On the asymptotic behavior of Elliptic integral near $k=1$ I was trying to prove that as $k\to1^-$,
$$
\int_0^1{\mathrm dx\over\sqrt{1-x^2}\sqrt{1-k^2x^2}}\sim\frac12\log{1\over1-k}
$$
Since the original integral is difficult to work with, I split it:
$$
\int_0^1{\mathrm dx\over\sqrt{1-x^2}\sqrt{1-k^2x^2}}=\int_0^k+\int_k^1
$$
in which it is evident that
$$
\begin{aligned}
\int_0^k{\mathrm dx\over\sqrt{1-x^2}\sqrt{1-k^2x^2}}
&=\int_0^k{\mathrm dx\over1-x^2}\cdot{\sqrt{1-x^2}\over\sqrt{1-k^2x^2}} \\
&=\frac12\int_0^k{\mathrm dx\over1-x}\cdot{\sqrt{1-x^2}\over\sqrt{1-k^2x^2}}+\mathcal O(1) \\
&=\frac12\int_0^k{\mathrm dx\over1-x}+\mathcal O\left(\int_0^k{1-k\over1-x}\mathrm dx\right)+\mathcal O(1) \\
&=\frac12\log{1\over1-k}+\mathcal O(1)
\end{aligned}
$$
I wonder how to show that $\int_k^1$ converges to zero as $k\to1^-$ so that I can complete the proof.
P.S. Alternative methods are welcome since I have been stuck on this for so much time :D

If $0<k<x<1$, then
\begin{align*}
\sqrt {1 - x^2 } \sqrt {1 - k^2 x^2 }  = \sqrt {1 - x} \sqrt {1 + x} \sqrt {1 - kx} \sqrt {1 + kx}   \ge \sqrt {1 - x} \sqrt {1 - k} .
\end{align*}
Hence,
$$
0 \leq \int_k^1 {\frac{{dx}}{{\sqrt {1 - x^2 } \sqrt {1 - k^2 x^2 } }}}  \le \frac{1}{{\sqrt {1 - k} }}\int_k^1 {\frac{{dx}}{{\sqrt {1 - x} }}}  = 2.
$$
This bound together with your estimate for the integral between $0$ and $k$ imply the desired asymptotics.
