Differentiability of Arithmetic-geometric Mean Let $\left(u_{n}\right)$ and $\left(v_{n}\right)$ classically defined by $u_{0}=a>0$, $v_{0}=b>0$ and $\forall n \in \mathbb{N}, u_{n+1}=\frac{1}{2}\left(u_{n}+v_{n}\right)$, $v_{n+1}=\sqrt{u_{n} v_{n}}$.
Both $\left(u_{n}\right)$ and $\left(v_{n}\right)$ converge toward a common limit that we denote by $f(a,b)$.
Is the function $f$ differentiable over $\mathbb{R}_+^* \times \mathbb{R}_+^*$? Analytic?
P.S.: This is probably a very well-know result, but I couldn't find it here.
 A: $f(a,b)$ has an expression in terms of a fairly simple integral,
$$ \frac{\pi}{2 f(a,b)} = \int_0^{\pi/2} \frac{d\theta}{\sqrt{a^2\cos^2{\theta}+b^2\sin^2{\theta}}} , $$
(originally proven by Gauss). The integrand is $C^{\infty}$ on $(0,\infty) \times (0,\infty) \times (0,\pi/2)$, so differentiation under the integral sign implies that the integral is as well (and hence, by the chain rule, so is $f$). It should be analytic too, since any multidirectional derivative exists and is continuous.
A: The AGM $\,f(a,b)\,$ is a homogeneous mean which implies that
$\, f(a,b) = a f(1,b/a) \,$ while the OEIS sequence A060691 has  generating function
$$\,f(b/a) = f(1,1-8x) = 1-4x-4x^2-16x^3-84x^4+O(x^5).$$
Thus $\,f(a,b)\,$ has a power series expansion in powers of
$\,x=(1-b/a)/8\,$ assuming that $\,|x|<1/8\,$ for convergence. Because
of the identity $$f(a,b) = f((a+b)/2,\sqrt{ab})$$ the region
of differentiability can be extended to all $\,a\,$ and $\,b\,$
nonzero.
Another way uses the Arithmetic-geometric mean formula
$$ f(a,b) = \frac{\pi}4 \cdot \frac{a+b}{K(\frac{a-b}{a+b})} $$
where $K(k)$ is the complete elliptic integral of the first kind. The function $K(k)$ is known to be differentiable.
