分母に1乗と2乗ルートの積分
分母に1乗と2乗ルートの積分
(1)
\[ \int\frac{1}{\left(z\pm1\right)\sqrt{z^{2}-1}}dz=\frac{\sqrt{z^{2}-1}}{\pm z+1}+C \](2)
\begin{align*} \int\frac{1}{\left(z\pm1\right)\sqrt{z^{2}+1}}dz & =-\sqrt{2}\tanh^{\bullet}\pm\frac{\pm z+1-\sqrt{z^{2}+1}}{\sqrt{2}z}+C\\ & =-\frac{1}{\sqrt{2}}\tanh^{\bullet}\left(\frac{1\mp z}{\sqrt{2}\sqrt{z^{2}+1}}\right) \end{align*}(1)
\begin{align*} \int\frac{1}{\left(z+1\right)\sqrt{z^{2}-1}}dz & =\int\frac{1}{\left(\cosh t+1\right)\sqrt{\cosh^{2}t-1}}\sinh tdt\cnd{z=\cosh t}\\ & =\int\frac{1}{\left(\cosh t+1\right)\sqrt{\sinh^{2}t}}\sinh tdt\\ & =\frac{\sinh t}{\sqrt{\sinh^{2}t}}\int\frac{1}{\cosh t+1}dt\\ & =\frac{\sinh t}{\sqrt{\sinh^{2}t}}\int\frac{1}{2\cosh^{2}\frac{t}{2}}dt\\ & =\frac{\sinh t}{\sqrt{\sinh^{2}t}}\tanh\frac{t}{2}+C\\ & =\frac{2\sinh^{2}\frac{t}{2}}{\sqrt{\sinh^{2}t}}+C\\ & =\frac{\cosh t-1}{\sqrt{\sinh^{2}t}}+C\\ & =\frac{z-1}{\sqrt{\sinh^{2}\cosh^{\bullet}z}}+C\\ & =\frac{z-1}{\sqrt{\left(\sqrt{\frac{z-1}{z+1}}\left(z+1\right)\right)^{2}}}+C\\ & =\frac{z-1}{\sqrt{\left(z-1\right)\left(z+1\right)}}+C\\ & =\frac{\left(z-1\right)\left(z+1\right)}{\left(z+1\right)\sqrt{\left(z-1\right)\left(z+1\right)}}+C\\ & =\frac{\sqrt{z^{2}-1}}{z+1}+C \end{align*} \begin{align*} \int\frac{1}{\left(z-1\right)\sqrt{z^{2}-1}}dz & =\int\frac{1}{\left(\left(-z\right)+1\right)\sqrt{\left(-z\right)^{2}-1}}d\left(-z\right)\\ & =\frac{\sqrt{\left(-z\right)^{2}-1}}{\left(-z\right)+1}+C\\ & =\frac{\sqrt{z^{2}-1}}{-z+1}+C \end{align*} これより、\begin{align*} \int\frac{1}{\left(z\pm1\right)\sqrt{z^{2}-1}}dz & =\frac{\sqrt{z^{2}-1}}{\pm z+1}+C \end{align*}
(2)
\begin{align*} \int\frac{1}{\left(z+1\right)\sqrt{z^{2}+1}}dz & =\int\frac{1}{\left(\tan t+1\right)\sqrt{\tan^{2}t+1}}\frac{1}{\cos^{2}t}dt\cnd{z=\tan t}\\ & =\frac{\sqrt{\cos^{2}t}}{\cos t}\int\frac{1}{\left(\tan t+1\right)\cos t}dt\\ & =\frac{\sqrt{\cos^{2}t}}{\cos t}\int\frac{1}{\left(\sin t+\cos t\right)}dt\\ & =-\frac{\sqrt{\cos^{2}t}}{\cos t}\frac{2}{\sqrt{2}}\tanh^{\bullet}\frac{1-\tan\frac{t}{2}}{\sqrt{2}}+C\\ & =-\frac{\sqrt{\cos^{2}t}}{\cos t}\frac{2}{\sqrt{2}}\tanh^{\bullet}\frac{1-\frac{\sin t}{1+\cos t}}{\sqrt{2}}+C\\ & =-\frac{\sqrt{\cos^{2}t}}{\cos t}\frac{2}{\sqrt{2}}\tanh^{\bullet}\frac{1+\cos t-\sin t}{\sqrt{2}\left(1+\cos t\right)}+C\\ & =-\sqrt{2}\tanh^{\bullet}\frac{\frac{1-z}{\sqrt{z^{2}+1}}+1}{\sqrt{2}\left(\frac{1}{\sqrt{z^{2}+1}}+1\right)}+C\\ & =-\sqrt{2}\tanh^{\bullet}\frac{1+\sqrt{z^{2}+1}-z}{\sqrt{2}\left(1+\sqrt{z^{2}+1}\right)}+C\\ & =-\sqrt{2}\tanh^{\bullet}\frac{-z^{2}-z\left(1-\sqrt{z^{2}+1}\right)}{\sqrt{2}\left(-z^{2}\right)}+C\\ & =-\sqrt{2}\tanh^{\bullet}\frac{z+1-\sqrt{z^{2}+1}}{\sqrt{2}z}+C \end{align*} または、\[ \int\frac{1}{\left(z+1\right)\sqrt{z^{2}+1}}dz=-\frac{1}{\sqrt{2}}\tanh^{\bullet}\left(\frac{1-z}{\sqrt{2}\sqrt{z^{2}+1}}\right) \] でもよい(略)。
\begin{align*} \int\frac{1}{\left(z-1\right)\sqrt{z^{2}+1}}dz & =\int\frac{1}{\left(\left(-z\right)+1\right)\sqrt{\left(-z\right)^{2}+1}}d\left(-z\right)\\ & =-\sqrt{2}\tanh^{\bullet}\frac{\left(-z\right)+1-\sqrt{\left(-z\right)^{2}+1}}{\sqrt{2}\left(-z\right)}+C\\ & =-\sqrt{2}\tanh^{\bullet}-\frac{-z+1-\sqrt{z^{2}+1}}{\sqrt{2}z}+C \end{align*} または、
\[ \int\frac{1}{\left(z-1\right)\sqrt{z^{2}+1}}dz=-\frac{1}{\sqrt{2}}\tanh^{\bullet}\left(\frac{1+z}{\sqrt{2}\sqrt{z^{2}+1}}\right) \] でもよい(略)。
これより、
\[ \int\frac{1}{\left(z\pm1\right)\sqrt{z^{2}+1}}dz=-\sqrt{2}\tanh^{\bullet}\pm\frac{\pm z+1-\sqrt{z^{2}+1}}{\sqrt{2}z}+C \] または(略)、
\[ \int\frac{1}{\left(z\pm1\right)\sqrt{z^{2}+1}}dz=-\frac{1}{\sqrt{2}}\tanh^{\bullet}\left(\frac{1\mp z}{\sqrt{2}\sqrt{z^{2}+1}}\right) \]
ページ情報
タイトル | 分母に1乗と2乗ルートの積分 |
URL | https://www.nomuramath.com/s119x15e/ |
SNSボタン |
イータ関数の導関数がでてきます
\[
\int_{0}^{\infty}\frac{\log x}{1+e^{x}}dx=?
\]
sinの3乗をxの2乗で割った定積分
\[
\int_{0}^{\infty}\frac{\sin^{3}x}{x^{2}}dx=?
\]
分母に正接がある関数の定積分
\[
\int_{0}^{\frac{\pi}{2}}\frac{x}{\tan x}dx=?
\]
ガンマ関数を2つ含む定積分でカタラン定数が出てきます
\[
\int_{0}^{\frac{1}{2}}\Gamma\left(1-x\right)\Gamma\left(1+x\right)dx=?
\]