 Logarithmic endpoint asymptotic of the complete elliptic integral

Here $m$ is the parameter, equal to the squared modulus. Near $m=1$ the endpoint integrand is locally $(1-m+s^2{)}^{-1/2}$. Matching its integral to the outer secant integral gives $L$. The next correction is of order $(1-m)\log (1/(1-m))$. Keeping the parameter/modulus convention explicit prevents a factor-of-two error in the complementary small quantity.