Limit of an elliptic integral

Question

Exercise: $$F(y):=\int_{0}^{y}\frac{1}{\sqrt{(1-x^2)(1-k^2x^2)}} dx$$ where $k$ is a constant in the range $0 ≤ k < 1$.Show that the limit $$L:=\lim \limits_{y \to 1^-} F(y)$$exists and that $$L<\frac{2}{\sqrt{1-k^2}}$$

I thought that if it is possible to show that the function is strictly increasing in the interval $(-1,1)$ and is bounded from above, then the existence of the limit can be proved according to the axiom of completeness.

Take any $x_1,x_2$ such that $-1<x_1<x_2<1$ then $$F(x_2)-F(x_1)=\int_{x_1}^{x_2}\frac{1}{\sqrt{(1-x^2)(1-k^2x^2)}} dx >0$$

But I couldn't proceed because couldn't find an upper bound for this integral. Could anybody help me, please.

Answer 1

The function $F(y)$ is increasing hence the limit (proper or improper) exists. We have $$F(y)\le {1\over \sqrt{1-k^2}}\int\limits_0^y{1\over \sqrt{1-x^2}}\,dx\\ ={1\over \sqrt{1-k^2}}\arcsin y\le {\pi\over 2}{1\over \sqrt{1-k^2}}$$ Therefore $$L\le {\pi\over 2}{1\over \sqrt{1-k^2}}$$

Answer 2

Let $x = \sin\theta$. Then, by using that $\frac{d}{d\theta} \sin\theta = \cos\theta$, $$ L = \int_0^{\pi/2} \frac{1}{\sqrt{1-k^2\sin^2\theta}} d\theta \le \frac{\pi}{2} \frac{1}{\sqrt{1-k^2}} < \frac{2}{\sqrt{1-k^2}},$$ because $\pi < 4$.