alpha+1}}{\alpha+1}\Tan^{\bullet}z-\frac{1}{\alpha+1}\frac{z^{\alpha+2}}{\alpha+2}F\left(1,\frac{\alpha+2}{2};\frac{\alpha+2}{2}+1;-z^{2}\right)+C\\ & =\frac{1}{\alpha+1}\left(z^{\alpha+1}\Tan^{\bullet}z-\frac{z^{\alpha+2}}{\alpha+2}F\left(1,\frac{\alpha}{2}+1;\frac{\alpha}{2}+2;-z^{2}\right)\right)+C \end{align*}巾関数と逆双曲線関数の積の積分
(1)
\[ \int z^{\alpha}\Sinh^{\bullet}zdz=\frac{1}{\alpha+1}\left(z^{\alpha+1}\Sinh^{\bullet}z-\frac{z^{\alpha+2}}{\alpha+2}F\left(\frac{1}{2},\frac{\alpha}{2}+1;\frac{\alpha}{2}+2;-z^{2}\right)\right)+C \](2)
\[ \int z^{\alpha}\Cosh^{\bullet}zdz=\frac{1}{\alpha+1}\left(z^{\alpha+1}\Cosh^{\bullet}z-\frac{\sqrt{1-z^{2}}}{\sqrt{z-1}\sqrt{z+1}}\frac{z^{\alpha+2}}{\alpha+2}F\left(\frac{1}{2},\frac{\alpha}{2}+1;\frac{\alpha}{2}+2;z^{2}\right)\right)+C \](3)
\[ \int z^{\alpha}\Tanh^{\bullet}zdz=\frac{1}{\alpha+1}\left(z^{\alpha+1}\Tanh^{\bullet}z-\frac{z^{\alpha+2}}{\alpha+2}F\left(1,\frac{\alpha}{2}+1;\frac{\alpha}{2}+2;z^{2}\right)\right)+C \]-
\(F\)は一般化超幾何関数(1)
\begin{align*} \int z^{\alpha}\Sinh^{\bullet}zdz & =\frac{1}{\alpha+1}z^{\alpha+1}\Sinh^{\bullet}z-\frac{1}{\alpha+1}\int z^{\alpha+1}\frac{1}{\sqrt{1+z^{2}}}dz\\ & =\frac{1}{\alpha+1}z^{\alpha+1}\Sinh^{\bullet}z-\frac{1}{\alpha+1}\int z^{\alpha+1}F\left(\frac{1}{2};;-z^{2}\right)dz\\ & =\frac{1}{\alpha+1}z^{\alpha+1}\Sinh^{\bullet}z-\frac{1}{\alpha+1}\frac{z^{\alpha+2}}{\alpha+2}F\left(\frac{1}{2},\frac{\alpha+2}{2};\frac{\alpha+2}{2}+1;-z^{2}\right)+C\\ & =\frac{1}{\alpha+1}\left(z^{\alpha+1}\Sinh^{\bullet}z-\frac{z^{\alpha+2}}{\alpha+2}F\left(\frac{1}{2},\frac{\alpha}{2}+1;\frac{\alpha}{2}+2;-z^{2}\right)\right)+C \end{align*}(2)
\begin{align*} \int z^{\alpha}\Cosh^{\bullet}zdz & =\frac{1}{\alpha+1}z^{\alpha+1}\Cosh^{\bullet}z-\frac{1}{\alpha+1}\int z^{\alpha+1}\frac{1}{\sqrt{z-1}\sqrt{z+1}}dz\\ & =\frac{1}{\alpha+1}z^{\alpha+1}\Cosh^{\bullet}z-\frac{1}{\alpha+1}\frac{\sqrt{1-z^{2}}}{\sqrt{z-1}\sqrt{z+1}}\int z^{\alpha+1}\frac{1}{\sqrt{1-z^{2}}}dz\\ & =\frac{1}{\alpha+1}z^{\alpha+1}\Cosh^{\bullet}z-\frac{1}{\alpha+1}\frac{\sqrt{1-z^{2}}}{\sqrt{z-1}\sqrt{z+1}}\int z^{\alpha+1}F\left(\frac{1}{2};;z^{2}\right)dz\\ & =\frac{1}{\alpha+1}z^{\alpha+1}\Cosh^{\bullet}z-\frac{1}{\alpha+1}\frac{\sqrt{1-z^{2}}}{\sqrt{z-1}\sqrt{z+1}}\frac{z^{\alpha+2}}{\alpha+2}F\left(\frac{1}{2},\frac{\alpha+2}{2};\frac{\alpha+2}{2}+1;z^{2}\right)+C\\ & =\frac{1}{\alpha+1}\left(z^{\alpha+1}\Cosh^{\bullet}z-\frac{\sqrt{1-z^{2}}}{\sqrt{z-1}\sqrt{z+1}}\frac{z^{\alpha+2}}{\alpha+2}F\left(\frac{1}{2},\frac{\alpha}{2}+1;\frac{\alpha}{2}+2;z^{2}\right)\right)+C \end{align*}(3)
\begin{align*} \int z^{\alpha}\Tanh^{\bullet}zdz & =\frac{z^{\alpha+1}}{\alpha+1}\Tanh^{\bullet}z-\frac{1}{\alpha+1}\int\frac{z^{\alpha+1}}{1-z^{2}}dz\\ & =\frac{z^{\alpha+1}}{\alpha+1}\Tanh^{\bullet}z-\frac{1}{\alpha+1}\int z^{\alpha+1}F\left(1;;z^{2}\right)dz\\ & =\frac{z^{\alpha+1}}{\alpha+1}\Tanh^{\bullet}z-\frac{1}{\alpha+1}\frac{z^{\alpha+2}}{\alpha+2}F\left(1,\frac{\alpha+2}{2};\frac{\alpha+2}{2}+1;z^{2}\right)+C\\ & =\frac{1}{\alpha+1}\left(z^{\alpha+1}\Tanh^{\bullet}z-\frac{z^{\alpha+2}}{\alpha+2}F\left(1,\frac{\alpha}{2}+1;\frac{\alpha}{2}+2;z^{2}\right)\right)+C \end{align*}
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三角関数と双曲線関数のn乗積分
\[
\int\sin^{2n+m_{\pm}}xdx=\frac{\Gamma\left(n+\frac{1}{2}+\frac{m_{\pm}}{2}\right)}{\Gamma\left(n+1+\frac{m_{\pm}}{2}\right)}\left\{ -\frac{1}{2}\sum_{k=0}^{n-1}\left(\frac{\Gamma\left(k+1+\frac{m_{\pm}}{2}\right)}{\Gamma\left(k+\frac{3}{2}+\frac{m_{\pm}}{2}\right)}\cos x\sin^{2k+1+m_{\pm}}x\right)+\frac{\Gamma\left(1+\frac{m_{\pm}}{2}\right)}{\Gamma\left(\frac{1}{2}+\frac{m_{\pm}}{2}\right)}\int\sin^{m_{\pm}}xdx\right\}
\]
三角関数の還元(負角・余角・補角)公式
\[
\sin(-x)=-\sin x
\]
三角関数と双曲線関数の実部と虚部
\[
\sin z=\sin\left(\Re\left(z\right)\right)\cosh\left(\Im\left(z\right)\right)+i\cos\left(\Re\left(z\right)\right)\sinh\left(\Im\left(z\right)\right)
\]
三角関数と双曲線関数の実部と虚部
\[
\tan z=\frac{\sin\left(2\Re z\right)+i\sinh\left(2\Im z\right)}{\cos\left(2\Re z\right)+\cosh\left(2\Im z\right)}
\]