Properties of such functions?

Question

Suppose we want to calculate such integral $\displaystyle \int_{0}^{1}\frac{\log(1-a^2x^2)}{x^2\sqrt{1-x^2}}dx$, where $|a| \le 1$. It's easy to calculate if we consider it as function of $F(a)$ and take first derivate. But why can we do that?

Actually how to prove that function is differentiable and $F'(a)$ is continious.

About second part of the question , I think we need to prove that function is uniform convergent to something (use Weierstrass-test). But what about the former part?

EDIT : I guess we just should prove uniform convergence of $F(a)$ and $F'(a)$ on $0 < x < 1$ and $|a| \le 1$.

EDIT2 : I guess the first part is : $\frac{\log(1-a^{2}x^{2})}{x^{2}\sqrt{1-x^2}} \le \frac{x^2}{x^2\sqrt{1-x^2}} = \frac{1}{\sqrt{1-x^2}}$(integral of the last function converges on $(0,1)$ ), for all $0 \le x \le 1$ and $|a| \le 1$. So Weierstrass test is appropriate. Am I right?

Answer 1

For any $x\in(-1,1)$ we have $-\log(1-x)=\sum_{n\geq 1}\frac{x^n}{n}$, hence $$\frac{\log(1-a^2 x^2)}{x} = -\sum_{n\geq 1}\frac{a^{2n} x^{2n-1}}{n}\tag{A}$$ and due to $\int_{0}^{1}\frac{x^{2n-1}}{\sqrt{1-x^2}}\,dx=\int_{0}^{\pi/2}\left(\sin\theta\right)^{2n-1}\,d\theta=\frac{4^n}{2n\binom{2n}{n}} $ we have: $$ \int_{0}^{1}\frac{\log(1-a^2 x^2)}{x\sqrt{1-x^2}}\,dx = -\sum_{n\geq 1}\frac{(2a)^{2n}}{2n^2 \binom{2n}{n}}=\color{blue}{-\arcsin^2(a)}\tag{B} $$ since the Taylor series of the squared arcsine function is well-known (see here and here).
The exchange of $\sum_{n\geq 1}$ and $\int_{0}^{1}(\ldots)\,dx$ is allowed by the dominated (or just monotone) convergence theorem. In a similar way we have $$\int_{0}^{1}\frac{\log(1-a^2 x^2)}{x^2\sqrt{1-x^2}}\,dx = -\sum_{n\geq 1}\frac{\pi a^{2n}\binom{2n-2}{n-1}}{2n\, 4^n}=\color{blue}{\pi\left(-1+\sqrt{1-a^2}\right)}.\tag{C} $$