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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 $$K(k) = \frac12 \left|\log (1-k^2)\right| + O(1) = \frac12 \left|\log \epsilon\right| + O(1).$$

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  • $\begingroup$ @bro: thanks, I will fix it. $\endgroup$ – Fabian Apr 24 '18 at 5:14
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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 $$K(k) \sim \int_{k'A}^B \frac{dw}{w} \sim - \ln k' = -\frac{1}{2}\ln(2\epsilon - \epsilon^2) \sim -\frac{1}{2} \ln \epsilon.$$

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Using two point Taylor expansion around $x=(0 ,\frac{\pi}{2})$ series you could get $$\frac{\pi ^5 k^4}{2560}-\frac{\pi ^5 k^2}{5760}+\frac{\pi }{4 \sqrt{1-k^2}}+\frac{\pi }{4}$$ it has a better Cuadraric error than the other series near $k=1$.

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  • $\begingroup$ What is "Cuadraric error"? Do you mean quadratic? $\endgroup$ – Xander Henderson Apr 14 '18 at 22:25
  • $\begingroup$ Yes quadratic error $\endgroup$ – Clerk Apr 15 '18 at 8:36

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