definite integral formula

[Grigorios Kostakos]

Prove that \[I_{n}=\displaystyle\int_{0}^{2\alpha}{x^{n}\sqrt{2\alpha{x}-x^2}\;dx}=\frac{\pi\,\alpha^{n+2}}{2}\mathop{\prod}\limits_{k=2}^{n}{\frac{2k+1}{k+2}}\,,\quad n\in\mathbb{N}\,,\;\alpha>0\,.\]

Hint

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Prove that \((n+2)\,I_{n}=\alpha\,(2n+1)\,I_{n-1}\,.\)

[Papapetros Vaggelis]

\(\displaystyle{\forall n\in\mathbb{N}:I_{n}>0}\).

For \(\displaystyle{n=1}\) we have,

\(\displaystyle{I_{1}=\int_{0}^{2\alpha}x\cdot \sqrt{2\alpha x-x^2}\, dx=\int_{0}^{2\alpha}x\cdot \sqrt{\alpha^2-(\alpha-x)^2}\, dx}\).

By the substitution \(\displaystyle{\alpha-x=\alpha\sin t,}\), we have \(\displaystyle{dx=-\alpha\cos t dt}\)

and, if \(\displaystyle{x=0}\), then \(\displaystyle{t=\frac{\pi}{2}}\), if\(\displaystyle{x=2\alpha}\), then \(\displaystyle{t=-\frac{\pi}{2}}\).

So,

\(\displaystyle{I_{1}=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\left(\alpha-\alpha\sin t\right)\cdot \sqrt{\alpha^2-\alpha^2\cos^2 t} (\alpha\cos t)dt=}\)

\(\displaystyle{=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\left(\alpha^3\cos^2 t-\alpha^2\sin t\cos^2 t\right)dt}\)

\(\displaystyle{=\left[\frac{\alpha^3}{4}\sin (2t)+\frac{\alpha^3}{2}t+\frac{\alpha^3}{3}\cos^3 t\right]_{-\frac{\pi}{2}}^{\frac{\pi}{2}}}\)

\(\displaystyle{=\frac{\pi\alpha^3}{2}}\).

For all \(\displaystyle{n\geq 2}\),

\(\displaystyle{I_{n}=\int_{0}^{2\alpha}x^{n}\sqrt{2\alpha x-x^2} dx=}\)

\(\displaystyle{=-\frac{1}{2}\int_{0}^{2\alpha}(-2x)x^{n-1}\sqrt{2\alpha x-x^2} dx}\)

\(\displaystyle{=-\frac{1}{2}\int_{0}^{2\alpha}\left[\left(2\alpha-2x\right)-2\alpha\right]x^{n-1}\sqrt{2\alpha x-x^2} dx}\)

\(\displaystyle{=-\frac{1}{2}\left[\int_{0}^{2\alpha}\left(2\alpha-2x\right)x^{n-1}\sqrt{2\alpha x-x^2}dx-2\alpha\int_{0}^{2\alpha}x^{n-1}\sqrt{2\alpha x-x^2}dx\right]}\)

\(\displaystyle{=-\frac{1}{2}\left[\frac{2}{3}\left[(\sqrt{2\alpha x-x^2})^3\cdot x^{n-1}\right]_{0}^{2\alpha}-\frac{2n-2}{3}\int_{0}^{2\alpha}x^{n-2}(\sqrt{2\alpha x-x^2})^3-2\alpha I_{n-1}\right]}\)

\(\displaystyle{=\frac{n-1}{3}\int_{0}^{2\alpha}x^{n-2}(2\alpha x-x^2)\sqrt{2\alpha x-x^2}dx+\alpha I_{n-1}}\)

\(\displaystyle{=\frac{n-1}{3}\int_{0}^{2\alpha}\left(2\alpha x^{n-1}\sqrt{2\alpha x-x^2}-x^n\sqrt{2\alpha x-x^2}\right)dx+\alpha I_{n-1}}\)

\(\displaystyle{=\frac{n-1}{3}\cdot 2\alpha I_{n-1}-\frac{n-1}{3}I_{n}+\alpha I_{n-1}}\)

\(\displaystyle{\Rightarrow I_{n}+\frac{n-1}{3}I_{n}=\frac{2\alpha(n-1)I_{n-1}+3\alpha I_{n-1}}{3}}\)

\(\displaystyle{\Rightarrow (n+2)I_{n}=\alpha(2n+1)I_{n-1}}\).

Therefore,
\begin{align*}
I_{n}&=\alpha\frac{2n+1}{n+2}I_{n-1}=\alpha\frac{2n+1}{n+2}\alpha \frac{2(n-1)+1}{(n-1)+2}I_{n-2}=...\\
&=\alpha^{n-1}\prod_{k=2}^n \frac{2k+1}{k+2}I_{1}=\alpha^{n-1}\prod_{k=2}^n\frac{2k+1}{k+2}\frac{\pi\alpha^3}{2}=\frac{\pi\alpha^{n+2}}{2}\prod_{k=2}^n\frac{2k+1}{k+2}
\end{align*}

[Grigorios Kostakos]

A 2nd solution using Beta and Gamma functions:

\begin{align*}
I_{n}&=\displaystyle\int_{0}^{2\alpha}{x^{n}\sqrt{2\alpha{x}-x^2}\;dx}=\int_{0}^{2\alpha}{x^{n+\frac{1}{2}}\,({2\alpha-x})^{\frac{1}{2}}\,dx}\\
&\mathop{=\!=\!=\!=\!=\!=}\limits^{\begin{subarray}{c}
{t\,=\,\frac{x}{2\alpha}} \\
{2\alpha\,dt\,=\,dx}
\end{subarray}}\,\int_{0}^{1}{({2\alpha})^{n+\frac{1}{2}}\,t^{n+\frac{1}{2}}\,({2\alpha-2\alpha{t}})^{\frac{1}{2}}\,2\alpha\,dt}=({2\alpha})^{n+2}\int_{0}^{1}{t^{n+\frac{1}{2}}\,({1-t})^{\frac{1}{2}}\,2\alpha\,dt}\\
&=({2\alpha})^{n+2}\int_{0}^{1}{t^{n+\frac{3}{2}-1}\,({1-t})^{\frac{3}{2}-1}\,dt}=({2\alpha})^{n+2}\,{\rm{B}}\bigl({n+\tfrac{3}{2},\tfrac{3}{2}}\bigr)\\
&\stackrel{(1)}{=}({2\alpha})^{n+2}\,\frac{\Gamma\bigl({n+\tfrac{3}{2}}\bigr)\,\Gamma\bigl({\tfrac{3}{2}}\bigr)}{\Gamma(n+3)}\stackrel{(1)}{=}2^{n+2}\alpha^{n+2}\,\frac{\displaystyle\frac{\sqrt{\pi}}{2^{n+1}}\mathop{\prod}\limits_{k=1}^{n}{(2k+1)}\,\frac{\sqrt{\pi}}{2}}{2\displaystyle\mathop{\prod}\limits_{k=1}^{n}{(k+2)}}\\
&=\frac{\pi\,\alpha^{n+2}}{2}\mathop{\prod}\limits_{k=1}^{n}{\frac{2k+1}{k+2}}=\frac{\pi\,\alpha^{n+2}}{2}\mathop{\prod}\limits_{k=2}^{n}{\frac{2k+1}{k+2}}\,,\;n\in\mathbb{N}\,.\qquad\square
\end{align*}
We use the formulas:
\begin{align*}
&{\rm{B}}({x,y})=\dfrac{\Gamma(x)\,\Gamma(y)}{\Gamma(x+y)} \quad&(1)\\
&\left\{{\begin{array}{rl}
\Gamma\bigl({n+\tfrac{3}{2}}\bigr)\!&=\displaystyle\frac{\sqrt{\pi}}{2^{n+1}}\mathop{\prod}\limits_{k=1}^{n}{(2k+1)}\\
\Gamma\bigl({\tfrac{3}{2}}\bigr)\!&=\dfrac{\sqrt{\pi}}{2}\\
\Gamma({n+3})\!&=\displaystyle2\mathop{\prod}\limits_{k=1}^{n}{(k+2)}
\end{array}}\right\}\quad&(2)
\end{align*}