ho.history overview - Historical question in analytic number theory

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

$$frac{zeta_{2}(-1)}{zeta_{2}(2)} = frac{1}{4}cdotfrac{6cdot 2}{pi^2} = frac{3cdot 1!}{1cdot pi^2}, frac{zeta_{2}(-3)}{zeta_{2}(4)} = -frac{1}{8}cdotfrac{30cdot24}{7pi^4} = -frac{15cdot 3!}{7cdot pi^4},$$
$$frac{zeta_{2}(-5)}{zeta_{2}(6)} = frac{1}{4}cdot frac{42cdot 6!}{31pi^6} = frac{63 cdot 5!}{31cdot pi^6}, frac{zeta_{2}(-7)}{zeta_{2}(8)} = -frac{17}{16}cdot frac{30cdot 8!}{127cdot pi^8} = -frac{255cdot 7!}{127pi^8}.$$

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

$$frac{zeta_{2}(1-n)}{zeta_{2}(n)} = -Gamma(n)frac{2^n-1}{(2^{n-1}-1)pi^n} cosleft(frac{pi n}{2}right).$$
Writing $zeta_{2}(n)$ as $(1 - 2^{1-n})zeta(n)$ gives the
asymmetric functional
equation
$$frac{zeta(1-n)}{zeta(n)} = frac{2}{(2pi)^n} Gamma(n)cosleft(frac{pi n}{2}right).$$
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
   

   $$(1 - 2^{n+1})zeta(-n) = left.left(ufrac{d}{du}right)^{n}rightvert_{u=1}left(frac{u}{1+u} right) = left.left(frac{d}{dx}right)^{n}rightvert_{x=0} left(frac{e^x}{1+e^x}right).$$




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.

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