If x is an integer, then in (1.2) reduces to the Hurwitz zeta function in (1.5). Since for all , throughout this argument, it is supposed that is defined at , (x being fixed) and .
Recall Poisson’s summation formula [[19], p.287] (see also [[21], pp.7-8]):
(2.1)
where the prime ′ denotes that is omitted from the summation. The summation on the left refers to the integral values of n given by ; but, when either α or β is an integer, the corresponding term is halved. On the right, the summation means . It is supposed that
-
(p)
and are continuous in , the obvious one-sided continuity only being required at or ;
-
(q)
and are such that the integrals and converge, and is an integral of .
If either α or β is infinite, say , further condition is required that and as .
Applying Poisson’s summation formula (2.1) to a function
we find that
where, for convenience, n being an integer,
(2.3)
Integrating by parts yields
(2.4)
It is observed that the integral in (2.4) now converges for , and so is analytic for . Integrating by parts again in (2.4), we get
(2.5)
It is also observed that the integral in (2.5) now converges , and so is analytic for . Continuing in this way, it is found that can be continued to an entire function of s.
Integrating by parts yields
(2.6)
Here we have
(2.7)
which converges for every fixed x with . Likewise, it is seen that
It is noted that the last integral in (2.8) converges for , and the second summation in (2.6) converges for . Therefore in (2.6) is analytic for .
Integrating by parts in (2.6) and considering (2.7), we get
(2.9)
The integral in (2.9) converges for . The expression of in (2.9), proved for , shows that is analytic for since then the general term is uniformly in s for s bounded and for every . Employing integration by parts repeatedly, we observe that can be continued analytically to the whole s plane.
It is found from (2.2), (2.3), (2.4), (2.6), and (2.7) that
Suppose now that . Then Poisson’s summation formula (2.1) gives
which, upon integrating by parts two involved integrals and using (2.7), yields
(2.11)
Adding (2.10) to (2.11) with the restriction of , we get
which yields
(2.12)
Applying the following integral formulas to (2.12):
(2.13)
and
(2.14)
we obtain, for ,
(2.15)
It is noted that (2.15) still holds for since the two involved series converge uniformly in s for (every ). Finally, we get, for ,
(2.16)
which, upon replacing s by , yields (1.6).