Calcular el $\int_{0}^{1}\dfrac{x^{2n}}{\sqrt{1-x^2}}\mathrm{d}x$

Question

Para un entero $n\ge0$, Calcular: $$\int_{0}^{1}\dfrac{x^{2n}}{\sqrt{1-x^2}}\mathrm{d}x.$$

Me gustaría recibir sugerencias sobre cómo calcularlo? Debo expandir $(1-x^2)^{-1/2}$ como una serie?

Gracias.

Answer 1

$\newcommand{\+}{^{\daga}} \newcommand{\ángulos}[1]{\left\langle\, nº 1 \,\right\rangle} \newcommand{\llaves}[1]{\left\lbrace\, nº 1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, nº 1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, nº 1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\piso}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\mitad}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\a la derecha\vert\,} \newcommand{\cy}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left (\, nº 1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\parcial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\raíz}[2][]{\,\sqrt[#1]{\vphantom{\large Un}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, nº 1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ \begin{align} \int_{0}^{1}{x^{2n} \over \root{1 - x^{2}}}\,\dd x& =\int_{0}^{1}{x^{n} \over \root{1 - x}}\,\half\,x^{-1/2}\,\dd x =\half\int_{0}^{1}x^{n - 1/2}\pars{1 - x}^{-1/2}\,\dd x \\[3mm]&=\half\,{\Gamma\pars{n + 1/2}\Gamma\pars{1/2} \over \Gamma\pars{n + 1}} ={\root{\pi} \over 2\,n!}\,\Gamma\pars{n + \half}\tag{1} \end{align} $\ds{\Gamma\pars{z}}$ es la Función Gamma ${\bf\mbox{6.1.1}}$.

Es de alguna manera relacionados con Wallis Fórmula ${\bf\mbox{6.1.49}}$ desde $$ \int_{0}^{1}{x^{2n} \over \raíz{1 - x^{2}}}\,\dd x =\int_{0}^{\pi/2}\sin^{2n}\pars{\theta}\,\dd\theta = {\pi \más de 2^{2n + 1}} {2n \elegir n}\etiqueta{2} $$

$\pars{1}$ $\pars{2}$ están relacionados a través de Gamma Duplicación De La Fórmula ${\bf\mbox{6.1.18}}$.

Answer 2

Sustituto $x=\sin{t}$. Entonces $t=\arcsin{x}$, $dt=\frac{dx}{\sqrt{1-x^2}}$, y:

$$I(n)=\int_{0}^{1}\dfrac{x^{2n}}{\sqrt{1-x^2}}\mathrm{d}x = \int_{0}^{\pi/2}\sin^{2n}{t}\,\mathrm{d}t.$$