More detail on the last stages of my earlier proof.
If we have
then we also have
and therefore if
Therefore from theory of elliptic functions,
So in our case
If we have
then we have
which is a very small positive number.
We also have
which has roots at
Two of these are imaginary and two are real.
The real ones are only at t less than zero when
which is roughly 1.00227
Therefore the only term which can be decreasing at t == 0 is the first term and since the derivative is zero at t == 0 (even function) this shows that there cannot be a derivative of zero before hand since the negative gradient of the first term must equal all the others at the origin only (all the others are increasing, the first term is decreasing).
from the Hadamard product form of the zeta function where the roots, pi are the roots of the Zeta(s+1/2) function.
Therefore the two sided Laplace transform of our function does indeed represent a product form of the Zeta function.
A graph of
Shows that function being even and having only a derivative of 0 at t=0, and being positive decreasing function are plausible.
And a graph of the first 4 terms.
Shows that only the first term (its the one on the left) having a negative gradient for t<0 is plausible.
© Aron Palmer 20 February 2006