The analytic continuation and functional equation for the Riemann zeta function were proved in Riemann's 1859 memoir "On the number of primes less than a given magnitude." What is the earliest reference for the analytic continuation and functional equation of Dirichlet L-functions? Who first proposed that they might satisfy a Riemann hypothesis? Dirichlet did none of these things; his paper dates from 1837, and as far as I know he only considered his L-functions as functions of a real variable.

  • I would imagine it is Hecke. But I think Dirichlet himself did make use of the fact that the L-function has a simple pole at $s=1$. So in a sense he was complex analytic. – Anweshi Jan 27 '10 at 14:45
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    I think it was earlier than Hecke, but my knowledge of the 19th century literature is very poor. Dirichlet's L-functions do not have any poles. You mean that the zeta function has a pole? Euler and everyone after him knew that it diverged like 1/(s-1) as s--->1 for s real, which is weaker than knowing anything about a pole, but sufficient for some applications ("Dirichlet density" of primes in arithmetic progressions). – David Hansen Jan 27 '10 at 14:48
  • I mean the Dedekind zeta function of the cyclotomic field, which is a product of the Dirichlet $L$-functions for various characters of its Galois group which is isomorphic to $(\mathbb{Z}/n\mathbb{Z})^*$, has a simple pole at $s = 1$. This is a significant step in the proof of Dirichlet's theorem. The answer to your question could be Dedekind, since the name occurs to me, and the zeta function of the number fields are named after him. – Anweshi Jan 27 '10 at 14:56
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    Dedekind did not get analytic continuation of "his" zeta-functions beyond Re(s) > 1. The first person to prove analytic continuation of Dedekind zeta-functions (in general) a bit to the left of Re(s) = 1 was Landau, in 1903 I believe. He got continuation as far as Re(s) > 1 - 1/[K:Q], where K is the number field whose zeta-function you're dealing with. This is treated in Lang's Algebraic Number Theory. Before Landau, the density business involved limits as s approaches 1 from the right, as David writes. That one-sided limit does not imply anything about complex-analyticity around s = 1. – KConrad Jan 28 '10 at 5:24
  • I had the impression that Dedekind knew how to meromorphically continue just a teensy bit, even before Landau, but I may be mistaken. – paul garrett Jul 23 '11 at 18:31
up vote 21 down vote accepted

Riemann was the first person who brought complex analysis into the game, but if you ask just about functional equations then he was not the first. In the 1840s, there were proofs of the functional equation for the $L$-function of the nontrivial character mod 4, relating values at $s$ and $1-s$ for real $s$ between 0 and 1, where the $L$-function is defined by its Dirichlet series. In particular, this happened before Riemann's work on the zeta-function. The proofs were due independently to Malmsten and Schlomilch. Eisenstein had a proof as well (unpublished) which was found in his copy of Gauss' Disquisitiones. It involves Poisson summation. Eisenstein's proof is dated 1849 and Weil suggested that this might have motivated Riemann in his work on the zeta-function.

For more on Eisenstein's proof, see Weil's "On Eisenstein's Copy of the Disquisitiones" pp. 463--469 of "Alg. Number Theory in honor of K. Iwasawa" Academic Press, Boston, 1989.

To extend on Matt's comment about Euler, here is something I wrote up some years ago about Euler's discovery of the functional equation only at integral points. I hope there are no typos.

Although Euler never found a convergent analytic expression for $\zeta(s)$ at negative numbers, in 1749 he published a method of computing values of the zeta function at negative integers by a precursor of Abel's Theorem applied to a divergent series. The computation led him to the asymmetric functional equation of $\zeta(s)$.

The technique uses the function $$ \zeta_{2}(s) = \sum_{n \geq 1} \frac{(-1)^{n-1}}{n^s} = 1 - \frac{1}{2^s} + \frac{1}{3^s} - \frac{1}{4^s} + \dots. $$ This looks not too different from $\zeta(s)$, but has the advantage as an alternating series of converging for all positive $s$. For $s > 1$, $\zeta_2(s) = (1 - 2^{1-s})\zeta(s)$. Of course this is true for complex $s$, but Euler only worked with real $s$, so we shall as well.

Disregarding convergence issues, Euler wrote $$ \zeta_{2}(-m) = \sum_{n \geq 1} (-1)^{n-1}n^m = 1 - 2^m + 3^m - 4^m + \dots, $$ which he proceeded to evaluate as follows. Differentiate the equation $$ \sum_{n \geq 0} X^n = \frac{1}{1-X} $$ to get $$ \sum_{n \geq 1} nX^{n-1} = \frac{1}{(1-X)^2}. $$ Setting $X = -1$, $$ \zeta_{2}(-1) = \frac{1}{4}. $$ Since $\zeta_{2}(-1) = (1-2^2)\zeta(-1)$, $\zeta(-1) = -1/12$. Notice we can't set $X = 1$ in the second power series and compute $\sum n = \zeta(-1)$ directly. So $\zeta_2(s)$ is nicer than $\zeta(s)$ in this Eulerian way.

Multiplying the second power series by $X$ and then differentiating, we get $$ \sum_{n \geq 1} n^2X^{n-1} = \frac{1+X}{(1-X)^3}. $$ Setting $X = -1$, $$ \zeta_{2}(-2) = 0. $$ By more successive multiplications by $X$ and differentiations, we get $$ \sum_{n \geq 1} n^3X^{n-1} = \frac{X^2+4X+1}{(1-X)^4}, $$ and $$ \sum_{n \geq 1} n^4X^{n-1} = \frac{(X+1)(X^2+10X+1)}{(1-X)^5}. $$ Setting $X = -1$, we find $\zeta_{2}(-3) = -1/8$ and $\zeta_{2}(-4) = 0$. Continuing further, with the recursion $$ \frac{d}{dx} \frac{P(x)}{(1-x)^n} = \frac{(1-x)P'(x) + nP(x)}{(1-x)^{n+1}}, $$ we get $$ \sum_{n \geq 1} n^5X^{n-1} = \frac{X^4+26X^3+66X^2 + 26X +1}{(1-X)^6}, $$ $$ \sum_{n \geq 1} n^6X^{n-1} = \frac{(X+1)(X^4 + 56X^3 + 246X^2 + 56X+1)} {(1-X)^7}, $$ $$ \sum_{n \geq 1} n^7X^{n-1} = \frac{X^6 + 120X^5 + 1191X^4 + 2416X^3 + 1191X^2 + 120X + 1}{(1-X)^8}. $$ Setting $X = -1$, we get $\zeta_{2}(-5) = 1/4, \ \zeta_{2}(-6) = 0, \ \zeta_{2}(-7) = -17/16$.

Apparently $\zeta_{2}$ vanishes at the negative even integers, while $$ \frac{\zeta_{2}(-1)}{\zeta_{2}(2)} = \frac{1}{4}\cdot\frac{6\cdot 2}{\pi^2} = \frac{3\cdot 1!}{1\cdot \pi^2}, \ \ \ \ \frac{\zeta_{2}(-3)}{\zeta_{2}(4)} = -\frac{1}{8}\cdot\frac{30\cdot24}{7\pi^4} = -\frac{15\cdot 3!}{7\cdot \pi^4}, $$ $$ \frac{\zeta_{2}(-5)}{\zeta_{2}(6)} = \frac{1}{4}\cdot \frac{42\cdot 6!}{31\pi^6} = \frac{63 \cdot 5!}{31\cdot \pi^6}, \ \ \ \ \frac{\zeta_{2}(-7)}{\zeta_{2}(8)} = -\frac{17}{16}\cdot \frac{30\cdot 8!}{127\cdot \pi^8} = -\frac{255\cdot 7!}{127\pi^8}. $$

The numbers $1, 3, 7, 15, 31, 63, 127, 255$ are all one less than a power of 2, so Euler was led to the observation that for $n \geq 2$, $$ \frac{\zeta_{2}(1-n)}{\zeta_{2}(n)} = \frac{(-1)^{n/2+1}(2^n-1)(n-1)!}{(2^{n-1}-1)\pi^n} $$ if $n$ is even and $$ \frac{\zeta_{2}(1-n)}{\zeta_{2}(n)} = 0 $$ if $n$ is odd. Notice how the vanishing of $\zeta_{2}(s)$ at negative even integers nicely compensates for the lack of knowledge of $\zeta_2(s)$ at positive odd integers $> 1$ (which is the same as not knowing $\zeta(s)$ at positive odd integers $> 1$).

Euler interpreted the $\pm$ sign at even $n$ and the vanishing at odd $n$ as the single factor $-\cos(\pi n/2)$, and with $(n-1)!$ written as $\Gamma(n)$ we get $$ \frac{\zeta_{2}(1-n)}{\zeta_{2}(n)} = -\Gamma(n)\frac{2^n-1}{(2^{n-1}-1)\pi^n} \cos\left(\frac{\pi n}{2}\right). $$ Writing $\zeta_{2}(n)$ as $(1 - 2^{1-n})\zeta(n)$ gives the asymmetric functional equation $$ \frac{\zeta(1-n)}{\zeta(n)} = \frac{2}{(2\pi)^n} \Gamma(n)\cos\left(\frac{\pi n}{2}\right). $$ Euler applied similar ideas to $L(s,\chi_4)$ and found its functional equation. You can work this out yourself in Exercise 2 below.


  1. Show that Euler's computation of zeta values at negative integers can be put in the form $$ (1 - 2^{n+1})\zeta(-n) = \left.\left(u\frac{d}{du}\right)^{n}\right\vert_{u=1}\left(\frac{u}{1+u} \right) = \left.\left(\frac{d}{dx}\right)^{n}\right\vert_{x=0} \left(\frac{e^x}{1+e^x}\right). $$

  2. To compute the divergent series
    $$ L(-n,\chi_4) = \sum_{j \geq 0} (-1)^{j}(2j+1)^n = 1 - 3^n + 5^n - 7^n - 9^n + 11^n - \dots $$ for nonnegative integers $n$, begin with the formal identity $$ \sum_{j \geq 0} X^{2j} = \frac{1}{1-X^2}. $$ Differentiate and set $X = i$ to show $L(0,\chi_4) = 1/2$. Repeatedly multiply by $X$, differentiate, and set $X = i$ in order to compute $L(-n,\chi_4)$ for $0 \leq n \leq 10$. This computational technique is not rigorous, but the answers are correct. Compare with the values of $L(n,\chi_4)$ for positive $n$, if you know those, to get a formula for $L(1-n,\chi_4)/L(n,\chi_4)$. Treat alternating signs like special values of a suitable trigonometric function.

  • Thanks! I always wondered, but was apprehensive of looking into the collected works of Euler. – Anweshi Jan 29 '10 at 21:14
  • Euler also “numerically checked” the equation at one non-integer value $s=3/2$. This part is a bit trickier due to the non-convergence at $s=-1/2. But he rewrote the series so it took longer and longer to start diverging and got a numerical estimate that way. – Noah Snyder Jun 30 at 2:45
  • Also this paper is in French instead of Latin, so it’s much more accessible. – Noah Snyder Jun 30 at 2:46

Davenport (Chapter 9 in Multiplicative Number Theory) claims that the functional equation for Dirichlet L-functions was first given by Hurwitz in 1882 (Werke I, pp.72-88), though only for quadratic characters. The proof uses what we now call the Hurwitz zeta function.

I was told just yesterday that some people refer to the Riemann Hypothesis for Dirichlet L-functions as the Piltz Hypothesis. This is confirmed in the wikipedia article.

  • Ah, I suspected that it may have been Hurwitz, and that Davenport would have something to say, but did not have the book handy. :) BTW, once I was skimming Hardy's collected works and found a sentence where he blithely asserted that RH for Dirichlet L-functions will be proven "within a week" of the original RH... – David Hansen Jan 27 '10 at 15:22
  • As a grad student I ran across a paper of McCurley that referenced Piltz for GRH. Or maybe it was this one:… The reference to Piltz (missing from the above wikipedia advert) is A. Piltz, Uber die Haufigkeit der Primzahlen in arithmetischen Progressionen und uber verwandte Gesetze, A. Neuenhahn, Jena, 1884. (also on GoogleBooks). This was his dissertation, and he also conjectures that $p_n - p_{n-1} < p^\alpha$ for all $\alpha > 0$. – Junkie Apr 26 '10 at 1:48

According to Wikipedia, "an equivalent relationship [equivalent to the functional equation] was conjectured by Euler in 1749". I've seen mention of this in other places too, but of course, that doesn't prove anything.

  • Euler knew how to evaluate zeta at negative integers via abel summation, and of course also how to compute it at positive even integers. He observed some form of the functional equation relating the values, but didn't have an overall notion of zeta as a function of a complex variable, as far as I know. It seems possible that Riemann was also influenced by these ideas of Euler; does anyone know whether this is the case? – Emerton Jan 29 '10 at 15:17
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    Knowing the functional equation, at all the positive integers is enough to recover the functional equation for all complex z :) But the proof uses machinery well beyond Euler times, and a bound on the growth of the Riemann zeta function usually derived from the functional equation. In any case, by a theorem of Carlson, two entire functions, that don't grow faster than exp(c*|z|) (c < pi/2) (which is the case for (z-1)zeta(z) and hence a fortiori the functional equation term (z-1)2^z*pi^(z-1)*(blah blah)) and that agree on all positive integers, must be equal for all complex z. – maks Jan 29 '10 at 18:43
  • Btw, in Carlson's theorem you just need c < pi (and that is optimal, because of the function sin(pi*z)), not c < pi/2, as I wrote earlier. – maks Jan 29 '10 at 18:46

Concerning the statement "An equivalent relationship [equivalent to the functional equation] was conjectured by Euler in 1749". This is discussed in Weil's book "Basic number theory." It concerns only the values at integral points: Euler understood $\zeta(1-2k)$ by a simple regularization, and noticed the relation to $\zeta(2k)$.

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