How can a ring fail to be finite over the subring of invariants under a finite group action? Let $R$ be a commutative, unital ring with an action of a finite group $G$.
If $G$ fixes a subring elementwise over which $R$ is finitely generated (e.g. if $R$ is a f.g. $k$-algebra, for $k$ some ring, and $G$'s action is by $k$-algebra automorphisms), then $R$ is finite over $R^G$, since under all circumstances it is integral over $R^G$, and in the present situation it is also finitely generated.
On the other hand, it seems plausible to me that in general, $R$ needn't be finite over $R^G$.

(1) Is this true?
(2) If "yes," how much pathology in $R$ is needed for it to happen? Can it be an integrally closed noetherian integral domain?
(3) Here is a toy example: $R=k[x_1,y_1,x_2,y_2,\dots,x_i,y_i,\dots]$; $G = \mathbb{Z}/2\mathbb{Z}$, with the generator interchanging $x_i$ and $y_i$ for all $i$. In this specific case, is $R$ finite over $R^G$?

 A: Mohan's comment shows that the answer to (1) is "yes" because the answer to (3) is "no". (2) is still open but I'll be satisfied with answers to (1) and (3) and mark the question answered.
A: Leaving a more elaborate answer as to why $R^G \subset R$ is not finite in the given example. (From the comments, it may not be immediately obvious that $R^G$ is generated by the elements written.)
I shall always assume below that $k$ has characteristic different from two.
First, consider the finite polynomial ring: $R_n = k[x_1, y_1, \ldots, x_n, y_n]$. For all the rings considered, $\sigma$ will denote the automorphism that swaps $x_i \leftrightarrow y_i$.
They key ingredient will be Noether's bound, which says:

Suppose $G$ is a finite group acting on $S = k[X_1, \ldots, X_n]$ by graded $k$-algebra automorphisms, and $|G|!$ is invertible in $K$. Then, $S^G$ is generated, as a $k$-algebra, by elements of degree $\leqslant |G|$.

Since $\sigma$ is a graded group action of order $2$ and $2!$ is invertible in $k$, it comes down to computing the homogeneous polynomials of degrees $1$ and $2$ that are fixed by $\sigma$. From this, it is easy to see that
$$R_n^G = k[x_{i} + y_{i},\, x_{i} y_{i},\, x_{i} x_{j} + y_{i} y_{j} : 1 \le i,\, j \le n].$$
(Note that intermediate computations might make it look like we also need $x_{i}^{2} + y_{i}^{2}$ and $x_{i} y_{j} + x_{j} y_{i}$, but algebraic identities reduce it to the above.)
Now, using the above, we can claim that the "same" thing holds in infinitely many variables as well, i.e.,
$$R^{G} = k[x_{i} + y_{i},\, x_{i} y_{i},\, x_{i} x_{j} + y_{i} y_{j} : i,\, j \geqslant 1].$$
To see this, note that if $f \in R^G$, then $f$ only involves finitely many variables and hence, $f$ is actually in some $R_n^G$, and then we are done.
Now it is easy to see that $R$ is not a finite $R^G$-algebra. Indeed, it suffices to show that
$$R \neq R^G[x_1, y_1, \ldots, x_n, y_n]$$
for any $n \geqslant 1$.
But it should be clear now that $x_{n + 1}$ isn't in the algebra on the right. (If it were, then a degree argument shows that $x_{n + 1}$ would be a $k$-linear combination of $\{x_i, y_i : i \leqslant n\} \cup \{x_i + y_i : i > n\}$.)
