Let $f:[-1,1]\rightarrow \mathbb{R}$ be continuous such that $f(x) = \sqrt{1-x^2} + x^2f(x^2)$ for $\forall x\in [-1,1]$. Find $ \int_{-1}^{1} f(x) \,dx $.
A have an answer that is an infinite series of beta functions. I would like to know if there is a closed form answer.
Since the function is even, we need to find $ \int_{0}^{1} f(x) \,dx $. First, we have $$\int_0^1 x^af(x)\,dx = \int_0^1x^a\sqrt{1-x^2}\,dx + \frac{1}{2}\int_0^1 x^{a+1}f(x^2)\,d(x^2) = \frac{1}{2}B\left(\frac{a+1}{2},\frac{3}{2}\right) + \frac{1}{2}\int_0^1x^\frac{a+1}{2}f(x)\,dx$$ Doing this recursively, we get $$\int_0^1 f(x)\,dx = \sum_{n=1}^\infty \frac{B\left(1-\frac{1}{2^n}, 3/2\right)}{2^n}$$ Is the sum equal to something meaningful? Alternatively, is there a way to solve the problem without a series expansion?