Paramétrico Problema De Integración.

Question

Si $$I_n = \int_0^1{x^n\sqrt{1-x^2}}\, dx$$ then find $\lim_{n\to\infty}\dfrac{I_n}{I_{n-2}}$.

Answer 1

Poner a $x=\sin t$ tenemos $$I_{n} =\int_{0}^{\pi/2}\sin^{n}t\cos^{2}t\,dt$$ and using integration by parts we can show that $$(n+2)I_{n}=(n-1)I_{n-2}$$ and the desired limit is $1$.

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Más en general podemos demostrar que si $$J_{m, n} = \int_{0}^{\pi/2}\cos^{m}x\sin^{n}x\,dx$$ then $$(m + n)J_{m, n} = (m - 1)J_{m - 2, n} = (n - 1)J_{m, n - 2}$$ Tenemos \begin{align} (m + 1)J_{m, n} &=-\int_{0}^{\pi/2}\sin^{n - 1}\frac{d}{dx}(\cos^{m + 1}x)\,dx\notag\\ &= -[\sin^{n - 1}x\cos^{m + 1}x]_{0}^{\pi/2} + \int_{0}^{\pi/2}(n - 1)\sin^{n - 2}x\cos^{m + 2}x\,dx\notag\\ &= (n - 1)\int_{0}^{\pi/2}\sin^{n - 2}x\cos^{m}x(1 - \sin^{2}x)\,dx\notag\\ &= (n - 1)J_{m, n - 2} - (n - 1)J_{m, n}\notag \end{align} Y así llegamos $$(m + n)J_{m, n} = (n - 1)J_{m, n - 2}$$ The other reduction formula is available by interchanging the roles of $m, n$ and noting that this does not affect $J_{m,n}$. For the current question we can see that $I_{n} = J_{2, n}$.