xlog(1+x)/√(1-x^2)[0,1]などの定積分

\begin{alignat}{2}
&(1)  \displaystyle\int_0^1 \frac{\log (1+x)}{\sqrt{1-x^2}}dx=-\frac{π}{2}\log 2+2G\\
&(2)  \displaystyle\int_0^1 \frac{\log (1-x)}{\sqrt{1-x^2}}dx=-\frac{π}{2}\log 2-2G\\
&(3)  \displaystyle\int_0^1 \frac{x\log (1+x)}{\sqrt{1-x^2}}dx=\frac{π}{2}-1\\
&(4)  \displaystyle\int_0^1 \frac{x\log (1-x)}{\sqrt{1-x^2}}dx=-\frac{π}{2}-1\\
&(5)  \displaystyle\int_{-a}^a \frac{\log (1+bx)}{\sqrt{a^2-x^2}}dx=π\log \frac{1+\sqrt{1-a^2b^2}}{2}  (ab \lt 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^π \log (a+b \cos x)dx=π\log \frac{a+\sqrt{a^2-b^2}}{2}  (a \gt b \gt 0)\\
\end{alignat}






\((1)\) から \((4)\) まで \(x=\sin t\) と置きます。\((dx= \cos tdt)\)

\begin{alignat}{2}
(1)  \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\\
&\\
(2)  \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}





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







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






\((5)\) \(x=a \cos t\) と置きます。\((dx=-a \sin tdt)\)
\begin{alignat}{2}
\displaystyle\int_{-a}^a \frac{\log (1+bx)}{\sqrt{a^2-x^2}}dx&=\displaystyle\int_π^0 \frac{\log (1+ab \cos t)}{\sqrt{a^2-a^2 \cos^2t}}\cdot (-a \sin t)dt\\
&=\displaystyle\int_0^π \log (1+ab \cos t)dt=π\log \frac{1+\sqrt{1-a^2b^2}}{2}
\end{alignat}以上より$$\displaystyle\int_{-a}^a \frac{\log (1+bx)}{\sqrt{a^2-x^2}}dx=π\log \frac{1+\sqrt{1-a^2b^2}}{2}$$

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