Landen's transformation

Remarks

Part 2 is part 1 at $\varphi = \pi/2$, where $\varphi_1 = \pi$ and $F(\pi, k_1) = 2K(k_1)$. Proof idea for part 1: differentiate both sides in $\varphi$ and check, with $\tan(\varphi_1 - \varphi) = k'\tan\varphi$, that the integrands agree; both sides vanish at $\varphi = 0$. Check at small $k$: then $k' \approx 1$ and $k_1 \approx k^2/4 \approx 0$, so $\varphi_1 \approx 2\varphi$ and the right side is about $\frac12 \cdot 2\varphi = \varphi = F(\varphi, 0)$. The modulus shrinks fast: $k_1 < k^2$ for $0 < k < 1$, since $k_1 = k^2/(1 + k')^2$. Iterating part 2 sends the modulus to $0$ quadratically and gives $K(k) = \frac{\pi}{2}\prod_{n \ge 1} (1 + k_n)$, where $k_{n+1}$ is obtained from $k_n$ as $k_1$ from $k$. In terms of the integral $I(a, b) = \int_0^{\pi/2} d\theta / \sqrt{a^2\cos^2\theta + b^2\sin^2\theta}$, the transformation reads $I(a, b) = I\big(\frac{a + b}{2}, \sqrt{ab}\big)$, which is the step behind the arithmetic–geometric mean formula.

Going deeper

For a student

The complete form $K(k) = (1 + k_1)K(k_1)$ is equivalent to the identity $K(k) = \pi / (2M(1, k'))$ with the arithmetic–geometric mean $M$. Indeed $k_1' = 2\sqrt{k'}/(1 + k')$, and the homogeneity $M(\lambda a, \lambda b) = \lambda M(a, b)$ gives $M(1, k_1') = \frac{2}{1 + k'} M(1, k')$. Read backwards, with $k$ in place of $k_1$, it is the ascending Landen transformation, which pushes the modulus toward $1$. For the theta-function side, Landen's transformation corresponds to replacing the nome $q$ by $q^2$, which doubles the ratio $K'/K$.