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*}
ページ情報
| タイトル | 巾関数と逆三角関数・逆双曲線関数の積の積分 |
| URL | https://www.nomuramath.com/c4qt8mvx/ |
| SNSボタン |
三角関数と双曲線関数の積分
\[
\int\cos xdx=\sin x
\]
1±itan(z)など
\[
1\pm i\tan z=\frac{1}{\cos\left(2\Re z\right)+\cosh\left(2\Im z\right)}\left(e^{\pm2i\Re z}+e^{\mp2\Im z}\right)
\]
三角関数の積
\[
\prod_{k=1}^{n-1}\sin\frac{k\pi}{n}=\frac{n}{2^{n-1}}
\]
三角関数を正接の半角、双曲線関数を双曲線正接の半角で表す。
\[
\sin z=\frac{2\tan\frac{z}{2}}{1+\tan^{2}\frac{z}{2}}
\]