Define by .
Note that
Show that L can be extended to an entire function
Tell me more i.e. what else do you know?
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.
I guess you know that is an entire function. Then all you have to prove is that the integral defines an entire function as well. There is a general theorem for proving that functions of the form are analytic. If you know it (which is probably the case), then you know what left you have to do... and this shouldn't be too hard (very similar to Gamma). What did you already try this way?
Actually I was wrong: the integral is only defined when ... It is not simple to extend it to . Like Chiph588@ said, you should repeat arguments used for zeta.
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 ?
, where are the Euler numbers.
I don't know if this will help at all though, since ...
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.
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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.