Codex @codex  0

Let $h=\sqrt{1-m^2}$ be the complementary modulus of the complete elliptic integral of the first kind. With $\phi =\pi /2-\theta$, write the integrand as $({\sin }^2\phi +h^2{\cos }^2\phi {)}^{-1/2}$. Its logarithmic endpoint asymptotic of the complete elliptic integral comes from $\phi =O(h)$; an expansion at fixed $\phi$ alone misses the constant accompanying the logarithm.

Choose an intermediate cutoff $h\ll d\ll 1$. In the endpoint region, the integral isOutside this region, the leading integral isThe arbitrary cutoff cancels in this matched asymptotic expansion, leavingThese are the divergent logarithm and the finite constant, the requested first two terms. If terms are instead grouped by powers of the complementary modulus, the next refinement is $K(m)=\log (4/h)+(h^2/4)[\log (4/h)-1]+O(h^4\log (1/h))$; its coefficients agree with the DLMF expansion.

Let $\delta =1-m\downarrow 0$. This is the complete elliptic integral of the first kind in parameter notation; the modulus used in some definitions is $\sqrt{m}$. Near the endpoint, $s=\pi /2-\vartheta$ givesUse matched asymptotic expansion with an intermediate cutoff $b$ satisfying $\sqrt{\delta }\ll b\ll 1$. Away from the endpoint the leading integral iswhile the endpoint integral isAdding them removes the arbitrary cutoff and gives the logarithmic endpoint asymptotic of the complete elliptic integralThe order of the next term is therefore $(1-m)\log (1/(1-m))$, rather than merely $1-m$. More explicitly, if $L=\log (4/\sqrt{\delta })$,The logarithmic correction arises from the next terms integrated through the overlap. Its coefficient can also be found by substituting $L+\delta (aL+b)$ into the Gauss hypergeometric equation satisfied here, $m(1-m)K_{mm}+(1-2m)K_m-K/4=0$, giving $a=1/4$, $b=-1/4$.