√x/(1-x)log(x/1-x)[0,1]の定積分

\begin{alignat}{2}
&(1) \displaystyle\int_0^1 \frac{1}{\sqrt{1-x}} \log \left(\frac{x}{1-x}\right)dx=4 \log 2\\
&(2) \displaystyle\int_0^1 \sqrt{\frac{x}{1-x}} \log \left(\frac{x}{1-x}\right)dx=π\\
&(3) \displaystyle\int_0^1 \frac{x}{\sqrt{1-x}} \log \left(\frac{x}{1-x}\right)dx=\frac{4}{3}(2 \log 2+1)\\
&(4) \displaystyle\int_0^1 \frac{x^2}{\sqrt{1-x}} \log \left(\frac{x}{1-x}\right)dx=\frac{8}{15}(4 \log 2+3)
\end{alignat}






<証明>

\((1)\) \(\sqrt{1-x}=t\) と置きます。\((1-x=t^2, dx=-2tdt)\)
\begin{alignat}{2}
&  \displaystyle\int_0^1 \frac{1}{\sqrt{1-x}} \log \left(\frac{x}{1-x}\right)dx\\
&=\displaystyle\int_1^0 \frac{1}{t} \log \left(\frac{1-t^2}{t^2}\right)(-2t)dt\\
&=2\displaystyle\int_0^1 \{\log (1-t^2)-2 \log t\}dt\\
&=2 \displaystyle\int_0^1 \{\log (1+t)+\log (1-t)-2 \log t\}dt\\
&=2[(1+t)\log (1+t)-t-(1-t) \log (1-t)-t-2t \log t +2t]_0^1\\
&=2 (2 \log 2-1-1+2)=4 \log 2
\end{alignat}以上より$$\displaystyle\int_0^1 \frac{1}{\sqrt{1-x}} \log \left(\frac{x}{1-x}\right)dx=4 \log 2$$






\((2)\) \(\displaystyle \sqrt{\frac{x}{1-x}}=t\) と置くと
\begin{alignat}{2}
&\frac{x}{1-x}=t^2,   x=t^2(1-x),  x=t^2-t^2x\\
&x+t^2x=t^2,  x(t^2+1)=t^2,  x=\frac{t^2}{t^2+1}\\
&x=1-\frac{1}{t^2+1},  dx=\frac{2t}{(t^2+1)^2}dt\\
\end{alignat}
となるので
\begin{alignat}{2}
&\displaystyle\int_0^1 \sqrt{\frac{x}{1-x}} \log \left(\frac{x}{1-x}\right)dx=\displaystyle\int_0^{\infty} t \log t^2 \cdot \frac{2t}{(t^2+1)^2}dt\\
&                       =-2\displaystyle\int_0^{\infty} t \log t \cdot \frac{-2t}{(t^2+1)^2}dt\\
&                       =-2\displaystyle\int_0^{\infty} t \log t \left(\frac{1}{t^2+1}\right)’dt
\end{alignat}部分積分を行います。
\begin{alignat}{2}
&=-2\left\{\left[\frac{t \log t}{t^2+1}\right]_0^{\infty}-\displaystyle\int_0^{\infty} \frac{\log t+1}{t^2+1}dt\right\}\\
&=2\displaystyle\int_0^{\infty} \left(\frac{\log t}{t^2+1}+\frac{1}{t^2+1}\right)dt
\end{alignat}
このとき左の積分は下記のようになります。(詳細はこちらです)$$\displaystyle\int_0^{\infty} \frac{\log t}{t^2+1}dt=0$$
よって$$=2\displaystyle\int_0^{\infty}\frac{1}{t^2+1}dt=2 \cdot \frac{π}{2}=π$$
以上より$$\displaystyle\int_0^1 \sqrt{\frac{x}{1-x}} \log \left(\frac{x}{1-x}\right)dx=π$$






\((3)\) \(\sqrt{1-x}=t\) と置きます。\((1-x=t^2, dx=-2tdt)\)
\begin{alignat}{2}
& \displaystyle\int_0^1 \frac{x}{\sqrt{1-x}} \log \left(\frac{x}{1-x}\right)dx\\
&=\displaystyle\int_1^0 \frac{1-t^2}{t}\log \left(\frac{1-t^2}{t^2}\right)(-2t)dt\\
&=2\displaystyle\int_0^1 (1-t^2)\{\log (1-t^2)-2 \log t\}dt\\
&=2\displaystyle\int_0^1 (1-t^2)\log (1-t^2)dt -4 \displaystyle\int_0^1 (1-t^2)\log tdt\\
\end{alignat}左右の積分をそれぞれ計算します。

\((A)\) 左側の積分

先に不定積分を計算します。
\begin{alignat}{2}
&  \displaystyle\int (1-t^2)\log (1-t^2)dt\\
&=\left(t-\frac{1}{3}t^3\right)\log (1-t^2) -\displaystyle\int \left(t-\frac{1}{3}t^3\right) \cdot \frac{-2t}{1-t^2}dt\\
&=\left(t-\frac{1}{3}t^3\right)\log (1-t^2) -\displaystyle\int \left(\frac{2t^2}{1-t^2}-\frac{2}{3} \cdot \frac{t^4}{1-t^2}\right) dt\\
&=\left(t-\frac{1}{3}t^3\right)\log (1-t^2) -\displaystyle\int \left\{-2+\frac{2}{1-t^2}-\frac{2}{3}\left(-t^2-1+\frac{1}{1-t^2}\right)\right\} dt\\
&=\left(t-\frac{1}{3}t^3\right)\log (1-t^2) -\displaystyle\int \left(\frac{2}{3}t^2-\frac{4}{3}+\frac{4}{3}\cdot \frac{1}{1-t^2}\right) dt\\
&=\left(t-\frac{1}{3}t^3\right)\log (1-t^2)+\frac{2}{9}t^3-\frac{4}{3}t+\frac{2}{3}\log \frac{1+t}{1-t}\\
&=\left(t-\frac{1}{3}t^3\right)\log (1-t)(1+t)+\frac{2}{9}t^3-\frac{4}{3}t+\frac{2}{3}\log(1+t)-\frac{2}{3}\log (1-t)\\
&=\left(\frac{2}{3}+t-\frac{1}{3}t^3\right)\log (1+t) +\left(-\frac{2}{3}+t-\frac{1}{3}t^3\right)\log (1-t)+\frac{2}{9}t^3-\frac{4}{3}t\\
\end{alignat}となるので
\begin{alignat}{2}
&  \displaystyle\int_0^1 (1-t^2)\log (1-t^2)dt\\
&=\left[\left(\frac{2}{3}+t-\frac{1}{3}t^3\right)\log (1+t) +\left(-\frac{2}{3}+t-\frac{1}{3}t^3\right)\log (1-t)+\frac{2}{9}t^3-\frac{4}{3}t\right]_0^1\\
&=\left(\frac{2}{3}+1-\frac{1}{3}\right)\log 2+\frac{2}{9}-\frac{4}{3}=\frac{4}{3}\log 2-\frac{10}{9}
\end{alignat}
\((B)\) 右側の積分
\begin{alignat}{2}
&  \displaystyle\int_0^1 (1-t^2)\log tdt\\
&=\left[\left(t-\frac{1}{3}t^3\right)\log t\right]_0^1-\displaystyle\int_0^1 \left(t-\frac{1}{3}t^3\right) \cdot \frac{1}{t}dt\\
&=-\displaystyle\int_0^1 \left(1-\frac{1}{3}t^2\right)dt=-\left[t-\frac{1}{9}t^3\right]_0^1=-\left(1-\frac{1}{9}\right)=-\frac{8}{9}\\
\end{alignat}よって
\begin{alignat}{2}
&\displaystyle\int_0^1 \frac{x}{\sqrt{1-x}} \log \left(\frac{x}{1-x}\right)dx=2\left(\frac{4}{3}\log 2-\frac{10}{9}\right)-4\left(-\frac{8}{9}\right)\\
&                       =\frac{8}{3}\log 2 -\frac{20}{9}+\frac{32}{9}\\
&                       =\frac{8}{3}\log 2+\frac{4}{3}=\frac{4}{3}(2 \log 2+1)\\
\end{alignat}以上より$$\displaystyle\int_0^1 \frac{x}{\sqrt{1-x}} \log \left(\frac{x}{1-x}\right)dx=\frac{4}{3}(2 \log 2+1)$$







\((4)\) \(\sqrt{1-x}=t\) と置きます。\((1-x=t^2, dx=-2tdt)\)
\begin{alignat}{2}
& \displaystyle\int_0^1 \frac{x^2}{\sqrt{1-x}} \log \left(\frac{x}{1-x}\right)dx\\
&=\displaystyle\int_1^0 \frac{(1-t^2)^2}{t}\log \left(\frac{1-t^2}{t^2}\right)(-2t)dt\\
&=2\displaystyle\int_0^1 (1-t^2)^2\{\log (1-t^2)-2 \log t\}dt\\
&=2\displaystyle\int_0^1 (1-t^2)^2\log (1-t^2)dt -4 \displaystyle\int_0^1 (1-t^2)\log tdt\\
\end{alignat}左右の積分をそれぞれ計算します。

\((A)\) 左側の積分

先に不定積分を計算します。
\begin{alignat}{2}
&  \displaystyle\int (1-t^2)^2\log (1-t^2)dt\\
&=\displaystyle\int (1-2t^2+t^4)\log (1-t^2)dt\\
&=\left(t-\frac{2}{3}t^3+\frac{1}{5}t^5\right)\log (1-t^2) -\displaystyle\int \left(t-\frac{2}{3}t^3+\frac{1}{5}t^5\right) \cdot \frac{-2t}{1-t^2}dt\\
&=\left(t-\frac{2}{3}t^3+\frac{1}{5}t^5\right)\log (1-t^2) -\displaystyle\int \left(-\frac{2}{5}t^6+\frac{4}{3}t^4-2t^2\right)\cdot \frac{1}{1-t^2} dt\\
&=\left(t-\frac{2}{3}t^3+\frac{1}{5}t^5\right)\log (1-t^2) -\displaystyle\int \left(\frac{2}{5}t^4-\frac{14}{15}t^2+\frac{16}{15}-\frac{16}{15}\cdot \frac{1}{1-t^2}\right) dt\\
&=\left(t-\frac{2}{3}t^3+\frac{1}{5}t^5\right)\log (1-t^2) -\frac{2}{25}t^5+\frac{14}{45}t^3-\frac{16}{15}t+\frac{8}{15}\log \frac{1+t}{1-t}\\
&=\left(t-\frac{2}{3}t^3+\frac{1}{5}t^5\right)\log (1-t)(1+t)-\frac{2}{25}t^5+\frac{14}{45}t^3-\frac{16}{15}t+\frac{8}{15}\log (1+t)-\frac{8}{15} \log (1-t)\\
&=\left(\frac{8}{15}+t-\frac{2}{3}t^3+\frac{1}{5}t^5\right)\log (1+t)+\left(-\frac{8}{15}+t-\frac{2}{3}t^3+\frac{1}{5}t^5\right)\log (1-t)-\frac{2}{25}t^5+\frac{14}{45}t^3-\frac{16}{15}t\\
\end{alignat}となるので
\begin{alignat}{2}
&  \displaystyle\int_0^1 (1-t^2)^2\log (1-t^2)dt\\
&=\left[\left(\frac{8}{15}+t-\frac{2}{3}t^3+\frac{1}{5}t^5\right)\log (1+t)+\left(-\frac{8}{15}+t-\frac{2}{3}t^3+\frac{1}{5}t^5\right)\log (1-t)-\frac{2}{25}t^5+\frac{14}{45}t^3-\frac{16}{15}t\right]_0^1\\
&=\left(\frac{8}{15}+1-\frac{2}{3}+\frac{1}{5}\right)\log 2-\frac{2}{25}+\frac{14}{45}-\frac{16}{15}=\frac{16}{15}\log 2-\frac{188}{225}
\end{alignat}
\((B)\) 右側の積分
\begin{alignat}{2}
&\displaystyle\int_0^1 (1-t^2)^2\log tdt=\displaystyle\int_0^1 (1-2t^2+t^4)\log tdt\\
&                =\left[\left(t-\frac{2}{3}t^3+\frac{1}{5}t^5\right)\log t\right]_0^1-\displaystyle\int_0^1 \left(t-\frac{2}{3}t^3+\frac{1}{5}t^5\right) \cdot \frac{1}{t}dt\\
&                =-\displaystyle\int_0^1 \left(1-\frac{2}{3}t^2+\frac{1}{5}t^4\right)dt=-\left[t-\frac{2}{9}t^3+\frac{1}{25}t^5\right]_0^1\\
&                =-\left(1-\frac{2}{9}+\frac{1}{25}\right)=-\frac{184}{225}\\
\end{alignat}よって
\begin{alignat}{2}
&\displaystyle\int_0^1 \frac{x^2}{\sqrt{1-x}} \log \left(\frac{x}{1-x}\right)dx=2\left(\frac{16}{15}\log 2-\frac{188}{225}\right)-4\left(-\frac{184}{225}\right)\\
&                       =\frac{32}{15}\log 2 -\frac{376}{225}+\frac{736}{225}\\
&                       =\frac{32}{15}\log 2+\frac{8}{5}=\frac{8}{15}(4 \log 2+3)\\
\end{alignat}以上より$$\displaystyle\int_0^1 \frac{x^2}{\sqrt{1-x}} \log \left(\frac{x}{1-x}\right)dx=\frac{8}{15}(4 \log 2+3)$$

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