Where the celebrated “functional equation” for the Riemann zeta function comes from, as well as for the Hurwitz zeta function more generally. I believe the approach here is cleaner and better motivated than either the contour integration or theta function approaches people usually give, which seem to me a long mess of magic calculation. I think the approach here should make the functional equation even obvious, in hindsight.

Defining the zeta functions

Recall from our previous discussion of difference equations that there is a unique function \(F(p, x)\) such that we have: \(\newcommand{\nat}{\mathbb{N}}\)

  • \(\displaystyle \lim_{n \to +\infty} F(p, x + n) = 0\) whenever \(\Re(p) < -1\).
  • \(\frac{d}{dx} F(p, x) = (p + 1)F(p - 1, x)\).
  • \(\int_{x}^{x + 1} F(p, x) \; dx = x^{p + 1}\).

Furthermore, this \(F(p, x)\) is complex-differentiable with respect to \(p\) as well.

Note that the second and third conditions together entail that \(F(p, x) = (p + 1)x^p + F(p, x + 1)\). Together with the first condition, this entails that \(F(p, x) = (p + 1) \sum_{n = 0}^{\infty} (x + n)^p\) whenever \(\Re(p) < -1\).

We define the Hurwitz zeta function \(\zeta(-p, x)\) for \(p \neq -1\) as \(F(p, x)/(p + 1)\). The Riemann zeta function is defined as the special case \(\zeta(-p) = \zeta(-p, 1)\).

For the Hurwitz zeta function, the corollaries of the above properties are:

  • \(\displaystyle \lim_{n \to +\infty} \zeta(-p, x + n) = 0\) whenever \(\Re(p) < -1\).
  • \(\frac{d}{dx} \zeta(-p, x) = p \zeta(-(p - 1), x)\).
  • \(\int_{x}^{x + 1} \zeta(-p, x) \; dx = x^{p + 1}/(p + 1)\).
  • \(\zeta(-p, x) = x^{p + 1}/(p + 1) + \zeta(-p, x + 1)\).
  • \(\zeta(-p, x) = \sum_{n = 0}^{\infty} (x+n)^p\) whenever \(\Re(p) < -1\).

[To be pedantically unambiguous, in our domain of interest, we interpret \(x^p\) for arbitrary \(p\) via \(x^p = \exp(p \log(x))\), where \(\log\) is a logarithm assuming real values on positive inputs and then analytically continued to \(\Re(x) \geq 0\), \(x \neq 0\). (We could continue it to larger domains for \(x\), but will have no need for this. This punctured closed half-plane is the domain of bases of exponentiation for our purposes). When \(\Re(p) > 0\) or \(p = 0\), we also define \(0^p\) as \(\lim_{x \to 0} x^p\) (which is \(0\) when \(\Re(p) > 0\) and \(1\) when \(p = 0\)) and define \(\zeta(-p, 0)\) as \(\lim_{x \to 0} \zeta(-p, x) = \zeta(-p, 1) + 0^p\).]

Each of the conditions which together uniquely define \(F(p, x)\) is such that \(n^p \sum_{k = 0}^{n - 1} F(p, (x + k)/n)\) inherits the same condition. Thus, these two are equal. This is the “duplication/multiplication formula”. Similarly, \(\zeta(-p, x) = n^p \sum_{k = 0}^{n - 1} \zeta(-p, (x + k)/n)\).

Naive approach to the functional equation

\(\zeta(-p, x)\) naively equals \(\sum_{n \in \mathbb{N}} (x + n)^p\). For \(x \in (0, 1)\) (with some fuzziness about whether this includes the endpoints), this equals \(\sum_{n \in \mathbb{Z}} f(x + n)\), where \(f(x) = u(x) x^p\), where \(u\) is the Heaviside step function. So by naive Poisson summation, we get that \(\zeta(-p, 0)\) (or perhaps \(\zeta(-p, 1)\), or perhaps the halfway value between them) is \(\sum_{n \in \mathbb{Z}} \int_{0}^{\infty} x^p e^{-2 \pi i n x} \; dx\).

The difference between \(\zeta(-p, 0)\) and \(\zeta(-p, 1)\) is \(0^p\), which let’s ignore as it would vanish when \(\Re(p) > 0\). So we’ll take our Fourier series to sum to \(\zeta(-p, 1) = \zeta(-p)\).

Similarly, the \(n = 0\) term is \(\int_{0}^{\infty} x^p \; dx = \left( \infty^{p + 1} - 0^{p + 1} \right)/(p + 1)\). When \(\Re(p) < - 1\), the \(\infty^{p + 1}\) vanishes. When \(\Re(p) > -1\), the \(0^{p + 1}\) vanishes. So let’s just ignore both.

Also, by naive rescaling of variables from \(x\) to \(x/(-2 \pi i n)\) for nonzero \(n\) (even though this change of variables by an imaginary factor actually changes the integration path, going out now to a different infinity, so to speak), we get that \(\int_{0}^{\infty} x^p e^{-2 \pi i n x} \; dx = (-2 \pi i n)^{-1 - p} \int_{0}^{\infty} x^p e^{-x} \; dx\). For \(\Re(p) > -1\), this converges to \((-2 \pi i n)^{-1 - p} \Gamma(1 + p)\). So let’s just say that’s what it is, regardless of other incompatible constraints we’ve used on \(p\).

So we have \(\zeta(-p) = \sum_{n \in \mathbb{N}^+} \sum_{i^2 = -1} (2 \pi n)^{-1 - p} (-i)^{-1 - p} \Gamma(1 + p)\). Naively taking \(i^{\theta} = \left( e^{i \pi/2} \right)^{\theta} = e^{i \pi \theta/2} = \cos(\pi \theta/2) + \sin(\pi \theta/2) i\) (even though \(i\) has multiple logarithms), then \(\sum_{i^2 = -1} (-i)^{-1 - p} = 2 \cos(\pi (1 + p)/2) = -2 \sin(\pi p/2)\).

Thus, \(\zeta(-p, x) = -2 \sin(\pi p/2) (2 \pi)^{-1 - p} \Gamma(1 + p) \zeta(-(-1 - p))\).

Can we make this naive derivation work? Yes, with a bit more care:

Proper derivation of Fourier series for the Hurwitz zeta

Observe that, when \(\Re(p) > 0\), we have that \(\zeta(-p, 0) = \zeta(-p, 1)\). Thus, we can make a continuous periodic function \(x \mapsto \zeta(-p, x')\) on real \(x\), where \(x'\) is the unique value in \((0, 1]\) which differs from \(x\) by an integer. This function will furthermore be differentiable in both directions at every point (though these two derivatives won’t match at integer inputs). \(\newcommand{\ZetaCoeff}[3]{c_{#2}^{#1}(#3)}\) \(\newcommand{\FCoeff}[3]{d_{#2}^{#1}(#3)}\) \(\newcommand{\start}{\mathrm{start}}\)

Such bidifferentiable periodic functions admit Fourier series which converge everywhere to them. That is, extracting Fourier coefficients in the usual way as \(c_n = \int_{0}^{1} \zeta(-p, x) e^{-2 \pi i n x} \; dx\) and then stringing them together as the Fourier series1 \(f(x) = \sum_{n \in \mathbb{Z}} c_n e^{2 \pi i n x}\), we will have that \(\zeta(-p, x) = f(x)\).

The computation of \(c_0\) is straightforward: \(\int_{0}^{1} \zeta(-p, x) dx = 0^p/(p + 1) = 0\), given our presumption that \(\Re(p) > 0\) (indeed, this holds even if \(\Re(p) > -1\)).

[ For \(c_{\pm n}\) for positive integer \(n\), we can reason directly from the multiplication formula \(\zeta(-p, x) = n^p \sum_{k = 0}^{n - 1} \zeta(-p, (x + k)/n)\) that \(c_{\pm n} = c_{\pm 1} n^{-1 - p}\), for arbitrary \(p\). This makes the shape of the functional equation relating \(\zeta(-p)\) and \(\zeta(-(-1-p))\) apparent, leaving only the computation of \(c_{\pm 1}\). Similarly, the value \(c_{-1}\) for a given \(p\) is automatically the complex conjugate of the value for \(c_1\) for the complex conjugate of \(p\), by conjugation symmetry, so we only need to establish the value of \(c_1\).

we need not dwell on these ways of looking at them, as the method by which \(c_{1}\) is computed works just as well to directly compute \(c_{\pm n}\) and observe the same scaling/conjugation properties. But for now, the rest of this is written in terms of calculating \(c_1\). ]

Note that, when \(\Re(p) > 0\) so that \(0^p = 0\), we have by integration by parts that \(-2 \pi i \int_{0}^{1} \zeta(-p, x) e^{-2 \pi i x} \; dx = -p \int_{0}^{1} \zeta(-(p - 1), x) e^{-2 \pi i x} \; dx\). (This integration by parts amounts to the differentiation rule for Fourier series applied to the fact \(\frac{d}{dx} \zeta(-p, x) = p \zeta(-(p - 1), x)\).)

So we just need to determine the Fourier series for \(\Re(p) \in (-1, 0)\) and this will then extend by this recurrence to telling us the Fourier series for all \(\Re(p) > -1\) with \(\Re(p)\) not an integer. [We can presumably also get the values at \(\Re(p)\) an integer by continuity/meromorphic continuation arguments, if we want them.].

Now, let us observe in the other direction that when \(\Re(p) < 0\), we can evaluate \(c_1 = \int_{0}^{1} \zeta(-p, x) e^{-2 \pi i x} \; dx\) like so: Defining the function \(G(m) = \left( \int_{0}^{m} x^p e^{-2 \pi i x} \; dx \right) + \left( \int_{m}^{m + 1} \zeta(-p, x) e^{-2 \pi i x} \; dx \right)\), we see by differentiating with respect to \(m\) that \(G(m)\) is constant across all values of \(m\), while \(c_1 = G(0)\). Integration by parts turns \((-2 \pi i) \int_{m}^{m + 1} \zeta(-p, x) e^{-2 \pi i x} \; dx\) into \(-m^p + p \int_{m}^{m + 1} \zeta(-(p - 1), x) e^{-2 \pi i x} \; dx\). When \(\Re(p) < 0\), this vanishes in the limit as \(m \to \infty\), keeping in mind how \(\zeta(-(p - 1), x)\) vanishes in the limit and how it is being integrated against a function with bounded magnitude over a bounded region. Thus, we get that \(c_1 = G(0) = G(m) = \lim_{m \to \infty} G(m) = \int_{0}^{\infty} x^p e^{-2 \pi i x} \; dx\). Furthermore, by choosing \(0 < m < \infty\), we see this value \(G(m)\) is convergent for \(\Re(p) > -1\), as \(\int_{0}^{m} x^p e^{-2 \pi i x} \; dx\) is absolutely convergent for \(m < \infty\) and \(\Re(p) > -1\), while \(\int_{m}^{m + 1} \zeta(-p, x) e^{-2 \pi i x} \; dx\) is the integral of a continuous function on a closed interval for \(m > 0\) and thus convergent for such \(m\) regardless of \(p\).

Our goal now is to study \(\int_{0}^{\infty} x^p e^{-2 \pi i x} \; dx\) as a function of \(p\).

By Abel integration (the continuous analogue of Abelian summation, saying that if the Laplace transform of a function converges at 0, then its value there is the limit of its value at positive reals), where this is convergent (which we’ve seen occurs when \(\Re(p) > -1\)), this is equal to \(\lim_{s = 2 \pi i + \epsilon, \epsilon > 0, \epsilon \to 0} \int_{0}^{\infty} x^p e^{-sx} \; dx\).

By the rescaling properties of the Laplace transform (i.e., by the change of variables replacing \(x\) with \(x/s\)), we have that \(\int_{0}^{\infty} x^p e^{-sx} dx = s^{-1 - p} \int_{0}^{1} x^p e^{-sx} dx\) for positive \(s\). When \(\Re(p) > 0\), where this integral is absolutely convergent, this is \(s^{-1 - p} \Gamma(1 + p)\) by the definition of \(\Gamma(1 + p)\). By the analyticity of the Laplace transform through its region of absolute convergence, we get that \(\int_{0}^{\infty} x^p e^{-sx} dx = s^{-1 - p} \Gamma(1 + p)\) even for complex \(s\) with \(\Re(s) > 0\), again with the condition that \(\Re(p) > 0\). (Here, we interpret \(s^p = \exp(p \log(s))\) using the analytic continuation of the natural logarithm taking real values on the real axis, to the half-plane \(\Re(s) > 0\). In other words, \(s\) is treated as having argument in \((-\pi/2, \pi/2)\) for the purposes of determining \(\log(s)\).)

By the famous integration by parts under which the Gamma function acquires its hallmark recurrence relation, we have for \(\Re(p) > 0\) that \(\int_{0}^{\infty} x^p e^{-sx} dx = (p/s) \int_{0}^{\infty} x^{p - 1} e^{-sx} dx\). The \(s = 1\) case of this is the recurrence \(\Gamma(1 + p) = p \Gamma(1 + (p - 1))\), which is used to analytically continue the definition of \(\Gamma(1 + p)\) to all inputs other than negative integers \(p\). In particular, we now have that \(\int_{0}^{\infty} x^p e^{-sx} dx = s^{-1 - p} \Gamma(1 + p)\) for \(\Re(s) > 0\) and \(\Re(p) > -1\) (with conditional convergence, as opposed to the absolute convergence with \(\Re(p) > 0\)).

Thus, we can compute \(\lim_{s = 2 \pi i + \epsilon, \epsilon > 0, \epsilon \to 0} \int_{0}^{\infty} x^p e^{-sx} \; dx\) as the continuously varying value of \(s^{-1 - p} \Gamma(1 + p)\) at \(s = 2 \pi i\), and conclude that \(\int_{0}^{\infty} x^p e^{-2 \pi i x} \; dx = (2 \pi i)^{-1 - p} \Gamma(1 + p)\), with exponentiation with base \(\pm i\) treated as having argument \(\pm \pi/2\).

This gives us the Fourier series for \(\Re(p) \in (-1, 0)\) but it then automatically extends by our previous recurrence to give us the Fourier series for general \(\Re(p) > -1\) with \(\Re(p)\) not an integer.

In particular, as we noted before, where \(\Re(p) > 0\), the Fourier series will indeed converge to \(\zeta(-p)\) at the endpoints of the interval. Thus, in the region with \(\Re(p) > 0\) and \(\Re(p)\) not an integer, we can now conclude that \(\zeta(-p) = -2 \sin(\pi p/2) (2 \pi)^{-1 - p} \Gamma(1 + p) \zeta(-(-1 - p))\) as desired. As this is an equation beteween meromorphic functions on a region with an accumulation point (indeed, on an inhabited open set), it holds at all \(p\) by continuation.

TODO: Clean up the above

TODO: Also discuss the Hurwitz zeta function Fourier coefficients at all other powers \(p\) or over other unit intervals.

Footnotes:

  1. Where the infinite sum is interpreted in case of conditional convergence via partial sums extending equally into negative and positive indices. ↩