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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 $a=1/4$, $b=-1/4$.