Define by .
Show that L can be extended to an entire function
Are you familiar with the derivation of the analytic continuation of ? Because this series has a similar continuation.
If not, that's ok because I don't think you're asking for the functional equation itself, just that it exists.
Before I make any head way on this problem, what are your thoughts on how to approach it?
What I know is how to show that the Gamma Function is meromorphic on . Then by evaluating in a specific way, one can show that is also meromorphic on , and therefore has to be meromorphic as well.
I actually thought that can somehow be extended holomorphically to similarly to the Gamma function. But thats all I can think of.
We have , so if you know that the integral defining Gamma converges, then so does this one.Meanwhile I asked myself why the integral converges in the first place for ?
Ok here is the proof for Zeta being meromorphic everywhere.
where are the Bernoulli numbers.
Fix , (all terms even)
Now for Re(s) > 1:
is holomorphic at t = 0, has Taylor expansion:
hence near t = 0.
Integrand of near t = 0 is . So converges if Re(n+s-1) > -1, i.e. if Re(s) > -n.
For : , a finite sum of meromorphic functions on .
is meromorphic for Re(s) > -n, n arbitrary, hence is meromorphic on .
I didn't know this proof; it is rather simple, and adaptable to many situations.
Notice that you only use the fact that there exists an expansion when , regardless of the value of (or of the convergence of the series expansion).
Since is (even analytic), there is also such an expansion when (this is Taylor-Young's theorem). And you can transpose the proof seamlessly.
It will show you that is meromorphic on , so you also have to justify it has no poles... That may require the values of the coefficients ; I'll think about it.
Sidenote: there is a justification missing in your proof, but maybe it was an implicit reference to a proof you did for Gamma. An argument is indeed necessary to justify that is analytic; it is of the form where, for all , is analytic, but this is not sufficient to conclude that the integral itself is analytic. A "domination" is needed, like for the continuity or differentiability theorems under the integral sign. But it is simple to apply here, so I guess you were told it is just like for Gamma.
Addendum: in fact, the same method proves the analyticity on .
Indeed, we get an expression like , for , where is analytic. We know that , hence the first term defines an analytic function.
Furthermore, for , (usual functional equation of used several times) hence is just a polynomial, hence it is analytic.
The only piece that is not analytic for is the term , which comes from . (This proves that 1 is the only pole for ; it is simple with residue 1).
However, for , the term is 0 because is 0 at 0. Therefore, all terms are analytic. Thus is analytic on for all n, hence on . qed.