Math 254A. ζ′(0) or The Product of the Positive Integers

We will evaluate ζ′(0) , using the functional equation

ζ(s) = \frac{π^{(s−1)/2}Γ(\frac{1−s}{2})ζ(1 − s)}{π^{−s/2}Γ(\frac{s}{2})} .  (1)

First, let

ξ(s) = sζ(1 − s) ,

so that ξ is an entire function; then ξ(0) = −1 . Moreover, the Kronecker limit formula says that

ζ(s) = \frac{1}{s − 1} + γ + O(|s − 1|)

near s = 1 , where γ is Euler’s constant

γ = lim_{n→∞} (1 + \frac{1}{2} + \frac{1}{3} + ⋯ + \frac{1}{n} − log n) = 0.5772 ….

Thus we have

ξ(s) = −1 + γs + O(|s|) near s = 0 . (2)

Rewriting (1) with the definition of ξ gives

ζ(s) = \frac{π^{s−1/2}Γ(\frac{1−s}{2})ξ(s)}{2Γ(\frac{s}{2} + 1)} .

One can plug in s = 0 and the fact that Γ(\frac{1}{2}) = \sqrt{π} to again obtain the fact that ζ(0) = −\frac{1}{2} . Now take logarithmic derivatives at s = 0 :

\frac{ζ′(0)}{ζ(0)} = log π − \frac{1}{2} · \frac{Γ′(\frac{1}{2})}{Γ(\frac{1}{2})} + \frac{ξ′(0)}{ξ(0)} − \frac{1}{2} · \frac{Γ′(1)}{Γ(1)} .  (3)

We already know from (2) that

\frac{ξ′(0)}{ξ(0)} = −γ .

To compute Γ′(1)/Γ(1) , recall the defining integral for the gamma function:

Γ(s) = ∫_{0}^{∞} e^{−t}t^{s−1}dt .

Differentiating under the integral sign and substituting s = 1 gives

Γ′(1) = ∫_{0}^{∞} e^{−t}(log t)dt .

Looking this up in a table of definite integrals yields the value −γ; thus

\frac{Γ′(1)}{Γ(1)} = −γ .

Finally, Legendre’s duplication formula (Neukirch VIII Prop. 1.2) says that

Γ(s)Γ(s + \frac{1}{2}) = \frac{2\sqrt{π}}{2^{2s}} Γ(2s) .

Taking logarithmic derivatives at s = \frac{1}{2} again gives the relation

\frac{Γ′(\frac{1}{2})}{Γ(\frac{1}{2})} + \frac{Γ′(1)}{Γ(1)} = −2 log 2 + 2\frac{Γ′(1)}{Γ(1)} ;

\frac{Γ′(\frac{1}{2})}{Γ(\frac{1}{2})} = −γ − 2 log 2 .

Putting these values back into (3) then gives

\frac{ζ′(0)}{ζ(0)} = log π + log 2;
ζ′(0) = −\frac{1}{2} log 2π .

Since

\frac{d}{ds}(\frac{1}{n^{s}}) |_{s=0}= − log n · n^{−s} |_{s=0}= − log n ,

it follows that

ζ′(0) “=” ∑_{n=1}^{∞} − log n

and therefore

∏_{n=1}^{∞} n “=” \sqrt{2π} .