Jacobson radical of formal power series over an integral domain 
How can I calculate the Jacobson radical of R[[x]] when R is an integral domain?

It's easy to prove J(R[x])=0 when R is an integral domain.
 A: Consider the (surjective) ring homomorphism $\varepsilon\colon R[[x]]\to R$ that sends an element $f\in R[[x]]$ to its constant term. If $I$ is an ideal of $R[[x]]$, then $\varepsilon(I)$ is an ideal of $R$ and we have the induced surjective homomorphism
$$
\varepsilon_I\colon R[[x]]/I\to R/\varepsilon(I),\qquad \varepsilon_I(f+I)=\varepsilon(f)+\varepsilon(I)
$$
If $I$ is maximal, then $R[[x]]/I$ is a field, so $\varepsilon_I$ is an isomorphism. In particular, $f\in I$ if and only if $\varepsilon(f)\in \varepsilon(I)$ and $\varepsilon(I)$ is a maximal ideal.
Conversely, if $J$ is a maximal ideal of $R$, then $I=J+xR[[x]]$ is a maximal ideal of $R[[x]]$ and $J=\varepsilon(I)$ (prove it).
Thus $f$ belongs to every maximal ideals of $R[[x]]$ if and only if its constant term belongs to every maximal ideal of $R$, that is, to the Jacobson radical $J(R)$. Therefore
$$
J(R[[x]])=J(R)+xR[[x]]
$$
A: Consider the quotient $R[[x]]/(J(R)+xR[[x]])$. You should be able to see that it is Jacobson semisimple, so $J(R[[x]])\supseteq J(R)+xR[[x]]$.
We'd like to prove the converse. We could use the characterization of the radical as the set of all $x$ such that $1+xr$ is a unit for all $r$. As a lemma, recall  that the units of $R[[x]]$ are exactly the things with unit constant terms.
So, consider an arbitrary $r\in R[[x]]$ and an arbitrary element $j+xr'\in J(R)+xR[[x]]$. Can you see why $1+(j+xr')r$ a unit in $R[[x]]$?
