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I am wondering about how this value is derived.

The definition of the elliptic integral of first kind $K(k)$ is used from Wolfram Alpha (which wierdly goes against the wikipedia definition. $K(k)$ in Wolfram is equal to $K(k^{2})$ in Wikipedia, I think.)

$\begin{align} K(\frac{1}{2})&=\int_{0}^{\frac{\pi}{2}}\frac{1}{\sqrt{1-\frac{1}{2}\sin^{2}x}}dx\\ &=\frac{\pi^{3/2}}{2\Gamma^2(\frac{3}{4})}\approx1.854 \end{align}$

I have searched the internet and math.se for days and could not find a proof of this value. All answers to similarly related problems just say that $K(\frac{1}{2})$ is a standard value and "easily derived from the definition".

Also, $E(\frac{1}{2})$ can be derived from $K(\frac{1}{2})$ by Legendre's relation, so I only need to know how to prove the special value of $K(\frac{1}{2})$.

Could somebody please give a proof of this? Thank you so much!

Maths Addict
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1 Answers1

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(Not seen in a dedicated post, but as an intermediate result in many answers, such as this one.)$$ \newcommand{\eqsubst}[1]{\underset{#1}{\phantom{\big[}=\phantom{\big]}}} \int_0^{\pi/2}\frac{dx}{\sqrt{1-\frac12\sin^2 x}}\quad\eqsubst{\color{blue}{y=\cos x}}\quad\int_0^1\sqrt{\frac{2}{1+y^2}}\,\frac{dy}{\sqrt{1-y^2}} \\\eqsubst{t=y^4}\frac1{2\sqrt2}\mathrm{B}\left(\frac14,\frac12\right)\color{gray}{=\frac{\Gamma^2(1/4)}{4\sqrt\pi}=\frac{\pi^{3/2}}{2\Gamma^2(3/4)}} $$ and similarly for the elliptic $E$.

metamorphy
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  • I was going to add a comment (to the question) with a link to my answer but then noticed that you have already linked it in your answer. +1 – Paramanand Singh Mar 16 '23 at 01:21