If, as You have written,...
(1)
... and for all complex zeroes is , then all the terms of the sum contain the term ...
Kind regards
Lemma: If is a root of , is too. (This is not too hard to verify.)
So we have
(I'm a little scared about being able to change order of summation up above... maybe someone else should take a look at this line to confirm its validity.)
Since we have in our equation, we can drop the in the denominator.
Now split up the sum.
Since and we have in our denominator, the first sum gets absorbed into the term.
Thus we get .
- The sum over all equals the sum over the ones in the upper plane plus those in the lower plane.
All the ones in the lower plane have a unique one in the upper plane (it's conjugate). So, to hit all the ones in the lower plane we sum over all the conjugates of the ones in the upper.
So
But clearly this is equal to
- We just changed in the second sum where the conjugation occurs.
But, we can add these two together to get .
This is assuming everything converged.
I think it would be interesting to work it through assuming the Riemann Hypothesis is false. Recall if it's false, then instead of a conjugate pair of zeros, we have a pair of conjugate pair: and with .
. . . guess I need to check to make sure Von Mangoldt's expression for is valid independent of the placement of zeros in the critical strip. I think it is since it can be derived using the Residue Theorem (all the terms are sums of residues).