To calculate the residues associated with the singularities of the integrand
function , we can work with the functions
and
separately:
has only one singularity: a simple pole at 0 with residue 1, so the product function there has a residue of . To determine the residues of , we first consider the function Here denotes Euler's Gamma function. Note: Riemann would have used different notation, and stated where . This is now more often denoted by . We can also write where s = 1/2 + it, which reduces the Riemann Hypothesis to the assertion that all zeros of are real. Recall that this is the function with the simple functional equation and whose only zeros are the nontrivial zeros of . The trivial zeros are 'annihilated' by the Gamma function This is an integral function of order 1, meaning that its sequence of zeros {zn} is such that the sum of |zn| -a converges if and only if a > 1 (see Theorem 18, Ingham). We further note that (see Theorem 15, Ingham) and may now apply Hadamard's Factorisation Theorem. This gives us which then givesHere bo and b1 are constants which have yet to be determined, and the are the zeros of , or equivalently, the nontrivial zeros of . Taking logarithms and then deriving gives Letting s = 0, while observing that ,
(Euler's constant) and ,
we see that .
Note: The third term in this equation is expressible in terms of the digamma function (confusingly) notated . This is not related to Chebyshev's psi function encountered earlier. It is known to satisfyTherefore we have for a constant c. This is easily seen to have singularities at s = 1 (residue = -1)
Hence has the following
singularities and residues:
This leads directly to the explicit formula. To summarise, the steps for
computing the residues are as follows:
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