# Asymptotic expansion of the complete elliptic integral of the first kind

The complete elliptic integral of the first kind is defined as $$K(k) = \int_0^{\pi/2} \frac{d x}{\sqrt{1 – k^2 \sin^2 x}}.$$ I would like to derive (at least the first term of) the asymptotic expansion for $k = 1 -\epsilon$. This is certainly not trivial due to lack of uniform convergence which implies that I can’t use the Taylor expansion of the integrand. What is the best way to proceed?

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You can employ the substitution $y=1- k^2 \sin^2x$ such that the integral can be written as
$$K(k) = \int_{1-k^2}^1\!dy\,\frac{1}{2 \sqrt{y(1-y)(y-1 +k^2)}}.$$

Now, we can expand the integrand in $\epsilon = 1-k$ and obtain
$$K(k) = \int_{1-k^2}^1\!dy\frac{1}{2 y \sqrt{1-y}} + O(1). \tag{1}$$
The estimate of the error term follows from the fact that expanding in $\eta =1-k^2$, we have
$$K(k) = \sum_{n=0}^\infty c_n \eta^n \int_{\eta}^1\!dy\, y^{-1-n} (1-y)^{-1/2}$$
with $c_n$ some constants. Now, we have for $n>0$ that
$$\int_{\eta}^1\!dy\, y^{-1-n} (1-y)^{-1/2} = O(\eta^{-n})$$
such that only the $n=0$ term diverges for $\eta \to 0$ (which corresponds to $\epsilon \to 0$).

It thus remains to estimate the first term in (1) for $k \to 1$ which is not that difficult:

In fact due to the $1/y$ behavior close to $y=0$, the integral is logarithmically divergent and we have that
$$K(k) = \frac12 \log|1-k^2| + O(1) = \frac12 \log|\epsilon| + O(1).$$

Write $k’^2 = 1 – k^2$. Make the substitution $v = k’ \cot x$ to obtain
$$K(k) = \int_0^{+\infty} (1 + v^2)^{-1/2} (1 + k’^2 v^2)^{-1/2} \, dv.$$

Now it is easy to see that this integral is $O(1)$ on any bounded interval $[0,A]$. Since we’re only interested in the leading term, we can look at the integral only on $[A + \infty]$. On that interval, the $(1 + v^2)^{-1/2}$ term is very close to $1/v$ (to within a factor of $(1 + 1/A^2)^{1/2}$, which can be as close to $1$ as we like, if we take $A$ large enough). Therefore we can write
$$K(k) \sim \int_A^{+\infty} v^{-1} (1 + k’^2 v^2)^{-1/2} \, dv.$$
Making the substitution $w = k’v$, we find
$$K(k) \sim \int_{k’A}^{+\infty} \frac{dw}{w(1+w^2)^{1/2}} \sim \int_{k’A}^{B} \frac{dw}{w(1+w^2)^{1/2}},$$
where $B$ is small and fixed, since the integral on $[B,+\infty]$ is clearly $O(1)$. Since $B$ is small, the factor $(1 + w^2)^{1/2}$ can be assumed close to $1$, so as a result
$$K(k) \sim \int_{k’A}^B \frac{dw}{w} \sim – \ln k’ = -\frac{1}{2}\ln(2\epsilon – \epsilon^2) \sim -\frac{1}{2} \ln \epsilon.$$