Complete elliptic integral of the first kind $K(m)$ asymptotc expansion at $m = -\infty$ What is the asymptotic behavior of $K\left(-\frac{1}{\delta^2}\right), \delta > 0$ when $\delta$ tends to zero? Here
$$
K(m) = \int\limits_0^{\pi/2} \frac{d\theta}{\sqrt{1 - m\sin^2 \theta}},
$$
i.e. complete elliptic integral of the first kind in terms of the parameter $m = k^2$
Edit Using Mathematica I've obtained
$$
K\left(-\frac{1}{\delta^2}\right) = \delta \left(2\ln 2 - \ln \delta\right) + O(\delta^3),
$$
anyway I would like to see how it could be achieved.
 A: For $m=-\frac1{\delta^2}$,
$$
\begin{align}
&\int_0^{\pi/2}\frac1{\sqrt{1-m\sin^2(\theta)}}\,\mathrm{d}\theta\\
&=\delta\underbrace{\int_0^{\sqrt\delta}\frac1{\sqrt{\delta^2+t^2}}\frac{\mathrm{d}t}{\sqrt{1-t^2}}}_{\large t\,\mapsto\,\delta t}
+\delta\underbrace{\int_{\sqrt\delta}^1\frac1{\sqrt{\delta^2+t^2}}\frac{\mathrm{d}t}{\sqrt{1-t^2}}}_{\large t\,\mapsto\,1/t}\\
&=\delta\int_0^{1/\sqrt\delta}\frac1{\sqrt{t^2+1}}\frac{\mathrm{d}t}{\sqrt{1-\delta^2t^2}}
+\delta\int_1^{1/\sqrt\delta}\frac1{\sqrt{1+\delta^2t^2}}\frac{\mathrm{d}t}{\sqrt{t^2-1}}\\[12pt]
&=\delta\left(-\tfrac12\log(\delta)+\log\left(1+\sqrt{1+\delta}\right)-\tfrac12\log(\delta)+\log\left(1+\sqrt{1-\delta}\right)\right)+O\left(\delta^2\right)\\[12pt]
&=\delta\left(-\tfrac12\log(\delta)+\log(2)+\frac14\delta-\tfrac12\log(\delta)+\log(2)-\frac14\delta\right)+O\left(\delta^2\right)\\[12pt]
&=\delta\left(-\log(\delta)+2\log(2)\right)+O\left(\delta^2\right)
\end{align}
$$
using the expansions
$$
\frac1{\sqrt{1\pm\delta^2t^2}}=1+O\left(\delta^2t^2\right)
$$
If we use
$$
\frac1{\sqrt{1\pm\delta^2t^2}}=1\mp\frac12\delta^2t^2+O\left(\delta^4t^4\right)
$$
we get
$$
\bbox[5px,border:2px solid #C0A000]{\delta\left(-\log(\delta)+2\log(2)\right)+O\left(\delta^3\right)}
$$
We can get more of the asymptotic expansion by carrying these expansions further.
A: We have to estimate:
$$K\left(-\frac{1}{\delta^2}\right)=\delta\cdot\int_{0}^{\pi/2}\frac{d\theta}{\sqrt{\delta^2+\sin^2\theta}}=\delta\cdot\int_{0}^{1}\frac{dt}{\sqrt{(\delta^2+t^2)(1-t^2)}}$$
hence it is sufficient to map $t$ to $1-u$ and approximate the integrand function in a neighbourhood of $u=0$ by neglecting the smallest terms to recover:
$$ K\left(-\frac{1}{\delta^2}\right)\approx \delta\cdot\log\frac{4}{\delta} $$
as wanted.
A: Maple says it is 
$$ \left( 2\,\ln  \left( 2 \right) -\ln  \left( \delta \right) 
 \right) \delta+ \left( {\frac {1}{4}}-{\frac {\ln  \left( 2 \right) 
}{2}}+{\frac {\ln  \left( \delta \right) }{4}} \right) {\delta}^{3}+
 \left( -{\frac {21}{128}}+{\frac {9\,\ln  \left( 2 \right) }{32}}-{
\frac {9\,\ln  \left( \delta \right) }{64}} \right) {\delta}^{5}+O
 \left( {\delta}^{7} \right)
$$
