A language $L \subseteq X^{\ast}$ is recursively enumerable if there exists a Turing machine enumerating the strings in $L$, here after a word from $L$ is written we do not know if it will be the last one, and the machine runs forever without printing anything new (or just repeats words already written). Equivalently a lanuage is recursively enumerable if there exists a Turing machine which accepts inputs, and halts if the input word is in $L$, and has an unspecified behaviour otherwise. Or similar if there exists a partial computable function $f : L \to \{1\}$.
Given a recursively enumerable language, is there an effective way to decide if the language is finite?
I guess not, but how to prove this?
To try to determine whether a language is finite, we have to have some information about the language. So, to make the problem precise, it's necessary to specify how that information will be provided. Perhaps the most typical way to pose this sort of question is the following:
So we are asking whether that problem can be solved by a computable function:
The answer to that is "no". There is a general result, Rice's theorem, which applies. It shows that, when we ask questions like this one about the language accepted by a Turing machine (i.e. not questions about the machine itself), no nontrivial property of languages can be decided by a computable function that takes as input the machine $e$.
The general proof method is by reducing the halting problem to the question above. If we want to tell whether a particular Turing machine $e$ halts when started on an empty tape, we can make another Turing machine $e'$ which accepts a finite language if and only if $e$ halts. The function taking $e$ to $e'$ is computable. So if $f$ above was computable, we could solve the halting problem by taking input $e$ and asking whether $f(e') = 1$. That would let us solve the halting problem with a computable function, which is impossible. We conclude that $f$ is not computable, because the map taking $e$ to $e'$ is.