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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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4 Answers 4

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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, 2018 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 the domain $[A, +\infty]$. 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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$$\int_0^{\pi/2} \frac{d x}{\sqrt{1 - k^2 \sin^2 x}}=\frac{K\left(\frac{k^2}{k^2-1}\right)}{\sqrt{1-k^2}}$$ Around $k=1$, a quite good asymptotics is

$$-\frac{t}{2} -\frac{1+t}{4} (1-k)-\frac{7+5 t}{32} (1-k)^2-\frac{34+21t}{192} (1-k)^3\qquad \text{where} \qquad t=\log \left(\frac{1-k}{8}\right)$$ that is to say $$-\frac{1}{48} (31-19 k) (1-k)-\frac{1}{64} \left(17 k^2-50 k+65\right)\log \left(\frac{1-k}{8}\right)$$ whci seems to be decent even far away from $k=1$.

Some results $$\left( \begin{array}{ccc} k & \text{approximation} & \text{exact} \\ 0.400 & 1.63887 & 1.64000 \\ 0.450& 1.66465 & 1.66089 \\ 0.500 & 1.69303 & 1.68575 \\ 0.550 & 1.72489 & 1.71535 \\ 0.600 & 1.76142 & 1.75075 \\ 0.650 & 1.80428 & 1.79345 \\ 0.700 & 1.85583 & 1.84569 \\ 0.750 & 1.91977 & 1.91099 \\ 0.800 & 2.00224 & 1.99530 \\ 0.850 & 2.11476 & 2.10994 \\ 0.900 & 2.28325 & 2.28055 \\ 0.950 & 2.59091 & 2.59001 \\ 0.960 & 2.69376 & 2.69314 \\ 0.970 & 2.82837 & 2.82800 \\ 0.980 & 3.02117 & 3.02098 \\ 0.990 & 3.35666 & 3.35660 \\ 0.991 & 3.40813 & 3.40808 \\ 0.992 & 3.46579 & 3.46575 \\ 0.993 & 3.53129 & 3.53126 \\ 0.994 & 3.60707 & 3.60704 \\ 0.995 & 3.69689 & 3.69688 \\ 0.996 & 3.80708 & 3.80707 \\ 0.997 & 3.94947 & 3.94947 \end{array} \right)$$

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Using two point Taylor expansion around $x=(0 ,\frac{\pi}{2})$ series you could get the expression $$\frac{\pi ^5 k^4}{2560}-\frac{\pi ^5 k^2}{5760}+\frac{\pi }{4 \sqrt{1-k^2}}+\frac{\pi }{4}$$ This has a better quadratic 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, 2018 at 22:25
  • $\begingroup$ Yes quadratic error $\endgroup$
    – Clerk
    Apr 15, 2018 at 8:36

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