# Equality of two elliptic integrals

by wolfram 1 and wolfram 2

It's true that $$\displaystyle \int_0^{\frac{\pi}{2}}\int_0^{\frac{\pi}{2}} \frac{2}{\sqrt{1-\sin^2\theta\sin^2\phi}} d\phi d\theta =\int_0^{\frac{\pi}{2}}\int_0^{\frac{\pi}{2}} \frac{1}{\sqrt{\sin\theta\sin\phi}} d\phi d\theta$$ ?

If yes, I need a way to prove the equality of the following two integrals

I tried everything but I am unable to convert into a standard form so How do I solve this problem.

Addition 1: For the second integral $$\int_0^{\frac{\pi}{2}}\int_0^{\frac{\pi}{2}} \frac{1}{\sqrt{\sin\theta\sin\phi}} d\phi d\theta =\bigg(\int_0^{\frac{\pi}{2}}\frac{1}{\sqrt{\sin\theta}}d\theta\bigg)^2$$ and we can use that $$2 \int^{\frac{\pi}{2}}_{0}\frac{1}{\sqrt{\sin x}} \mathrm dx = \frac{\Gamma(1/2)\Gamma(1/4)}{\Gamma(3/4)} = \frac{\Gamma \left( \frac{1}{4}\right)^2}{\sqrt{2\pi}}$$

Addition 2: For the first integral

Let $$\displaystyle K(k)=\int_0^{\frac{\pi}{2}}\frac{1}{\sqrt{1-k^2\sin^2 t}}dt$$ ( Complete Elliptic Integral of the First Kind). we know that $$\displaystyle K(k)=\frac{\pi}{2}\sum_{n=0}^\infty \left(\frac{(2n)!}{2^{2n}(n!)^2}\right)^2k^{2n}$$. Then it's not difficult de show that $$\int_0^{\frac{\pi}{2}}\int_0^{\frac{\pi}{2}} \frac{1}{\sqrt{1-\sin^2\theta\sin^2\phi}} d\phi d\theta=(\pi /2 )^2 \sum _{n=0}^{\infty }(\frac{(2n)!}{4^n(n!)^2})^3$$.

The egality of the two integrals hold if we can calculte $$\sum _{n=0}^{\infty }(\frac{(2n)!}{4^n(n!)^2})^3$$ Wolfram gives

• Yes both integrals are $2K^2(1/2)$, where $K(m)$ is the complete elliptic integral of the first kind.
– user
Mar 5, 2021 at 23:11
• @User: Any proof of it?
– FDP
Mar 5, 2021 at 23:13
• @User Thank you for confirmation . Proof would be welcome Mar 5, 2021 at 23:17
• @user According to Mathematica, the first integral is equal to $2K^2 (1/2)$ whilst the second one is equal to $\frac{\sqrt{2}\pi^2\Gamma(5/4)}{\Gamma(3/4)^3}$. The 2 values are not equal but the difference between them is numerically small ($-1.7 *10^{-15}$
– NN2
Mar 5, 2021 at 23:19
• @NN2 Both values are equal. You are speaking about numerical noice.
– user
Mar 5, 2021 at 23:30

The equivalence of the elliptic integrals can be shown in the following way: \begin{align} \int\limits_0^{\frac{\pi}{2}}\int\limits_0^{\frac{\pi}{2}} \frac{d\phi\, d\theta}{\sqrt{1-\sin^2\theta\sin^2\phi}} & =\int\limits_0^{\frac{\pi}{2}}K(\sin\theta)d\theta\tag1\\ &=\int\limits_0^{1}\frac{K(r)dr}{\sqrt{1-r^2}}\tag2\\ &=\int\limits_0^{1}\frac{K\left(\frac{2\sqrt k}{1+k}\right)dk}{\sqrt{k}(1+k)}\tag3\\ &=\int\limits_0^{1}\frac{K\left(k\right)dk}{\sqrt{k}}\tag4\\ &=\int\limits_0^{1}\frac{dk}{\sqrt{k}} \int\limits_0^{\frac{\pi}{2}} \frac{d\phi}{\sqrt{1-k^2\sin^2\phi}}\tag5\\ &=\int\limits_0^{\frac{\pi}{2}}d\phi\int\limits_0^{1} \frac{dk}{\sqrt{k}\sqrt{1-k^2\sin^2\phi}}\tag6\\ &=\int\limits_0^{\frac{\pi}{2}}d\phi\int\limits_0^{\phi} \frac{d\theta}{\sqrt{\sin\theta\sin\phi}}\tag7\\ &=\frac12\int\limits_0^{\frac{\pi}{2}}\int\limits_0^{\frac{\pi}{2}} \frac{d\phi\, d\theta}{\sqrt{\sin\theta\sin\phi}}.\tag8\\ \end{align}

Explanation:

$$(1)$$: Definition of the complete elliptic integral of the first kind $$K(k)=\int\limits_0^{\frac{\pi}{2}}\frac{d\phi}{\sqrt{1-k^2\sin^2\phi}}$$.

$$(2)$$: $$\sin\theta\mapsto r$$.

$$(3)$$: $$r\mapsto\frac{2\sqrt k}{1+k}$$.

$$(4)$$: Landen's transformation $$K(k)=\frac1{1+k}K\left(\frac{2\sqrt k}{1+k}\right)$$.

$$(5)$$: Definition of the complete elliptic integral of the first kind.

$$(6)$$: Interchange of integration order.

$$(7)$$: $$k\mapsto\frac{\sin\theta}{\sin\phi}$$.

$$(8)$$: Use of the integrand symmetry.

• thank you user, it's perfect Mar 6, 2021 at 19:29
• You're welcome.
– user
Mar 6, 2021 at 19:32
• Thank you for your recent my question and for your answer. Apr 30, 2021 at 22:14

I wish i have found out the following solution (thanks go to Etanche and Jandri).

\begin{align}J&=\int_0^{\frac{\pi}{2}}\int_0^{\frac{\pi}{2}} \frac{1}{\sqrt{1-\sin^2(\theta)\sin^2 \varphi}}d\varphi d\theta\\ &\overset{z\left(\varphi\right)=\arcsin\left(\sin(\theta)\sin \varphi\right)}=\int_0^{\frac{\pi}{2}} \left(\int_0^ \theta\frac{1}{\sqrt{\sin(\theta-z)\sin(\theta+ z)}}dz\right)d\theta\tag1\\ &=\frac{1}{2}\int_0^{\frac{\pi}{2}} \left(\int_{u}^{\pi-u}\frac{1}{\sqrt{\sin u\sin v}}dv\right)du \tag2\\ &=\frac{1}{2}\int_0^{\frac{\pi}{2}} \left(\int_{u}^{\frac{\pi}{2}}\frac{1}{\sqrt{\sin u\sin v}}dv\right)du+\underbrace{\frac{1}{2}\int_0^{\frac{\pi}{2}} \left(\int_{\frac{\pi}{2}}^{\pi-u}\frac{1}{\sqrt{\sin u\sin v}}dv\right)du}_{w=\pi-v}\\ &=\int_0^{\frac{\pi}{2}} \left(\int_{u}^{\frac{\pi}{2}}\frac{1}{\sqrt{\sin u\sin v}}dv\right)du\\ &=\int_0^{\frac{\pi}{2}} \int_0^{\frac{\pi}{2}}\frac{1}{\sqrt{\sin u\sin v}}dudv-\int_0^{\frac{\pi}{2}} \left(\int_{0}^{u}\frac{1}{\sqrt{\sin u\sin v}}dv\right)du\\ &=\int_0^{\frac{\pi}{2}} \int_0^{\frac{\pi}{2}}\frac{1}{\sqrt{\sin u\sin v}}dudv-\int_0^{\frac{\pi}{2}} \frac{1}{\sqrt{\sin u}}\left(\int_0^{u}\frac{1}{\sqrt{\sin v}}dv\right)du\\ &=\int_0^{\frac{\pi}{2}} \int_0^{\frac{\pi}{2}}\frac{1}{\sqrt{\sin u\sin v}}dudv-\frac{1}{2}\left(\int_0^{\frac{\pi}{2}}\frac{1}{\sqrt{\sin v}}dv\right)^2\tag3\\ &=\boxed{\frac{1}{2}\int_0^{\frac{\pi}{2}} \int_0^{\frac{\pi}{2}}\frac{1}{\sqrt{\sin u\sin v}}dudv} \end{align}

$$(1)$$: $$\displaystyle dz=\dfrac{\sqrt{\sin^2\theta-\sin^2 z}}{\sqrt{1-\sin^2 z}}d\varphi$$, $$\sin^2 a-\sin^2 b=\sin(a-b)\sin(a+b)$$

$$(2)$$: Perform the change of variable $$u=\theta-z,v=\theta+z$$

$$(3)$$: $$\displaystyle\int_0^{\frac{\pi}{2}} f(x)f^\prime(x)dx=\frac{1}{2}\left(f^2\left(\frac{\pi}{2}\right)-f^2(0)\right)$$ and $$\displaystyle f(x)=\int_0^{x}\frac{1}{\sqrt{\sin u}}du$$

(Edit: problem fixed again)

• Is not the integral in (2) pure imaginary?
– user
Mar 7, 2021 at 7:59
• @User: i'm working on it to fix this problem.
– FDP
Mar 7, 2021 at 9:04
• FDP nice solution Mar 7, 2021 at 9:40
• Could you explain the integration limits in (2)? They do not seem to be correct after your edit.
– user
Mar 7, 2021 at 9:53
• @User: thanks. Fixed.
– FDP
Mar 7, 2021 at 20:39