Intrigued by this brilliant answer from Ron Gordon, I was attempting to find the Maclaurin series for $$f(x)=\frac{\arcsin x}{\sqrt{1-x^2}}=g(x)G(x)$$
with $g(x)=\frac{1}{\sqrt{1-x^2}}$ and $G(x)$ its primitive. So I attempted to multipy series, which yielded this:
$$f(x)=\sum_{n=0}^{\infty}x^{2n+1} (-1)^n\sum_{k=1}^{n}\frac{1}{k+1} { -\frac{1}{2}\choose n-k}{ -\frac{1}{2}\choose k},$$
which I'm unable to simplify further. How to proceed? Or is this approach doomed?
Note that $$\int_0^1\frac{dt}{1-x^2+x^2t^2}=\frac{1}{x\sqrt{1-x^2}}\arctan\left(\frac{x}{\sqrt{1-x^2}}\right)=\frac{\arcsin(x)}{x\sqrt{1-x^2}}$$ so we can write $$\frac{\arcsin(x)}{\sqrt{1-x^2}}=\sum_{n=0}^\infty\left(\int_0^1(1-t^2)^n\,dt\right)x^{2n+1}.$$ But $$\int_0^1(1-t^2)^n\,dt=\int_0^1\sum_{k=0}^n(-1)^k\binom{n}{k}t^{2k}\,dt=\sum_{k=0}^n\frac{(-1)^k\binom{n}{k}}{2k+1}=\frac{(2n)!!}{(2n+1)!!}.$$ Hence, $$\frac{\arcsin(x)}{\sqrt{1-x^2}}=\sum_{n=0}^\infty\frac{(2n)!!}{(2n+1)!!}x^{2n+1}.$$ Also see here for proof of $\sum_{k=0}^n\frac{(-1)^k\binom{n}{k}}{2k+1}=\frac{(2n)!!}{(2n+1)!!}$.
$ \frac{\arcsin(x)}{\sqrt{1-x^2}} = \sum_{n=0}^\infty\left(\int_0^1(1-t^2)^n,dt\right)x^{2n+1}. $
Where does the integral appear from and how does it relate to the previous one ?
– Victor Sep 27 '15 at 03:39Another but similar proof which does not need to use the summation formula above is this one. Start with defining
$$I(t)= \frac{1}{\sqrt{1-x^2}}\arctan{\frac{x\sin{t}}{\sqrt{1-x^2}}}$$
Then by the Fundamental theorem of calculus
$$\frac{\arcsin{x}}{\sqrt{1-x^2}}=I\left(\frac{\pi}{2}\right)-I(0)=\int_{0}^{\pi/2} \frac{\partial I}{\partial t}\mathrm{d}t=\int_{0}^{\pi/2}\frac{x\cos t}{1-x^2\cos^2 t }\mathrm{d}t$$
Ergo
$$\frac{\arcsin{x}}{\sqrt{1-x^2}}=\sum_{n=0}^{\infty}x^{2n+1}\int_0^{\pi/2}\cos^{2n+1}\! t\,\mathrm{d}t$$
Denote $J_n:=\int_0^{\pi/2}\cos^{2n+1}\! t\,\mathrm{d}t$, by per partes we have
$$J_n = \int_0^{\pi/2}\cos^{2n+1}\! t\,\mathrm{d}t = 2n\int_0^{\pi/2}\cos^{2n-1}\sin^2 t\!\,\mathrm{d}t=2n\left(J_{n-1}-J_{n}\right)$$
So $$J_n = \frac{2n}{2n+1}J_{n-1} =\frac{2n}{2n+1}\frac{2n-2}{2n-1}J_{n-2}=\dots = \frac{(2n)!!}{(2n+1)!!}J_0=\frac{(2n)!!}{(2n+1)!!}$$
since $J_0 = \int_0^{\pi/2}\cos\! t\,\mathrm{d}t =1$. Over all we get desired result
$$\frac{\arcsin{x}}{\sqrt{1-x^2}}=\sum_{n=0}^{\infty}\frac{(2n)!!}{(2n+1)!!}x^{2n+1}$$
Note: Similar integral would have been also...
$$\int_{0}^{\pi/2}\frac{\mathrm{d}t}{1-x\sin t}$$
Another proof is using that $f(x)={{\arcsin x}\over {\sqrt{1-x^2}}}$ satisfies $(1-x^2)f'=xf+1$, which can be easily verified.
It is clear that \begin{equation*} \frac{\arcsin x}{\sqrt{1-x^2}\,}=\frac12\bigl[(\arcsin x)^2\bigr]'. \end{equation*} Accordingly, it is sufficient to find out the power series expansion of $(\arcsin x)^2$ around $x=0$. The fact is \begin{equation*} (\arcsin x)^2=2!\sum_{n=0}^{\infty} [(2n)!!]^2 \frac{x^{2n+2}}{(2n+2)!}, \quad |x|<1. \end{equation*} A more general conclusion than the above power series expansion is \begin{equation}\label{arcsin-series-expansion-unify} \biggl(\frac{\arcsin x}{x}\biggr)^{k} =1+\sum_{m=1}^{\infty} (-1)^m\frac{Q(k,2m)}{\binom{k+2m}{k}}\frac{(2x)^{2m}}{(2m)!}, \end{equation} where \begin{equation}\label{Q(m-k)-sum-dfn} Q(k,m)=\sum_{\ell=0}^{m} \binom{k+\ell-1}{k-1} s(k+m-1,k+\ell-1)\biggl(\frac{k+m-2}{2}\biggr)^{\ell} \end{equation} for $k\in\mathbb{N}$ and $m\ge2$.
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