x^2(1-x)/(1+x)(1+x^2)logx[0,1]などの定積分

\begin{alignat}{2}
&(1) \displaystyle\int_0^1 \frac{1-x}{(1+x)(1+x^2)\log x}dx=-\frac{\log 2}{2}\\
&(2) \displaystyle\int_0^1 \frac{x^2(1-x)}{(1+x)(1+x^2)\log x}dx=\log \frac{2\sqrt{2}}{π}\\
\end{alignat}








<証明>

\((1)(2)\) の被積分関数に次の等式を代入します。$$\displaystyle\int_0^1 x^a da=\left[\frac{x^a}{\log x}\right]_0^1=\frac{x-1}{\log x}$$


\begin{alignat}{2}
&(1) \displaystyle\int_0^1 \frac{1-x}{(1+x)(1+x^2)\log x}dx=-\displaystyle\int_0^1 \displaystyle\int_0^1 \frac{x^a}{(1+x)(1+x^2)}dadx\\
&=-\displaystyle\int_0^1 \displaystyle\int_0^1 \frac{(1-x)x^a}{(1-x^2)(1+x^2)}dadx=\displaystyle\int_0^1 \displaystyle\int_0^1 \frac{x^{a+1}-x^a}{1-x^4}dxda\\
\end{alignat}\(x^4=t\) と置きます。\((4x^3dx=dt)\)
\begin{alignat}{2}
&=\displaystyle\int_0^1 \displaystyle\int_0^1 \frac{t^{\frac{a+1}{4}}-t^{\frac{a}{4}}}{1-t} \cdot \frac{1}{4t^{\frac{3}{4}}}dtda=\frac{1}{4}\displaystyle\int_0^1 \displaystyle\int_0^1 \frac{t^{\frac{a-2}{4}}-t^{\frac{a-3}{4}}}{1-t}dtda\\
&=\frac{1}{4}\displaystyle\int_0^1 \left\{ψ\left(\frac{a+1}{4}\right)-ψ\left(\frac{a+2}{4}\right)\right\}da\\
&=\left[\log Γ\left(\frac{a+1}{4}\right)-\log Γ\left(\frac{a+2}{4}\right)\right]_0^1\\
&=\log \frac{Γ\left(\frac{1}{2}\right)Γ\left(\frac{1}{2}\right)}{Γ\left(\frac{3}{4}\right)Γ\left(\frac{1}{4}\right)}=\log π \cdot \frac{\sin \frac{π}{4}}{π}=\log \frac{1}{\sqrt{2}}=-\frac{1}{2}\log 2
\end{alignat}以上より$$\displaystyle\int_0^1 \frac{1-x}{(1+x)(1+x^2)\log x}dx=-\frac{\log 2}{2}$$









\begin{alignat}{2}
&(2) \displaystyle\int_0^1 \frac{x^2(1-x)}{(1+x)(1+x^2)\log x}dx=-\displaystyle\int_0^1 \displaystyle\int_0^1 \frac{x^{a+2}}{(1+x)(1+x^2)}dadx\\
&=-\displaystyle\int_0^1 \displaystyle\int_0^1 \frac{(1-x)x^{a+2}}{(1-x^2)(1+x^2)}dadx=\displaystyle\int_0^1 \displaystyle\int_0^1 \frac{x^{a+3}-x^{a+2}}{1-x^4}dxda\\
\end{alignat}\(x^4=t\) と置きます。\((4x^3dx=dt)\)
\begin{alignat}{2}
&=\displaystyle\int_0^1 \displaystyle\int_0^1 \frac{t^{\frac{a+3}{4}}-t^{\frac{a+2}{4}}}{1-t} \cdot \frac{1}{4t^{\frac{3}{4}}}dtda=\frac{1}{4}\displaystyle\int_0^1 \displaystyle\int_0^1 \frac{t^{\frac{a}{4}}-t^{\frac{a-1}{4}}}{1-t}dtda\\
&=\frac{1}{4}\displaystyle\int_0^1 \left\{ψ\left(\frac{a+3}{4}\right)-ψ\left(\frac{a}{4}+1\right)\right\}da\\
&=\left[\log Γ\left(\frac{a+3}{4}\right)-\log Γ\left(\frac{a}{4}+1\right)\right]_0^1\\
&=\log \frac{Γ(1)Γ(1)}{Γ\left(\frac{5}{4}\right)Γ\left(\frac{3}{4}\right)}=\log \frac{1}{\frac{1}{4}Γ\left(\frac{1}{4}\right)Γ\left(\frac{3}{4}\right)}=\log 4 \cdot \frac{\sin \frac{π}{4}}{π}=\log \frac{2\sqrt{2}}{π}
\end{alignat}以上より$$\displaystyle\int_0^1 \frac{x^2(1-x)}{(1+x)(1+x^2)\log x}dx=\log \frac{2\sqrt{2}}{π}$$

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