x^{3n}/[3]√(1-x^3)[0,1]などの定積分

\begin{alignat}{2}
&(1)  \displaystyle\int_0^1 \frac{1}{\sqrt[3]{1-x^3}}dx=\frac{2π}{3\sqrt{3}}\\
&(2)  \displaystyle\int_0^1 \frac{x^{2n}}{\sqrt{1-x^2}}dx=\frac{(2n-1)!!}{(2n)!!}\cdot \frac{π}{2}\\
&(3)  \displaystyle\int_0^1 \frac{x^{2n+1}}{\sqrt{1-x^2}}dx=\frac{(2n)!!}{(2n+1)!!}\\
&(4)  \displaystyle\int_0^1 \frac{x^{3n}}{\sqrt[3]{1-x^3}}dx=\frac{1 \cdot 4 \cdot 7 \cdots (3n-2)}{3 \cdot 6 \cdot 9 \cdots (3n)}\cdot \frac{2π}{3\sqrt{3}}\\
&(5)  \displaystyle\int_0^1 \frac{x^{mn}}{\sqrt[m]{1-x^m}}dx=\frac{π}{n!m^{n+1} \sin\frac{π}{m}}\displaystyle\prod_{k=1}^n \{(k-1)m+1\}
\end{alignat}







<証明> 

\((1)\) と \((4)\) では \(x^3=t\) と置きます。\(\displaystyle \left(3x^2dx=dt, dx=\frac{1}{3}t^{-\frac{2}{3}}dt\right)\)
\((2)\) と \((3)\) では \(x^2=t\) と置きます。 \(\displaystyle \left(2xdx=dt, dx=\frac{1}{2}t^{-\frac{1}{2}}dt\right)\)
\begin{alignat}{2}
(1) \displaystyle\int_0^1 \frac{1}{\sqrt[3]{1-x^3}}dx&=\displaystyle\int_0^1 \frac{1}{(1-t)^{\frac{1}{3}}}\cdot \frac{1}{3}t^{-\frac{2}{3}}dt=\frac{1}{3}\displaystyle\int_0^1 t^{-\frac{2}{3}}(1-t)^{-\frac{1}{3}}dt\\
&=\frac{1}{3}B\left(\frac{1}{3},\frac{2}{3}\right)=\frac{1}{3}\cdot \frac{Γ\left(\frac{1}{3}\right) Γ\left(\frac{2}{3}\right)}{Γ(1)} \\
&=\frac{1}{3}Γ\left(\frac{1}{3}\right)Γ\left(1-\frac{1}{3}\right)=\frac{1}{3}\cdot \frac{π}{ \sin \frac{π}{3}}=\frac{2π}{3\sqrt{3}}
\end{alignat}以上より$$\displaystyle\int_0^1 \frac{1}{\sqrt[3]{1-x^3}}dx=\frac{2π}{3\sqrt{3}}$$







\begin{alignat}{2}
(2)  \displaystyle\int_0^1 \frac{x^{2n}}{\sqrt{1-x^2}}dx&=\displaystyle\int_0^1 \frac{t^n}{\sqrt{1-t}}\frac{1}{2}t^{-\frac{1}{2}}dt=\frac{1}{2}\displaystyle\int_0^1 t^{n-\frac{1}{2}}(1-t)^{-\frac{1}{2}}dt\\
&=\frac{1}{2}B\left(n+\frac{1}{2},\frac{1}{2}\right)=\frac{1}{2}\cdot \frac{Γ\left(n+\frac{1}{2}\right)Γ\left(\frac{1}{2}\right)}{Γ(n+1)}\\
&=\frac{\sqrt{π}}{2 \cdot n!}\cdot \frac{(2n-1)!!}{2^n}\sqrt{π}= \frac{(2n-1)!!}{(2n)!!}\cdot \frac{π}{2}\\
\end{alignat}以上より$$\displaystyle\int_0^1 \frac{x^{2n}}{\sqrt{1-x^2}}dx=\frac{(2n-1)!!}{(2n)!!}\cdot \frac{π}{2}$$







\begin{alignat}{2}
(3)  \displaystyle\int_0^1 \frac{x^{2n+1}}{\sqrt{1-x^2}}dx&=\displaystyle\int_0^1 \frac{t^{n+\frac{1}{2}}}{\sqrt{1-t}} \cdot \frac{1}{2}t^{-\frac{1}{2}}dt=\frac{1}{2}\displaystyle\int_0^1 t^n(1-t)^{-\frac{1}{2}}dt\\
&=\frac{1}{2}B\left(n+1, \frac{1}{2}\right)=\frac{1}{2} \cdot \frac{Γ(n+1)Γ\left(\frac{1}{2}\right)}{Γ\left(n+\frac{3}{2}\right)}\\
&=\frac{1}{2} \cdot n! \cdot \sqrt{π} \cdot \frac{2}{2n+1} \cdot \frac{2^n}{(2n-1)!!} \cdot \frac{1}{\sqrt{π}}=\frac{(2n)!!}{(2n+1)!!}\\
\end{alignat}以上より$$\displaystyle\int_0^1 \frac{x^{2n+1}}{\sqrt{1-x^2}}dx=\frac{(2n)!!}{(2n+1)!!}$$







\begin{alignat}{2}
(4)  \displaystyle\int_0^1 \frac{x^{3n}}{\sqrt[3]{1-x^3}}dx&=\displaystyle\int_0^1 \frac{t^n}{\sqrt[3]{1-t}}\cdot \frac{1}{3}t^{-\frac{2}{3}}dt=\frac{1}{3}\displaystyle\int_0^1 t^{n-\frac{2}{3}}(1-t)^{-\frac{1}{3}}dt\\
&=\frac{1}{3}B\left(n+\frac{1}{3},\frac{2}{3}\right)=\frac{1}{3}\cdot \frac{Γ\left(n+\frac{1}{3}\right)Γ\left(\frac{2}{3}\right)}{Γ(n+1)}\\
&=\frac{1}{3 \cdot n!}Γ\left(\frac{2}{3}\right)Γ\left(n+\frac{1}{3}\right)\\
&=\frac{1}{3 \cdot n!}Γ\left(\frac{2}{3}\right) \frac{3n-2}{3}\cdot \frac{3n-5}{3}\cdot \frac{3n-8}{3} \cdots \frac{4}{3} \cdot \frac{1}{3}Γ\left(\frac{1}{3}\right)\\
&=\frac{1}{3 \cdot n!}\cdot \frac{1 \cdot 4 \cdot 7 \cdots (3n-2)}{3^n} Γ\left(\frac{1}{3}\right) Γ\left(\frac{2}{3}\right)\\
&=\frac{1 \cdot 4 \cdot 7 \cdots (3n-2)}{3 \cdot 6 \cdot 9 \cdots (3n)}\cdot \frac{2π}{3\sqrt{3}}
\end{alignat}以上より$$\displaystyle\int_0^1 \frac{x^{3n}}{\sqrt[3]{1-x^3}}dx=\frac{1 \cdot 4 \cdot 7 \cdots (3n-2)}{3 \cdot 6 \cdot 9 \cdots (3n)}\cdot \frac{2π}{3\sqrt{3}}$$







\((5)\) \(x^m=t\) と置きます。\((mx^{m-1}dx=dt, x=t^{\frac{1}{m}}, x^{m-1}=t^{\frac{m-1}{m}})\)
\begin{alignat}{2}
\displaystyle\int_0^1 \frac{x^{mn}}{\sqrt[m]{1-x^m}}dx&=\displaystyle\int_0^1 \frac{t^n}{(1-t)^{\frac{1}{m}}}\cdot \frac{1}{mx^{m-1}}dt\\
&=\frac{1}{m}\displaystyle\int_0^1 t^n(1-t)^{-\frac{1}{m}}t^{-\frac{m-1}{m}}dt=\frac{1}{m}\displaystyle\int_0^1 t^{n-1+\frac{1}{m}}(1-t)^{-\frac{1}{m}}dt\\
&=\frac{1}{m}B\left(n+\frac{1}{m},1-\frac{1}{m}\right)=\frac{1}{m}\frac{Γ\left(n+\frac{1}{m}\right)Γ\left(1-\frac{1}{m}\right)}{Γ(n+1)}\\
&=\frac{1}{m \cdot n!}\cdot \frac{(n-1)m+1}{m}\cdot \frac{(n-2)m+1}{m}   \cdots   \frac{m+1}{m}\cdot \frac{1}{m}Γ\left(\frac{1}{m}\right)Γ\left(1-\frac{1}{m}\right)\\
&=\frac{1}{m\cdot n!}\cdot \frac{1}{m^n} \displaystyle\prod_{k=1}^n \{(k-1)m+1\}Γ\left(\frac{1}{m}\right)Γ\left(1-\frac{1}{m}\right)\\
&=\frac{π}{n!m^{n+1} \sin \frac{π}{m}}\displaystyle\prod_{k=1}^n \{(k-1)m+1\}
\end{alignat}以上より$$\displaystyle\int_0^1 \frac{x^{mn}}{\sqrt[m]{1-x^m}}dx=\frac{π}{n!m^{n+1} \sin\frac{π}{m}}\displaystyle\prod_{k=1}^n \{(k-1)m+1\}$$

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