MHB Expressing zeta(3) in terms of a Glaisher-Kinkelin-like constant

[polygamma]

In a previous thread I showed how to express $\zeta'(-1)$ in terms of the Glaisher-Kinkelin constant.

http://mathhelpboards.com/challenge-questions-puzzles-28/euler-maclaurin-summation-formula-riemann-zeta-function-7702.html

This thread is about expressing $\zeta(3)$ (sometimes referred to as Apery's constant) in terms of a constant similar to the Glaisher-Kinkelin constant.

Specifically, $$\zeta(3) = 4 \pi^{2} \log B$$ where $$\log B = \lim_{n \to \infty} \left[ \sum_{k=1}^{n} k^{2} \log k - \left(\frac{n^{3}}{3} + \frac{n^{2}}{2} + \frac{n}{6} \right) \log n + \frac{n^{3}}{9} - \frac{n}{12} \right] $$
Use the Euler-Maclaurin summation formula (or perhaps summation by parts) to show that the constant $B$ exists.Then using the representation of the Riemann zeta function derived in the other thread,

$$ \zeta(s) = \lim_{n \to \infty} \left( \sum_{k=1}^{n} k^{-s} - \frac{n^{1-s}}{1-s} - \frac{n^{-s}}{2} + \frac{s n^{-s-1}}{12} \right) \ \ \big(\text{Re}(s) > -3 \big) $$

show that

$$ \zeta'(-2) = - \log B $$Finally use the functional equation of the Riemann zeta function to show that $$ \zeta(3) = 4 \pi^{2} \log B $$

 

[polygamma]

Let $f(x) = x^{2} \ln x$.

Then

$$ \sum_{k=1}^{n-1} f(k) = \sum_{k=1}^{n} f(k) - n^{2} \ln n = \int_{1}^{n} f(x) \ dx + B_{1} \Big(f(n) -f(1) \Big) + \frac{B_{2}}{2!} \Big( f'(n) - f'(1) \Big) $$

$$ + \frac{1}{3!} \int_{1}^{n} B_{3} (x - \lfloor x \rfloor) f^{'''}(x) \ dx$$$$ = \frac{x^{3} \log x}{3} - \frac{x^{3}}{9} \Big|_{1}^{n} - \frac{1}{2} \Big(n^{2} \ln n -0 \Big) + \frac{1}{12} \Big(2n \ln n + n -1 \Big) + \frac{1}{6} \int_{1}^{n} B_{3} (x - \lfloor x \rfloor) \frac{2}{x} \ dx$$

$$ = \frac{n^{3} \log n}{3} - \frac{n^{3}}{9} + \frac{1}{9} - \frac{n^{2} \ln n}{2} + \frac{n \log n}{6} + \frac{n}{12}- \frac{1}{12} + \frac{1}{3} \int_{1}^{n} \frac{B_{3} (x - \lfloor x \rfloor)}{x} \ dx$$$$ \implies \sum_{k=1}^{n} k^{2} \ln k - \Big( \frac{n^{3}}{3} + \frac{n^{2}}{2} + \frac{n}{6} \Big) \ln n + \frac{n^{3}}{9} - \frac{n}{12} = \frac{1}{36} + \frac{1}{3} \int_{1}^{n} \frac{B_{3} (x - \lfloor x \rfloor)}{x} \ dx$$Now take the limit of both sides of the equation.

The integral $ \displaystyle \int_{1}^{\infty} \frac{B_{3} (x - \lfloor x \rfloor)}{x} \ dx$ converges (condtionally) by Dirichlet's convergence test for integrals.

Therefore,

$$ \log B = \lim_{n \to \infty} \left[ \sum_{k=1}^{n} k^{2} \log k - \left(\frac{n^{3}}{3} + \frac{n^{2}}{2} + \frac{n}{6} \right) \log n + \frac{n^{3}}{9} - \frac{n}{12} \right] $$

exists.From the other thread,$$ \zeta'(s) = \lim_{n \to \infty} \Bigg[- \sum_{k=1}^{n} k^{-s} \log k - \frac{-n^{1-s} (1-s) \log n +n^{1-s}}{(1-s)^{2}} + \frac{n^{-s} \log n}{2} $$

$$ + \frac{1}{12} \left(n^{-s-1}- sn^{-s-1} \log n \right) \Bigg] \ \ (\text{Re}(s) > -3)$$Plug in $s=-2$ to get

$$ \zeta'(-2) = \lim_{n \to \infty} \Bigg[- \sum_{k=1}^{n} k^{2} \log k - \frac{-3n^{3} \log n +n^{3}}{9} + \frac{n^{2} \log n}{2} + \frac{1}{12} \left(n+2n \log n \right) \Bigg] $$

$$ = \lim_{n \to \infty} \left[ -\sum_{k=1}^{n} k^{2} \log k + \left(\frac{n^{3}}{3} + \frac{n^{2}}{2} + \frac{n}{6} \right) \log n - \frac{n^{3}}{9} + \frac{n}{12} \right] = - \log B$$Next differentiate the functional equation of the Riemann zeta function.

$$ \zeta'(s) = \log (2) 2^{s} \pi^{s-1} \sin \left( \frac{\pi s}{2} \right) \Gamma(1-s) \zeta(1-s)+ \log(\pi) 2^{s} \pi^{s-1} \sin \left( \frac{\pi s}{2} \right) \Gamma(1-s) \zeta(1-s)$$

$$ + \frac{\pi}{2} 2^{s} \pi^{s-1} \cos \left( \frac{\pi s}{2} \right) \Gamma(1-s) \zeta(1-s) - 2^{s} \pi^{s-1} \sin \left( \frac{\pi s}{2} \right) \Gamma'(1-s) \zeta(1-s)$$

$$ -2^{s} \pi^{s-1} \sin \left( \frac{\pi s}{2} \right) \Gamma(1-s) \zeta'(1-s)$$At $s=-2$,

$$ \zeta'(-2) = \frac{\pi}{2} 2^{-2} \pi^{-3} \cos \left( -\pi \right) \Gamma(3) \zeta(3) = -\frac{\zeta(3)}{4 \pi^{2}}$$Which implies

$$ \zeta(3) = -4 \pi^{2}\zeta(-2) = 4 \pi^{2} \log B $$
As far as I know, the constant $B$ doesn't have a name. But it can be found in several papers.

I'm pretty sure you could express $\zeta(5)$ in terms of a similar constant. But that would require using a different representation of the Riemann zeta function.