Suppose $0<k<1$ and $\displaystyle K(k)=\int_0^1\frac{\mathrm{d}x}{\sqrt{(1-x^2)(1-k^2x^2)}}$. Let $\tilde{k}$ be $\tilde{k}^2=1-k^2$. Show that $$\displaystyle K(k)=\frac{2}{1+\tilde{k}}K\left(\frac{1-\tilde{k}}{1+\tilde{k}}\right)$$

There's a hint in Stein's Complex Analysis which is this change of variable : $x=\dfrac{2t}{1+\tilde{k}+(1-\tilde{k})t^2}$.

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    $\begingroup$ Seems like you can find your answer here math.stackexchange.com/questions/811444/…. Exact same question just now in the trigonometric form. $\endgroup$ May 29, 2014 at 13:11
  • $\begingroup$ That's a bit similar but not same ! $\endgroup$ May 30, 2014 at 15:30
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    $\begingroup$ It's actually a particular case of the hypergeometric identity $$_{2}F_{1} \Big( a, a + \frac{1}{2} - b,;b+ \frac{1}{2};z^{2}\Big) = (1+z)^{-2a} \ _{2}F_{1} \Big(a,b;2b;\frac{4z}{(1+z)^{2}} \Big)$$ $\endgroup$ May 30, 2014 at 17:43
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    $\begingroup$ Apart from two typos in your last expression, you are actually pretty close to prove the identity yourself. $$\sqrt{1+\left(\frac{1-\tilde{k}}{1+\tilde{k}}\right)^2t^4-\frac{ 2\tilde{k}^2t^2 + 2t^2}{\color{red}{(1+\tilde{k})^2}} } = \sqrt{(1-t^2)\left(1-\left(\frac{1-\tilde{k}}{1+\tilde{k}}\right)^2t^2\right)}$$ $$\sqrt{1+\left(\frac{1-\tilde{k}}{1+\tilde{k}}\right)^2t^4+\frac{ 2\tilde{k}^2t^2 - 2t^2}{\color{red}{(1+\tilde{k})^2}}} = 1-\left(\frac{1-\tilde{k}}{1-\tilde{k}}\right)t^2 $$ $\endgroup$ May 30, 2014 at 19:06
  • $\begingroup$ Oh my god ... thank you ... you saved me :) $\endgroup$ May 30, 2014 at 21:16

2 Answers 2


For me, the only way I can remember this sort of complicated identities is through the relationship between the complete elliptic integral of the first kind $K(k)$ and the corresponding arithmetic-geometric mean.

$$K(k) = \int_0^\frac{\pi}{2} \frac{d\theta}{\sqrt{1-k^2\sin^2\theta}} = \frac{\pi}{2\text{AGM}( 1, \sqrt{1-k^2})}\tag{*1}$$

Consider following integral

$$I(a,b) = \int_0^{\frac{\pi}{2}} \frac{d\theta}{\sqrt{a^2\cos^2\theta + b^2\sin^2\theta}}$$ Introduce $x = b\tan\theta$, we can rewrite it as

$$I(a,b) = \int_0^{\frac{\pi}{2}} \frac{d\tan\theta}{ \sqrt{(1+\tan^2\theta)(a^2+b^2\tan^2\theta})} = \int_0^\infty \frac{dx}{\sqrt{(x^2 + a^2)(x^2+b^2)}}\tag{*2}$$ Substitute $x$ by $\sqrt{ab} t$, we have $$I(a,b) = \frac{1}{\sqrt{ab}}\int_0^\infty \frac{dt}{\sqrt{t^4 + \left(\frac{a}{b}+\frac{b}{a}\right)t^2 + 1}} = \frac{1}{\sqrt{ab}}\int_0^\infty \frac{1}{\sqrt{ ( t - t^{-1})^2 + \frac{(a+b)^2}{ab}}}\frac{dt}{t} $$ Notice the last integrand is invariant under transform $\displaystyle\;t \leftrightarrow \frac{1}{t}$. If we introduce two more variables $s$ and $y$ such that $$s = \frac12 (t - t^{-1}) = \frac{y}{\sqrt{ab}}$$ and using the fact $$\frac{dt}{t} = \frac{d(t - t^{-1})}{t + t^{-1}} = \frac{ds}{\sqrt{s^2+1}}$$ We can rewrite $I(a,b)$ as $$ \frac{2}{\sqrt{ab}}\int_1^\infty \frac{1}{\sqrt{ ( t - t^{-1})^2 + \frac{(a+b)^2}{ab}}}\frac{dt}{t} = \frac{2}{\sqrt{ab}}\int_0^\infty \frac{ds}{\sqrt{\left(4s^2 + \frac{(a+b)^2}{ab}\right)(s^2+1)}}\\ = \int_0^\infty \frac{dy}{\sqrt{\left(y^2 + \left(\frac{a+b}{2}\right)^2\right)(y^2 + ab)}} $$ Compare this with $(*2)$, we obtain an important identity: $$I(a,b) = I\left(\frac{a+b}{2}, \sqrt{ab}\right)$$ This means $I(a,b)$ is invariant if we replace $(a,b)$ by their AM and GM.

Start with any pair of numbers $a,b$, it is well known if you repeat taking AM/GM of them, the pairs will ultimately converge to a single number. This is called the arithmetic geometric mean of $a$ and $b$ and usually denoted as $\text{AGM}(a,b)$. If one replace $a$, $b$ by this AGM in the definition of $I(a,b)$, we obtain

$$I(a,b) = \frac{\pi}{2\text{AGM}(a,b)}$$

Together with the obvious identity $K(k) = I(1,\sqrt{1-k^2})$, we immediately obtain $(*1)$.

Using these tools and notice $\text{AGM}(a,b)$ is homogenous. i.e.

$$\text{AGM}(\lambda a, \lambda b) = \lambda \text{AGM}(a,b) \quad\implies\quad I(\lambda a, \lambda b) = \frac{1}{\lambda} I(a,b),$$ the desired identity follows immediately.

$$ K(k) = I(1,\tilde{k}) = I\left(\frac{1+\tilde{k}}{2},\sqrt{\tilde{k}}\right) = \frac{2}{1+\tilde{k}} I\left(1,2\frac{\sqrt{\tilde{k}}}{1+\tilde{k}}\right)\\ = \frac{2}{1+\tilde{k}} K\left( \sqrt{1 - \frac{4\tilde{k}}{(1+\tilde{k})^2}} \right) = \frac{2}{1+\tilde{k}} K\left( \frac{1-\tilde{k}}{1+\tilde{k}} \right) $$

  • $\begingroup$ complicated but very very interesting .. thank you so much $\endgroup$ May 29, 2014 at 18:02
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    $\begingroup$ A nice description about Landen's transform! (+1) $\endgroup$ Jun 2, 2014 at 22:33
  • $\begingroup$ Very nice. See also: JM Borwein and PB Borwein ”The arithmetic-geometric mean and fast computation of elementary functions”, SIAM V26, 351 (1984) $\endgroup$
    – minmax
    Nov 16, 2021 at 22:39




Now by reparameterizing the Integral :

$\begin{align*} \displaystyle K(k)&=2\int_0^1\frac{1+\tilde{k}+(1-\tilde{k})t^2}{\sqrt{(1+\tilde{k})^2+(1-\tilde{k})^2t^4-2\tilde{k}^2t^2-2t^2}}. \frac{1+\tilde{k}+(1-\tilde{k})t^2}{\sqrt{(1+\tilde{k})^2+(1-\tilde{k})^2t^4-2k^2t^2}}\\ &\qquad.\frac{1+\tilde{k}-(1-\tilde{k})t^2}{[1+\tilde{k}+(1-\tilde{k})t^2]^2}.\mathrm{d}t\\ &=\frac{2}{1+\tilde{k}}\int_0^1\frac{1}{\sqrt{1+(\frac{1-\tilde{k}}{1+\tilde{k}})^2t^4-\frac{2\tilde{k}^2t^2+2t^2}{\color{red}{(1+\tilde{k})^2}} }}. \frac{1-(\frac{1-\tilde{k}}{1+\tilde{k}})t^2}{\sqrt{1+(\frac{1-\tilde{k}}{1+\tilde{k}})^2t^4+\frac{2\tilde{k}^2t^2-2t^2}{\color{red}{(1+\tilde{k})^2}}}}\mathrm{d}t \end{align*}$

Here is the Complete proof after achille hui's correction:

$\begin{align*} \displaystyle LHS&=\frac{2}{1+\tilde{k}}\int_0^1 \frac{1-(\frac{1-\tilde{k}}{1+\tilde{k}})t^2} {\sqrt{(1-t^2)\left(1-\frac{1-\tilde{k}}{1-\tilde{k}}t^2\right)}.\sqrt{\left(1-(\frac{1-\tilde{k}}{1+\tilde{k}})t^2\right)^2}}\mathrm{d}t\\ &=\displaystyle\frac{2}{1+\tilde{k}}\int_0^1 \frac{\mathrm{d}t} {\sqrt{(1-t^2)\left(1-\frac{1-\tilde{k}}{1-\tilde{k}}t^2\right)}}\\ &=\displaystyle\frac{2}{1+\tilde{k}}K\left(\frac{1-\tilde{k}}{1+\tilde{k}}\right)\tag{QED} \end{align*}$


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