Find the constant for $int_{0}^{1} {frac{mathrm{d}x}{sqrt{(1-x^2)(1-(kx)^4)}}} sim Cln(1-k)$

Question

I encounter a problem for Elliptic integral, to find the exact $C$ for
$$int_{0}^{1} {frac{mathrm{d}x}{sqrt{(1-x^2)(1-(kx)^4)}}} sim Cln(1-k)$$
as $kuparrow1;(0<k<1)$.
to establish such asymptotic behavior around $kuparrow1$ is nothing special, we have
$$begin{aligned}
int_{0}^{1} {frac{mathrm{d}x}{sqrt{(1-x^2)(1-(kx)^4)}}}
& = int_{0}^{1} {frac{mathrm{d}x}{sqrt{(1-x)(1+x)(1-kx)(1+kx)(1+(kx)^2)}}} \
& le int_{0}^{1} {frac{mathrm{d}x}{sqrt{(1-x)(1-kx)}}} = frac{2operatorname{artanh}(sqrt{k})}{sqrt{k}}
end{aligned}$$
and
$$begin{aligned}
int_{0}^{1} {frac{mathrm{d}x}{sqrt{(1-x^2)(1-(kx)^4)}}}
& ge frac1{2sqrt{2}} int_{0}^{1} {frac{mathrm{d}x}{sqrt{(1-x)(1-kx)}}} \
& ge frac1{2sqrt{2}} int_{0}^{1} {frac{mathrm{d}x}{sqrt{(1-kx)^2}}} = -frac1{2sqrt{2}}frac{ln(1-k)}{k}
end{aligned}$$
notice
$$frac{operatorname{artanh}(sqrt{k})}{sqrt{k}} sim -frac1{2}frac{ln(1-k)}{k} text{ as } kuparrow1$$
What I know is the original integral as well belonging to Elliptic integral of the first kind, namely
$$int_{0}^{1} {frac{mathrm{d}x}{sqrt{(1-x^2)(1-(kx)^4)}}} = frac1{sqrt{1+k^2}}Kleft(frac{2k^2}{1+k^2}right)$$
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.

Jack D'Aurizio · Answer

I will propose a creative approach based on Fourier-Legendre expansions. In $L^2(0,1)$ we have$^{(*)}$
$$ K(x)=sum_{ngeq 0}frac{2}{2n+1}P_n(2x-1),qquad -log(1-x)=1+sum_{ngeq 1}left(frac{1}{n}+frac{1}{n+1}right)P_n(2x-1) $$
so
$$ K(x)+frac{1}{2}log(1-x)=frac{3}{2}-sum_{ngeq 1}frac{P_n(2x-1)}{2n(n+1)(2n+1)} $$
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 $xto 1^-$.
A straightforward consequence is that
$$begin{eqnarray*} int_{0}^{1}frac{dx}{sqrt{(1-x^2)(1-k^4 x^4)}}=frac{1}{sqrt{1+k^2}}Kleft(frac{2k^2}{1+k^2}right)&sim& -frac{1}{2sqrt{2}}logleft(1-frac{2k^2}{k^2+1}right)\&sim&-frac{1}{2sqrt{2}}logleft(frac{1+k}{1+k^2}(1-k)right)end{eqnarray*} $$
so the wanted constant is $C=-frac{1}{2sqrt{2}}$. Keeping track of the first term of the regular part,
$$boxed{ int_{0}^{1}frac{dx}{sqrt{(1-x^2)(1-k^4 x^4)}} sim -frac{1}{2sqrt{2}}log(1-k)+sqrt{2}log(2)qquadtext{as }kto 1^-.}$$
(*)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.

Find the constant for $int_{0}^{1} {frac{mathrm{d}x}{sqrt{(1-x^2)(1-(kx)^4)}}} sim Cln(1-k)$

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