Past exam of the mathematics course of the University of Cambridge / 2005 / iii / Paper 79 / 1 / a

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.

There are two contributing regions: an endpoint of width $x^{-1}$ near $u=0$, and a very narrow peak near $u=1$. Both must be retained in the asymptotic expansion.

For the peak, set $a=e^{-x}$ and $u=1+av$. At fixed $v$, $\log u\sim av$ and $e^{-xu}\sim a$, so the integrand times $du$ approaches $dv/(1+v^2)$. This is the Lorentzian contribution from shrinking regularization. Its total contribution isFor example, cut off this region at $|u-1|=e^{-x/2}$: its scaled endpoints tend to infinity, while the changes in the exponential and logarithm inside the peak are exponentially small. The matched tails outside it are also exponentially small compared with the endpoint terms below.

At the endpoint, put $v=xu$ and $L=\log x$. To algebraic orders in $L^{-1}$, the regularizing denominator term is negligible there, giving the Laplace endpoint expansion with a logarithmic denominatorThis integral represents the endpoint expansion only; it is cut off before the original peak when making the argument precise. Since ${\Gamma }^{\prime }(1)=-{\gamma }_E$, with ${\gamma }_E$ the Euler--Mascheroni constant, the first two contributions to the original integral areThe endpoint correction refines to $[x{L}^2{]}^{-1}[1-2{\gamma }_E/L+3({\gamma }_E^2+{\pi }^2/6)/L^2+\cdots ]$, using derivatives of the gamma function. An application of Watson's lemma confined to $u=0$ would omit the leading peak contribution $\pi$.