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?
Asymptotics – Asymptotic Expansion of the Complete Elliptic Integral of the First Kind
asymptoticselliptic integrals
Related Solutions
I prefer to have the singular behaviour near $0$ rather than at $\frac{\pi}{2}$, so let's make the substitution $\varphi = \frac{\pi}{2} - \theta$. We obtain
$$K(k) = \int_0^{\frac{\pi}{2}} \frac{d\varphi}{\sqrt{1 - k^2\cos^2\varphi}}.$$
Split the integral at $\frac{\pi}{4}$. The part
$$\int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \frac{d\varphi}{\sqrt{1 - k^2\cos^2\varphi}}$$
remains harmless as $k \to 1$ and tends to
$$\int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \frac{d\varphi}{\sin \varphi}.$$
For the other part, write $1 = \sin^2 \varphi + \cos^2 \varphi$ to obtain $1 - k^2\cos^2\varphi = \sin^2 \varphi + (1-k^2)\cos^2 \varphi$. Let $\varepsilon = \sqrt{1-k^2}$. Then
\begin{align} \int_0^{\frac{\pi}{4}} \frac{d\varphi}{\sqrt{1 - k^2\cos^2 \varphi}} &= \int_0^{\frac{\pi}{4}} \frac{\cos^2 \varphi + \sin^2 \varphi}{\sqrt{\cos^2 \varphi + \sin^2 \varphi}\cdot \sqrt{\sin^2 \varphi + \varepsilon^2 \cos^2\varphi}}\,d\varphi\\ &= \int_0^{\frac{\pi}{4}} \frac{1 + \tan^2\varphi}{\sqrt{1 + \tan^2 \varphi}\cdot \sqrt{\varepsilon^2 + \tan^2 \varphi}}\,d\varphi \tag{$t = \tan \varphi$}\\ &= \int_0^1 \frac{dt}{\sqrt{1+t^2}\cdot \sqrt{\varepsilon^2 + t^2}}\\ &= \int_0^1 \frac{dt}{\sqrt{\varepsilon^2 + t^2}} - \int_0^1 \Biggl(1 - \frac{1}{\sqrt{1+t^2}}\Biggr) \frac{dt}{\sqrt{\varepsilon^2 + t^2}}. \end{align}
Since
$$1 - \frac{1}{\sqrt{1+t^2}} = \frac{\sqrt{1+t^2}-1}{\sqrt{1+t^2}} = \frac{t^2}{\sqrt{1+t^2}\cdot (1 + \sqrt{1+t^2})},$$
the last integral remains bounded and tends to
$$\int_0^1 \frac{t \,dt}{1 + t^2 + \sqrt{1+t^2}}$$
as $\varepsilon \to 0$.
And the substitution $t = \varepsilon u$ gives us
\begin{align} \int_0^1 \frac{dt}{\sqrt{\varepsilon^2 + t^2}} &= \int_0^{\frac{1}{\varepsilon}} \frac{du}{\sqrt{1+u^2}}\\ &= \operatorname{Ar sinh} \frac{1}{\varepsilon}\\ &= \log \biggl(\frac{1}{\varepsilon} + \sqrt{1 + \frac{1}{\varepsilon^2}}\biggr)\\ &= \log \frac{1}{\varepsilon} + \log 2 + \log \frac{1 + \sqrt{1+\varepsilon^2}}{2}. \end{align}
Thus we have
$$K(k) = \log \frac{1}{\sqrt{1-k^2}} + O(1) = \frac{1}{2}\log \frac{1}{1-k} - \frac{1}{2}\log (1+k) + O(1) = \frac{1}{2}\log \frac{1}{1-k} + O(1).$$
In our specific case, with $k = \sin \bigl(\frac{\pi}{2} - \frac{\delta}{2}\bigr) = \cos \frac{\delta}{2}$, we have $\varepsilon = \sqrt{1-k^2} = \sin \frac{\delta}{2} = \frac{\delta}{2} + O(\delta^3)$, so
$$\log \frac{1}{\varepsilon} = \log \frac{2}{\delta} + O(\delta^2)$$
and overall
$$K\bigl(\cos \tfrac{\delta}{2}\bigr) = \log \frac{1}{\delta} + O(1),$$
where the $O(1)$ term is not only bounded, it in fact converges as $\delta \to 0$. We have the relevant terms:
$$K\bigl(\cos \tfrac{\delta}{2}\bigr) = \log \frac{1}{\delta} + 2\log 2 + \int_0^1 \frac{t\,dt}{1+t^2+\sqrt{1+t^2}} + \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \frac{d\varphi}{\sin \varphi} + o(1).$$
We approximate the general integral $$\int_0^{\frac{\pi}{2}}\left(1-m^2 \sin^2 \theta \right)^\alpha d\theta \tag{1}$$
By Taylor developpement $$ \left(1-m^2 \sin^2 \theta \right)^\alpha = 1-\alpha \sin^2 (\theta). m^2 +\mathcal{O}(m^4) $$
As $\left(1-m^2 \sin^2 \theta \right)^\alpha >0$, we can change the order of integration and limit $$ \begin{align} \int_0^{\frac{\pi}{2}}\left(1-m^2 \sin^2 \theta \right)^\alpha d\theta &= \int_0^{\frac{\pi}{2}}\left(1-\alpha \sin^2 (\theta). m^2 +\mathcal{O}(m^4) \right) d\theta \\ &=\frac{\pi}{2}-\alpha \frac{\pi}{4} m^2 + \mathcal{O}(m^4) \tag{2} \end{align} $$
Replace $\alpha = -\frac{1}{2}$ and $\frac{1}{2}$ to $(2)$, you will get the results.
Best Answer
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 $$K(k) = \frac12 \left|\log (1-k^2)\right| + O(1) = \frac12 \left|\log \epsilon\right| + O(1).$$