What is the Molien series of the SO(2)-invariant ring on the plane (sometimes written C[X]^{SO(2)} )? Let SO(2) be the group of rotations in the plane. What is the Molien series (sometimes called the Hilbert-Poincare series) of the SO(2)-invariant ring of polynomials?
N.B. The main goal being to calculate the generating set of the SO(2)-invariant ring of polynomials on the plane.
 A: Write $z = x + iy, \bar{z} = x - iy$ as usual, where $x, y \in \mathbb{C}[x, y]$ are regarded as complex-valued polynomial functions on the plane. The action of $SO(2)$ diagonalizes as
$$z \mapsto e^{i \theta} z, \bar{z} \mapsto e^{-i \theta} \bar{z}$$
so a monomial $z^n \bar{z}^m$ is sent to $e^{i(n-m)\theta} z^n \bar{z}^m$. So the invariant polynomials are given by $z^n \bar{z}^n = (x^2 + y^2)^n$; they are generated by $x^2 + y^2$ and the Molien series is $\frac{1}{1 - t^2}$.
Alternatively we can compute the Molien series as an integral
$$\int_{S^1} \frac{1}{\det(1 - tr_{\theta})} \, d \theta$$
where $r_{\theta} = \left[ \begin{array}{cc} \cos \theta & \sin \theta \\ - \sin \theta & \cos \theta \end{array} \right]$ is the rotation by $\theta$. The denominator of the integrand is $(1 - t \cos \theta)^2 + t^2 \sin^2 \theta = t^2 - 2t \cos \theta + 1$, which gives
$$\frac{1}{2\pi} \int_0^{2\pi} \frac{1}{t^2 + 1 - 2t \cos \theta} \, d \theta.$$
WolframAlpha suggests this integral can be evaluated by a tangent half-angle substitution but at this point it seems like more trouble than it's worth.
