The asymptotic behaviour of Elliptic integral near k=1

[julian]

TL;DR

I'm looking at a proof of the asymptotic expression for the Elliptic function of the first kind and I'm having trouble understanding a step in the proof.

I'm looking at a proof of the asymptotic expression for the Elliptic function of the first kind

https://math.stackexchange.com/ques...ptotic-behavior-of-elliptic-integral-near-k-1

and I'm having trouble understanding this step in the proof:
$$
\begin{align*}
\frac{1}{2} \int_0^k \dfrac{dx}{1 - x} \dfrac{ \sqrt{1 - x^2} }{ \sqrt{1 - k^2 x^2} } + \mathcal{O} (1) = \frac{1}{2} \int_0^k \dfrac{dx}{1 - x} + \mathcal{O} \left( \int_0^k \dfrac{1-k}{1-x} dx \right) + \mathcal{O} (1)
\end{align*}
$$
I've written
$$
\begin{align*}
\frac{1}{2} \int_0^k \dfrac{dx}{1 - x} \dfrac{ \sqrt{1 - x^2} }{ \sqrt{1 - k^2 x^2} } &= \frac{1}{2} \int_0^k \dfrac{dx}{1 - x} + \frac{1}{2} \int_0^k \frac{dx}{1 - x} \left( \dfrac{ \sqrt{1 - x^2} }{ \sqrt{1 - k^2 x^2} } - 1 \right)
\nonumber \\
&= \frac{1}{2} \int_0^k \dfrac{dx}{1 - x} - \frac{1}{2} \int_0^k \dfrac{1-k}{1-x} dx + \frac{1}{2} \int_0^k \frac{dx}{1 - x} \left( \dfrac{ \sqrt{1 - x^2} }{ \sqrt{1 - k^2 x^2} } - k \right)
\end{align*}
$$
But not sure where to go from here.

 

[Delta2]

Work towards proving that the last integral is less or equal than $\int_0^k \frac{1-k}{1-x}dx$. I think it is sufficient to show that for "proper" values of $x$ and $k$ it is $$\frac{\sqrt{1-x^2}}{\sqrt{1-k^2x^2}}\leq 1$$

 

[julian]

So for $0 \leq k < 1$ we have for $0 \leq x \leq k$ that

\begin{align*}
\dfrac{ \sqrt{1 - x^2} }{ \sqrt{1 - k^2 x^2} } - k < 1 - k .
\end{align*}

Huzzah.