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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.

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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.

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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.

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    $\begingroup$ Would you mind elaborating on the approximation you've used? $\endgroup$ – uranix May 8 '15 at 8:22
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Maple says it is $$ \left( 2\,\ln \left( 2 \right) -\ln \left( \delta \right) \right) \delta+ \left( {\frac {1}{4}}-{\frac {\ln \left( 2 \right) }{2}}+{\frac {\ln \left( \delta \right) }{4}} \right) {\delta}^{3}+ \left( -{\frac {21}{128}}+{\frac {9\,\ln \left( 2 \right) }{32}}-{ \frac {9\,\ln \left( \delta \right) }{64}} \right) {\delta}^{5}+O \left( {\delta}^{7} \right) $$

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