Exercise 15.2.12

Proof. Consider the integral

where $\gamma ,\delta$ are such that $[\gamma ,\delta ]\subset ]0,\pi [$ and $𝜃\mapsto f(𝜃)=\frac{1}{\sqrt{1-k^2{\sin <!--\; FUNCTION\; APPLICATION\; -->}^2𝜃}}$ is defined (and continuous) on $[\gamma ,\delta ]$:

if the modulus $k$ is real and positive, this requires $[\gamma ,\delta ]\subset ]-\arcsin \left(\frac{1}{k}\right),\arcsin \left(\frac{1}{k}\right)[$.

Write $\alpha =\sin <!--\; FUNCTION\; APPLICATION\; -->(\gamma ),\beta =\sin <!--\; FUNCTION\; APPLICATION\; -->(\delta )$, and consider $\psi =\arcsin :[-1,1]\to [0,\pi ]$ (so that $t=\sin <!--\; FUNCTION\; APPLICATION\; -->𝜃⟺𝜃=\psi (t)$ if $-1<t<1$ and $𝜃\in [0,\pi ])$).

Then $\psi$ is continuously differentiable, and is strictly increasing, thus $\psi ([\alpha ,\beta ])=[\psi (\alpha ),\psi (\beta )]$, and $\psi$ induces a bijection $[\alpha ,\beta ]\to [\psi (\alpha ),\psi (\beta )]=[\gamma ,\delta ]$. The Theorem of Integration by Substitution gives

where

Therefore, if ${\int }_{\gamma }^{\delta }\frac{1}{\sqrt{1-k^2{\sin <!--\; FUNCTION\; APPLICATION\; -->}^2𝜃}}d𝜃$ make sense,

Suppose now that $k=i$. Then, for all $r\in ]-1,1[$,

Therefore, for all $r\in ]-1,1[$, and for all $s\in ]-\frac{\varpi }{2},\frac{\varpi }{2}[$,

Therefore, for all $s\in ]-\frac{\varpi }{2},\frac{\varpi }{2}[$, $\phi (s)=\sin <!--\; FUNCTION\; APPLICATION\; -->\mathrm{am}(s)=\mathrm{sn}(s)$, for the modulus $k=i$. By continuity, this is also true for $s=\pm \frac{\varpi }{2}$:

If we know the properties of symmetry (15.9) and periodicity of $\mathrm{sn}$, we can conclude $\phi =\mathrm{sn}$ for the modulus $k=i$. □