When counting the number of integers $n(x)$ below a certain non-integer number $x$, the following series could be used:

$$n(x) = x-\frac12 + \sum_{n=1}^{\infty} \left(\frac{e^{x \mu_n}} {\mu_n}+\frac{e^{x \overline{\mu_n}}} {\overline{\mu_n}}\right)$$

where $\mu_n = 2\pi n i$ which are the zeros of the function $\xi_i(s) = \frac{2}{s}\sinh\left(\frac{s}{2}\right)$ that has the simple Hadamard product:

$$\displaystyle \xi_i(s) = \prod_{n=1}^\infty \left(1- \frac{s}{2 \pi ni} \right) \left(1- \frac{s}{{-2 \pi ni}} \right)$$

Note that $\xi_i(0)=1$ just like $\xi(0)=1$ in the Hadamard product of the non-trivial zeros of Riemann $\xi$-function when ignoring its probably superfluous factor $\frac12$.

Summing the powers of these paired zeros as follows yields ($B_r$ = Bernoulli number):

$$\hat{\sigma}_r = \sum_{n=1}^\infty \left(\frac{1}{(2\pi ni)^r}+ \frac{1}{(-2\pi ni)^r}\right) = -\frac{B_{r}}{r\,\Gamma(r)} \qquad r \in \mathbb{N}, r \gt 1\tag{1}$$

The domain of the series could be expanded as follows:

$$\hat{\sigma}_s = \frac{1}{(2\pi i)^s}\,\left(1+e^{\pi s i}\right)\sum_{n=1}^\infty \frac{1}{n^s}\qquad s \in \mathbb{C}, \Re(s) \gt 1 \tag{2}$$

$$\hat{\sigma}_s = 2^{1-s}\,\pi^{-s}\cos\left(\frac{\pi s}{2}\right)\sum_{n=1}^\infty\frac{1}{n^s}\qquad s \in \mathbb{C}, \Re(s) \gt 1 \tag{3}$$

Transferring the $\Gamma(r)$ from the RHS of (1) and $r \mapsto s$ gives:

$$2^{1-s}\,\pi^{-s}\cos\left(\frac{\pi s}{2}\right)\,\Gamma(s)\,\zeta(s) = \,\,? \tag{4}$$

which is 5/6-th of the famous functional equation. We know through various proofs (e.g. 7 different ones are listed in Titchmarsh book on the Zeta functions) that the ? $= \zeta(1-s)$ and that this provides the full analytic continuation of $\zeta(s)$ towards $s \in \mathbb{C}\,\, /\,\, {1}$.

Question: (I hope not too trivial...)

I know the Euler product reflects the multiplicative structure of the integers, whereas the functional equation reflects the additive structure, but is there an intuitive explanation on why the functional equation should emerge from summing the powers of the zeros required for the oscillating term to count the integers?


I read this interesting discussion, but could not derive the answer from it.

  • $\begingroup$ I don't understand your question. The coefficients of the Fourier series of $x-\lfloor x\rfloor$ are (up to minor modification) the same as the coefficients of $\zeta(1-s)=\sum_n a_n n^s$, this is the functional equation. That this sequence also appears in the poles of $\frac1{e^x-1}=\sum_n n a_n e^{-nx}$ is more or less equivalent to the Poisson summation formula (we care of its poles because the functional equation appears when applying the residue theorem to $\int_C \frac{x^s}{e^x-1} dx$ with $C$ a contour enclosing the real axis) $\endgroup$ – reuns Jan 23 at 0:58
  • $\begingroup$ @Reuns, this question has everything to do about what the $a_n$ explicitly are. Your comment amounts to "I can parametrize the problem with $a_n$ and show that their are several generating functions for $a_n$ from which the target identity follows." Agno is asking about the details for one application of that standard mathematical algorithm with explicit differing reps of the coefficients of the generating functions. $\endgroup$ – Tom Copeland Jan 23 at 2:14
  • $\begingroup$ @TomCopeland As usual no idea of what you mean. After reading again I think Agno just randomly started with the Fourier series of $x-\lfloor x\rfloor$. I don't know how he obtained $(1)$, it can be obtained either from the mittag leffler expansion of $\frac{x}{e^x-1}$ or from the functional equation, the two are intricately related. $\endgroup$ – reuns Jan 23 at 2:44
  • $\begingroup$ $\zeta(s)=\chi(s)\zeta(1-s)$. Once we know that $F(s)=\zeta(1-s)\chi(s)$ is a Dirichlet series then $F(2n)= \zeta(2n),n\ge 1$ implies immediately that $F(s)=\zeta(s)$. $\endgroup$ – reuns Jan 23 at 2:57
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    $\begingroup$ @Reuns, an appropriate, i.e. non-troll, response to a question you explicitly say you "don't understand" is to either be silent or ask a question of some part of it you don't understand rather than propose your own pet perspective as the only 'right' question and, therefore, 'right' answer. If you do think you understand the question, then give an answer rather than lengthy comments. (To note a phrase you've directed to another user that seems to have the quality of self-projection "much nonsense has been written on this.") $\endgroup$ – Tom Copeland Jan 23 at 3:01

The intermediary seems to be the Bernoulli number sequence which was originally birthed in summing up powers of the integers and in turn eventually gave birth, via the midwife the Mellin transform, to the Riemann and Hurwitz zeta functions. The MO-Q to which you link on motivating derivations of the functional equation for the Riemann zeta has an analytic continuation of the coefficients of the e.g.f. for the Bernoullis (the AC in fact gives the Riemann zeta function) with the numbers expressed two different ways, from which the FE of the Riemann zeta falls out. Your Eqn. 1 could be used to replace one of those reps for the Bernoullis--the one containing $\cos(\frac{\pi n}{2})$--giving the same end result, the FE. (Another perspective on the AC of the Bernoulli numbers to the Hurwitz and Riemann zeta functions is presented in this MO-Q.)

If you take the derivative of your initial equation, you obtain the Dirac delta function/operator comb on the left and a sum of cosines of the right, giving the core Poisson summation identity. The Mellin transform of the Dirac comb will give you the Riemann zeta function. For more on this, see "The Correspondence Principle" by Hughes and Ninham.

Edit 1/23-4/21:

Let me be elaborate on the last paragraph.

As you depict in your associated MSE-Q, a doubly infinite staircase function is obtained by adding $x$ to the Fourier series rep of the sawtooth wave. For $x > 0$, you can write the piecewise continuous semi-infinite staircase function as

$$H(x) \; n(x) = \sum_{n \geq 1} H(x-n) = H(x) [ \; x - \frac{1}{2} + 2 \sum_{n \geq 1} \frac{\sin(2 \pi n x)}{2 \pi n} \; ],$$

where $H(x)$ is the Heaviside step function (Heaviside knew all this).

Taking the derivative of both sides gives, for $x > 0 $, half of the core of the Poisson-summation distribution formula

$$ \sum_{n \geq 1} \delta(x-n) = H(x) [\;1 + 2 \sum_{n \ge 1} \cos(2 \pi n x) \;],$$

and, since

$$ \int_{0^{+}}^{\infty} x^{s-1} \delta(x-n) \; dx = n^{s-1}$$


$$ 2 \;\int_{0^{+}}^{\infty} x^{s-1} \cos(2\pi n x) dx = 2 \; (2\pi n)^{-s} \int_{0}^{\infty} x^{s-1} \cos(x) \; dx$$

$$= 2\; (2\pi n)^{-s} \; (s-1)!\; \cos(\frac{\pi}{2}s)$$

for $0 < Re(s) < 1$, taking the RHS as the analytic continuation for all $s$, we have a rudimentary form of the zeta FE crystallizing.

The Mellin transform term by term of the Dirac comb gives the Riemann zeta function series rep

$$ \zeta(1-s) = \sum_{n \ge 1} \frac{1}{n^{1-s}}$$

for $Re(s) < 0$. However, the $n =0$ term, i.e., the constant term, in the cosine series poses a problem in the term by term Mellin transformation of the series. Tossing it out--regularizing through the Hadamard finite-part scheme, justified by an inverse Mellin transform rep just as for the AC of the integral for the Euler gamma function--and equating the analytically continued Mellin transforms of the two reps gives the Riemann zeta functional symmetry equation

$$\zeta(1-s) = 2 \; (2\pi)^{-s} \; (s-1)! \; \cos(\frac{\pi}{2}s) \; \zeta(s).$$

Note how Mellin interpolation (MI) of the coefficients of an e.g.f. (a.k.a. Ramanujan's favorite Master Formula) underlies these transforms:

$$ \cos(2\pi n x) = \sum_{k \ge 0} \cos(\pi \frac{k}{2}) (2\pi n)^k \frac{x^k}{k!} = \sum_{k \ge 0} c_k \frac{x^k}{k!} = e^{c. x} ,$$

so to MI the coefficients, apply the normalized Mellin transform to the e.g.f. with the argument negated (in this case negation returns the same function)

$$\int_{0}^{\infty} e^{-c.x} \; \frac{x^{s-1}}{(s-1)!} \; dx = (c.)^{-s} = c_{-s} $$

$$ = \int_{0}^{\infty} \frac{x^{s-1}}{(s-1)!} \; \cos(-2\pi n x) \; dx = \cos(\pi \frac{k}{2}) (2\pi n)^k \; |_{k \to -s}. $$

For completeness, playing fast and loose with Dirac delta function/op reps, we can again apply MI via

$$ \int_{0}^{\infty} \frac{x^{s-1}}{(s-1)!} \; \delta(x-n) \; dx =\int_{0}^{\infty} \frac{x^{s-1}}{(s-1)!} \; \frac{1}{n} \delta(1-\frac{x}{n}) \; dx $$

$$ =\int_{0}^{\infty} \frac{x^{s-1}}{(s-1)!} \; \frac{1}{n} \frac{(1-\frac{x}{n})^{-1}}{(-1)!} \; dx = \int_{0}^{\infty} \frac{x^{s-1}}{(s-1)!} \; \sum_{k \geq 0}(-1)^k \frac{1}{n^{k+1}} \; \frac{1}{(-k-1)!} \; \frac{x^k}{k!} \; dx$$

$$ =\frac{1}{n^{k+1}} \; \frac{1}{(-k-1)!} \; |_{k \to -s} = \frac{1}{(s-1)!} \; n^{s-1} .$$

This is consistent with the limiting case of $H(1-x) \; \frac{(1-x)^{\omega}}{\omega!}$ as $\omega$ tends to $-1$ for the analytically continued integral rep of the Euler beta function, with $H(x)$ the Heaviside step function, and, therefore, fractional calculus. Being cautiously semi-conservative, one could look at the inverse Mellin transform rep of $\delta(x-n)$.

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    $\begingroup$ Thanks Tom. I wish I could reward you with an additional +1 for the "midwife". You have 'delivered' your insights very well ;-) $\endgroup$ – Agno Jan 23 at 15:48
  • $\begingroup$ Haha, had a friend, whom I taught how to dance country swing, who was a lawyer, a singer, and a midwife, so natural for the term to pop out. // Btw, your question is in the spirit of the Langlands program. $\endgroup$ – Tom Copeland Jan 23 at 16:31
  • $\begingroup$ It's a good question. The most likely reason I can think of why it hasn't received more upvotes is the concern I once heard two young directors/screenwriters of a movie who were worried about its reception gave--comparability--precisely the novelty of the movie. $\endgroup$ – Tom Copeland Jan 24 at 21:29
  • $\begingroup$ Thanks for the encouraging comment. The upvotes clearly indicate that your answer is much better than the question ;-) $\endgroup$ – Agno Jan 24 at 23:06
  • $\begingroup$ Let me provide a bit more context on where my question originates from. It started when I learned about Li's criterion, which implies that the RH can be expressed in terms of the growth of the coefficients of power series expansions for (mapped versions of) logs of log derivatives of functions involving $\zeta(s)$. Since primes and integers are directly related through the Euler product, I decided to explore for the integers whether an equivalent of Li's criterion does exist. I ended up making a full comparison and this is what I found.. (..drumroll...) $\endgroup$ – Agno Jan 24 at 23:06

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