2

I work on Complete elliptic integral of the second kind and I want to show this :

$$E(2)=-i\Big(E(2)-\sqrt{\frac{2}{\pi}}\Gamma\Big(\frac{3}{4}\Big)^2\Big)$$ Where $E(k)$ denotes the Complete Elliptic Integral of the Second Kind with parameter $m=k^2$ And $\Gamma(x)$ denotes the gamma function an $i$ the imaginary unit .

To prove it I compare two integrals like this :

$$\int_{0}^{\frac{\pi}{2}}\sqrt{(\sin^2(x)-\cos^2(x))} \quad and \int_{0}^{\frac{\pi}{2}}\sqrt{(\cos^2(x)-\sin^2(x))}$$

And in fact with a numerical approach it sems to be :

$$\int_{0}^{\frac{\pi}{2}}\sqrt{(\sin^2(x)-\cos^2(x))} = \int_{0}^{\frac{\pi}{2}}\sqrt{(\cos^2(x)-\sin^2(x))}$$ Wich gives the result

Remains to evaluate this two integrals wich is not so hard .

So I would like to know if there is an other way to prove it .

Thanks a lot

Miss and Mister cassoulet char
  • 3,511

1 Answers1

5

A few standard identities between elliptic integrals will help here. Let's use definitions based on parameter $m$ so that $$K(m) =\int_{0}^{\pi/2}\frac{dx}{\sqrt{1-m\sin^2x}},E(m)=\int_{0}^{\pi/2}\sqrt{1-m\sin^2x}\,dx\tag{1}$$ Also note the Legendre's identity $$K(m) E(1-m)+K(1-m)E(m)-K(m)K(1-m)=\frac{\pi}{2}\tag{2}$$ Putting $m=1/2$ in the above identity we get $$2K(1/2)E(1/2)-K^2(1/2)=\frac{\pi}{2}$$ Also it is well known (and easily computed from the definitions $(1)$) that $$K(1/2)=\frac{\Gamma^2(1/4)}{4\sqrt{\pi}}$$ and therefore $$E(1/2)=\frac{\pi^{3/2}}{\Gamma^2(1/4)}+\frac{\Gamma^2(1/4)}{8\sqrt{\pi}}$$ Next use the identity $$E(1/m)=\frac{E(m) +iE(1-m)-(1-m)K(m) - imK(1-m)} {\sqrt{m}} \tag{3}$$ with $m=1/2$ and you are done.

Paramanand Singh
  • 87,309