3
$\begingroup$

In all proofs of the Weil-Guinand explicit formula, there's this step (this is from Paul Garrett's notes):

enter image description here

Now consider this:

(1) $\frac{\zeta^\prime(s)}{\zeta(s)}$ has poles at $s=1$ and $s=\frac{1}{2} + i \gamma$, so if we move the contour from $\Re(s) = 1+\epsilon$, to $\Re(s) = \frac{1}{2}$, we'll pick up these residues.

(2) Yet if we write out the Dirichlet series of $\frac{\zeta^\prime(s)}{\zeta(s)}$, then $p^{-ms}$ has no pole between $\Re(s) = 1+\epsilon$ and $\Re(s) = \frac{1}{2}$, so the proof claims we can move the contour to $\Re(s) = \frac{1}{2}$ without picking up any residues. Is this really justified?

So the question is: does moving the contour involve any residues? (1) says yes. (2) says no. This is strange.

$\endgroup$

1 Answer 1

4
$\begingroup$

The integration and summation are interchanged before the contour is moved...

Yes, it is strange that such manipulations affect the outcome, but, for example, the analytic continuation of the sum is not the sum of the analytic continuations... :)

Yes, there is an air of danger! :)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.