sin^{-1}x/x(1+px^2)[0,1]などの定積分

\begin{alignat}{2}
&(1)  \displaystyle\int_0^1 \frac{\sin^{-1} x}{x}dx=\frac{π}{2}\log 2\\
&(2)  \displaystyle\int_0^1 \frac{\cot^{-1} x}{1+x}dx=-\frac{π}{2}\log 2+2G\\
&(3)  \displaystyle\int_0^1 \frac{\cot^{-1} x}{1-x}dx=\frac{π}{2}\log 2+2G\\
&(4)  \displaystyle\int_0^1 \frac{\sin^{-1} x}{x(1+px^2)}dx=\frac{π}{2}\log \frac{1+\sqrt{1+p}}{\sqrt{1+p}}  (p \gt -1)\\
\end{alignat}





<証明>

次の定積分の結果を用います。[詳細はこちらです。(A)(B)(C)]
\begin{alignat}{2}
&(A)  \displaystyle\int_0^{\frac{π}{2}} \log ( 1+\sin x) dx=-\frac{π}{2} \log 2 +2G\\
&(B)  \displaystyle\int_0^{\frac{π}{2}} \log ( 1- \sin x) dx=-\frac{π}{2} \log 2 -2G\\
&(C)  \displaystyle\int_0^{\frac{π}{2}} \log (1+a \sin^2 x)dx=π\log \frac{1+\sqrt{1+a}}{2}  (a \gt -1)\\
\end{alignat}






\((1)\) から \((3)\) まで部分積分を行ってから

\(x=\sin t\) と置きます。\((dx=\cos tdt)\)

\begin{alignat}{2}
(1)  \displaystyle\int_0^1 \frac{\sin^{-1} x}{x}dx&=[\log x \sin^{-1} x]_0^1-\displaystyle\int_0^1 \frac{\log x}{\sqrt{1-x^2}}dx=-\displaystyle\int_0^1 \frac{\log x}{\sqrt{1-x^2}}dx\\
&=-\displaystyle\int_0^{\frac{π}{2}} \frac{\log (\sin t)}{\sqrt{1-\sin^2 t}} \cdot \cos t dt=-\displaystyle\int_0^{\frac{π}{2}} \log (\sin t)dt=\frac{π}{2}\log 2\\
\end{alignat}以上より$$\displaystyle\int_0^1 \frac{\sin^{-1} x}{x}dx=\frac{π}{2}\log 2$$







\begin{alignat}{2}
(2)  \displaystyle\int_0^1 \frac{\cot^{-1} x}{1+x}dx&=[\log (1+x) \cos^{-1} x]_0^1 +\displaystyle\int_0^1 \frac{\log (1+x)}{\sqrt{1-x^2}}dx=\displaystyle\int_0^1 \frac{\log (1+x)}{\sqrt{1-x^2}}dx\\
&=\displaystyle\int_0^{\frac{π}{2}} \frac{\log (1+\sin t)}{\sqrt{1-\sin^2 t}} \cdot \cos tdt=\displaystyle\int_0^{\frac{π}{2}} \log (1+\sin t)dt=-\frac{π}{2} \log 2 +2G\\
\end{alignat}以上より$$\displaystyle\int_0^1 \frac{\cot^{-1} x}{1+x}dx=-\frac{π}{2}\log 2+2G$$








\begin{alignat}{2}
(3)  \displaystyle\int_0^1 \frac{\cot^{-1} x}{1-x}dx&=[-\log (1-x) \cos^{-1} x]_0^1 -\displaystyle\int_0^1 \frac{\log (1-x)}{\sqrt{1-x^2}}dx=-\displaystyle\int_0^1 \frac{\log (1+x)}{\sqrt{1-x^2}}dx\\
&=-\displaystyle\int_0^{\frac{π}{2}} \frac{\log (1-\sin t)}{\sqrt{1-\sin^2 t}} \cdot \cos tdt=-\displaystyle\int_0^{\frac{π}{2}} \log (1-\sin t)dt=\frac{π}{2} \log 2 +2G\\
\end{alignat}以上より$$\displaystyle\int_0^1 \frac{\cot^{-1} x}{1-x}dx=\frac{π}{2}\log 2+2G$$







\((4)\) 予め、次の積分を計算しておきます。$$\displaystyle\int \frac{1}{x(1+px^2)}dx=\displaystyle\int \left(\frac{1}{x}-\frac{px}{1+px^2}\right)dx=\log |x|-\frac{1}{2}\log |1+px^2|=\frac{1}{2}\log \left|\frac{x^2}{1+px^2}\right|$$この積分を用いて、部分積分を行います。
\begin{alignat}{2}
\displaystyle\int_0^1 \frac{\sin^{-1} x}{x(1+px^2)}dx&=\left[\sin^{-1}x \cdot \frac{1}{2}\log \left|\frac{x^2}{1+px^2}\right|\right]_0^1 -\displaystyle\int_0^1 \frac{1}{\sqrt{1-x^2}} \cdot \frac{1}{2}\log \left|\frac{x^2}{1+px^2}\right|dx\\
&=\frac{π}{4}\log \frac{1}{1+p}-\frac{1}{2}\displaystyle\int_0^1 \frac{2\log x- \log (1+px^2)}{\sqrt{1-x^2}}dx\\
&=\frac{π}{4}\log \frac{1}{1+p}-\displaystyle\int_0^1 \frac{\log x}{\sqrt{1-x^2}}dx+\frac{1}{2}\displaystyle\int_0^1 \frac{\log (1+px^2)}{\sqrt{1-x^2}}dx\\
\end{alignat}\(x=\sin t\) と置きます。\((dx=\cos tdt)\)
\begin{alignat}{2}
&=\frac{π}{4}\log \frac{1}{1+p} -\displaystyle\int_0^{\frac{π}{2}} \frac{\log (\sin t)}{\sqrt{1-\sin^2 t}} \cdot \cos tdt+\frac{1}{2}\displaystyle\int_0^{\frac{π}{2}} \frac{\log (1+p \sin^2 t)}{\sqrt{1-\sin^2 t}} \cdot \cos tdt\\
&=\frac{π}{4}\log \frac{1}{1+p} -\displaystyle\int_0^{\frac{π}{2}}\log (\sin t)dt+\frac{1}{2}\displaystyle\int_0^{\frac{π}{2}} \log (1+p \sin^2 t)dt\\
&=\frac{π}{4}\log \frac{1}{1+p}+\frac{π}{2}\log 2 +\frac{π}{2}\log \frac{1+\sqrt{1+p}}{2}\\
&=\frac{π}{2}\left(\log \frac{1}{\sqrt{1+p}}+\log 2+\log \frac{1+\sqrt{1+p}}{2}\right)=\frac{π}{2}\log \frac{1+\sqrt{1+p}}{\sqrt{1+p}}
\end{alignat}以上より$$\displaystyle\int_0^1 \frac{\sin^{-1} x}{x(1+px^2)}dx=\frac{π}{2}\log \frac{1+\sqrt{1+p}}{\sqrt{1+p}}  (p \gt -1)$$

コメントを残す

メールアドレスが公開されることはありません。 * が付いている欄は必須項目です