Find the constant for $\int_{0}^{1} {\frac{\mathrm{d}x}{\sqrt{(1-x^2)(1-(kx)^4)}}} \sim C\ln(1-k)$ I encounter a problem for Elliptic integral, to find the exact $C$ for
$$\int_{0}^{1} {\frac{\mathrm{d}x}{\sqrt{(1-x^2)(1-(kx)^4)}}} \sim C\ln(1-k)$$
as $k\uparrow1\;(0<k<1)$.
to establish such asymptotic behavior around $k\uparrow1$ is nothing special, we have
$$\begin{aligned}
\int_{0}^{1} {\frac{\mathrm{d}x}{\sqrt{(1-x^2)(1-(kx)^4)}}}
& = \int_{0}^{1} {\frac{\mathrm{d}x}{\sqrt{(1-x)(1+x)(1-kx)(1+kx)(1+(kx)^2)}}} \\
& \le \int_{0}^{1} {\frac{\mathrm{d}x}{\sqrt{(1-x)(1-kx)}}} = \frac{2\operatorname{artanh}(\sqrt{k})}{\sqrt{k}}
\end{aligned}$$
and
$$\begin{aligned}
\int_{0}^{1} {\frac{\mathrm{d}x}{\sqrt{(1-x^2)(1-(kx)^4)}}}
& \ge \frac1{2\sqrt{2}} \int_{0}^{1} {\frac{\mathrm{d}x}{\sqrt{(1-x)(1-kx)}}} \\
& \ge \frac1{2\sqrt{2}} \int_{0}^{1} {\frac{\mathrm{d}x}{\sqrt{(1-kx)^2}}} = -\frac1{2\sqrt{2}}\frac{\ln(1-k)}{k}
\end{aligned}$$
notice
$$\frac{\operatorname{artanh}(\sqrt{k})}{\sqrt{k}} \sim -\frac1{2}\frac{\ln(1-k)}{k} \text{ as } k\uparrow1$$
What I know is the original integral as well belonging to Elliptic integral of the first kind, namely
$$\int_{0}^{1} {\frac{\mathrm{d}x}{\sqrt{(1-x^2)(1-(kx)^4)}}} = \frac1{\sqrt{1+k^2}}K\left(\frac{2k^2}{1+k^2}\right)$$
as little knowledge I have for Elliptic integral, I can't figure out what exactly $C$ is for such behavior, and I am wondering if there is a relatively fundamental way to find it.
thanks in advance for any suggestion.
 A: I will propose a creative approach based on Fourier-Legendre expansions. In $L^2(0,1)$ we have$^{(*)}$
$$ K(x)=\sum_{n\geq 0}\frac{2}{2n+1}P_n(2x-1),\qquad -\log(1-x)=1+\sum_{n\geq 1}\left(\frac{1}{n}+\frac{1}{n+1}\right)P_n(2x-1) $$
so
$$ K(x)+\frac{1}{2}\log(1-x)=\frac{3}{2}-\sum_{n\geq 1}\frac{P_n(2x-1)}{2n(n+1)(2n+1)} $$
where the RHS is continuous and bounded on $[0,1]$ due to $|P_n(2x-1)|\leq 1$. This implies $K(x)\sim -\frac{1}{2}\log(1-x)$ as $x\to 1^-$.
A straightforward consequence is that
$$\begin{eqnarray*} \int_{0}^{1}\frac{dx}{\sqrt{(1-x^2)(1-k^4 x^4)}}=\frac{1}{\sqrt{1+k^2}}K\left(\frac{2k^2}{1+k^2}\right)&\sim& -\frac{1}{2\sqrt{2}}\log\left(1-\frac{2k^2}{k^2+1}\right)\\&\sim&-\frac{1}{2\sqrt{2}}\log\left(\frac{1+k}{1+k^2}(1-k)\right)\end{eqnarray*} $$
so the wanted constant is $C=-\frac{1}{2\sqrt{2}}$. Keeping track of the first term of the regular part,
$$\boxed{ \int_{0}^{1}\frac{dx}{\sqrt{(1-x^2)(1-k^4 x^4)}} \sim -\frac{1}{2\sqrt{2}}\log(1-k)+\sqrt{2}\log(2)\qquad\text{as }k\to 1^-.}$$
(*)The first identity is easily derived from the generating function for Legendre polynomials; the second identity can be proved by computing $\int_{0}^{1}\log(x) P_n(2x-1)\,dx$ through Rodrigues' formula and integration by parts.
