x^n/√(1-x)[0,1]などの定積分

\begin{alignat}{2}
&(1)  \displaystyle\int_0^1 \frac{x^n}{\sqrt{1-x}}dx=\frac{2\cdot(2n)!!}{(2n+1)!!}\\
&(2)  \displaystyle\int_0^1 \frac{x^{n-\frac{1}{2}}}{\sqrt{1-x}}dx=\frac{(2n-1)!!}{(2n)!!}π\\
&(3)  \displaystyle\int_0^1 \frac{x^{p+\frac{1}{2}}}{(1-x)^{p+\frac{1}{2}}}dx=\frac{(2p+1)π}{2 \cos pπ}\\
&(4)  \displaystyle\int_0^1 \frac{x^{p-\frac{1}{2}}}{(1-x)^{p+\frac{1}{2}}}dx=\frac{π}{ \cos pπ}
\end{alignat}ただし、全て \(\displaystyle n \in \mathrm{N},\,|p| \lt \frac{1}{2} \)








<証明> 

\begin{alignat}{2}
(1) \displaystyle\int_0^1 \frac{x^n}{\sqrt{1-x}}dx&=\displaystyle\int_0^1 x^n(1-x)^{-\frac{1}{2}}dx=B\left(n+1,\frac{1}{2}\right)\\
&=\frac{Γ(n+1)Γ\left(\frac{1}{2}\right)}{Γ\left(n+\frac{3}{2}\right)}=\frac{n!\sqrt{π}}{\left(n+\frac{1}{2}\right) Γ\left(n+\frac{1}{2}\right)}\\
&=\frac{n!\sqrt{π}}{n+\frac{1}{2}}\cdot \frac{2^n}{(2n-1)!!\sqrt{π}}\\
&=2\cdot \frac{2\cdot 4 \cdot 6 \cdots (2n)}{(2n+1)(2n-1)!!}= \frac{2\cdot(2n)!!}{(2n+1)!!}\\
\end{alignat}






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







\begin{alignat}{2}
(3)  \displaystyle\int_0^1 \frac{x^{p+\frac{1}{2}}}{(1-x)^{p+\frac{1}{2}}}dx&=\displaystyle\int_0^1 x^{p+\frac{1}{2}}(1-x)^{-p-\frac{1}{2}}=B\left(p+\frac{3}{2},-p+\frac{1}{2}\right)\\
&=\frac{Γ\left(p+\frac{3}{2}\right)Γ\left(-p+\frac{1}{2}\right)}{Γ(2)}\\
&=\left(p+\frac{1}{2}\right)Γ\left(p+\frac{1}{2}\right)Γ\left(\frac{1}{2}-p\right)\\
&=\frac{2p+1}{2}\cdot \frac{π}{ \cos pπ}=\frac{(2p+1)π}{2 \cos pπ}
\end{alignat}






\begin{alignat}{2}
(4) \displaystyle\int_0^1 \frac{x^{p+\frac{1}{2}}}{(1-x)^{p-\frac{1}{2}}}dx&=\displaystyle\int_0^1 x^{p-\frac{1}{2}}(1-x)^{-p-\frac{1}{2}}=B\left(p+\frac{1}{2},-p+\frac{1}{2}\right)\\
&=\frac{Γ\left(p+\frac{1}{2}\right)Γ\left(-p+\frac{1}{2}\right)}{Γ(1)}\\
&=Γ\left(p+\frac{1}{2}\right)Γ\left(\frac{1}{2}-p\right)=\frac{π}{ \cos pπ}
\end{alignat}

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