x^{3n}/[3]√(1-x^3)[0,1]などの定積分

\begin{alignat}{2}
&(1)  \displaystyle\int_0^1 \frac{x^{3n}}{\sqrt[3]{1-x^3}}dx=\frac{2π}{3\sqrt{3}}\cdot \frac{Γ\left(n+\frac{1}{3}\right)}{n!Γ\left(\frac{1}{3}\right)}\\
&(2)  \displaystyle\int_0^1 \frac{x^{3n-1}}{\sqrt[3]{1-x^3}}dx=\frac{1}{3}\cdot \frac{(n-1)!Γ\left(\frac{2}{3}\right)}{Γ\left(n+\frac{2}{3}\right)}\\
&(3)  \displaystyle\int_0^1 \frac{x^{3n-2}}{\sqrt[3]{1-x^3}}dx=\frac{1}{3}\cdot \frac{Γ\left(n-\frac{1}{3}\right)Γ\left(\frac{2}{3}\right)}{Γ\left(n+\frac{1}{3}\right)}\\
\end{alignat}ただし、全て \(n \in \mathrm{N}\)








<証明>

全て \(x^3=t\) と置きます。\((3x^2dx=dt)\)

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







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







\begin{alignat}{2}
&(2) \displaystyle\int_0^1 \frac{x^{3n-2}}{\sqrt[3]{1-x^3}}dx=\displaystyle\int_0^1 \frac{t^{n-\frac{2}{3}}}{\sqrt[3]{1-t}}\cdot \frac{1}{3t^{\frac{2}{3}}}dt=\frac{1}{3}\displaystyle\int_0^1 t^{n-\frac{4}{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\left(n+\frac{1}{3}\right)}
\end{alignat}以上より$$\displaystyle\int_0^1 \frac{x^{3n-2}}{\sqrt[3]{1-x^3}}dx=\frac{1}{3}\cdot \frac{Γ\left(n-\frac{1}{3}\right)Γ\left(\frac{2}{3}\right)}{Γ\left(n+\frac{1}{3}\right)}$$

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