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積分
\[
\int f(\cos x,\sin x)dx=\int f\left(\frac{1-t^{2}}{1+t^{2}},\frac{2t}{1+t^{2}}\right)\frac{2}{1+t^{2}}dt\cmt{t=\tan\frac{x}{2}}
\]
三角関数と双曲線関数の対数の積分
\[
\int\Log\sin^{\alpha}zdz=z\Log\sin^{\alpha}x+\frac{i\alpha}{2}z^{2}+\alpha z\Li_{1}\left(e^{2iz}\right)+\frac{i\alpha}{2}\Li_{2}\left(e^{2iz}\right)+\C{}
\]
偏角の3角関数
\[
\sin\Arg z=\frac{\Im z}{\left|z\right|}
\]
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。
\[
\sin z=\frac{2\tan\frac{z}{2}}{1+\tan^{2}\frac{z}{2}}
\]