Jacobi amplitude

Remarks

The map is strictly increasing because its derivative $1/\sqrt{1 - k^2\sin^2\varphi}$ lies between $1$ and $1/k'$, where $k' = \sqrt{1 - k^2}$; the lower bound $1$ makes it unbounded in both directions. So $\operatorname{am}(\cdot, k)$ is smooth, odd, and $\frac{d}{du}\operatorname{am}(u, k) = \sqrt{1 - k^2\sin^2\operatorname{am}(u, k)}$. Since the integrand has period $\pi$, $F(\varphi + \pi, k) = F(\varphi, k) + 2K$, with $K = K(k)$ the complete elliptic integral of the first kind; hence $\operatorname{am}(u + 2K, k) = \operatorname{am}(u, k) + \pi$. Example: for $k = 0$, $F(\varphi, 0) = \varphi$, so $\operatorname{am}(u, 0) = u$; and $\operatorname{am}(K, k) = \pi/2$ for every $k$. Non-example: for $k = 1$ the integral $F(\varphi, 1) = \int_0^\varphi d\theta/\cos\theta$ only makes sense for $\abs{\varphi} < \pi/2$, so the definition above does not apply. The inverse there is the Gudermannian function $\operatorname{am}(u, 1) = 2\arctan(e^u) - \pi/2$, a bijection from $\R$ onto $(-\pi/2, \pi/2)$. The Jacobi elliptic functions are $\operatorname{sn} u = \sin\operatorname{am} u$ and $\operatorname{cn} u = \cos\operatorname{am} u$.