Question

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.

Answer 1

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 Eisenstein had a proof as well (unpublished) which was found in his copy of Gauss' Disquisitiones.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.

Answer 2

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.

Exercises

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.

Answer 3

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.

Answer 4

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.

Answer 5

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 $\zeta(2k)$.