The Euler Maclaurin summation formula and the Riemann zeta function

[polygamma]

The Euler-Maclaurin summation formula and the Riemann zeta function

The Euler-Maclaurin summation formula states that if $f(x)$ has $(2p+1)$ continuous derivatives on the interval $[m,n]$ (where $m$ and $n$ are natural numbers), then

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

$$ + \frac{1}{(2p+1)!}\int_{m}^{n} B_{2p+1}(x-\lfloor x \rfloor ) f^{(2p+1)}(x) \ dx $$

where $B_{j}$ are the Bernoulli numbers and $B_{j}(x)$ are the Bernoulli polynomials.


You can derive the formula by first repeatedly integrating $ \displaystyle \int_{0}^{1} f(x) \ dx = \int_{0}^{1} B_{0}(x) f(x) \ dx$ by parts. Then replace $f(x)$ with $f(x+k)$ and sum both sides of equation from $m$ to $n-1$.By applying the Euler Macluarin summation formula to $ \displaystyle \sum_{k=n}^{\infty} {k^{-s}}$ show that for $\text{Re}(s) > -3$, $$ \zeta(s) = \lim_{n \to \infty} \left( \sum_{k=1}^{n} k^{-s} - \frac{n^{1-s}}{1-s} - \frac{n^{-s}}{2} + \frac{s n^{-s-1}}{12} \right) . $$Then use the representation to show that $ \displaystyle \zeta'(-1) = \frac{1}{12} - \log A$ where $A$ is the Glaisher-Kinkelin constant given by

$$A = \lim_{n \to \infty} \frac{\prod_{k=1}^{n} k^{k}}{n^{n^{2}+n/2+1/12} e^{-n^{2}/4}} . $$

 

[polygamma]

$$ \sum_{k=m}^{\infty} k^{-s} = \zeta(s) - \sum_{k=1}^{m-1} k^{-s} = \zeta(s) - \sum_{k=0}^{m} k^{-s} + m^{-s} $$

$$ = \int_{m}^{\infty} x^{-s} \ dx - \frac{1}{2} \Big( 0 - m^{-s} \Big) + \frac{1/6}{2!} \Big( 0 - sm^{-s-1} \Big) - \frac{s(s+1)(s+2)}{3!}\int_{m}^{\infty} B_{3}(x-\lfloor x \rfloor ) x^{-s-3} \ dx$$

$$ = \frac{m^{1-s}}{s-1} + \frac{m^{-s}}{2} + \frac{sm^{-s-1}}{12} - \frac{s(s+1)(s+2)}{3!}\int_{m}^{\infty} B_{3}(x-\lfloor x \rfloor ) x^{-s-3} \ dx$$

$$ \implies \zeta(s) = \sum_{k=1}^{m} k^{-s} + \frac{m^{1-s}}{s-1} - \frac{m^{-s}}{2} + \frac{sm^{-s-1}}{12} - \frac{s(s+1)(s+2)}{3!}\int_{m}^{\infty} B_{3}(x-\lfloor x \rfloor ) x^{-s-3} \ dx $$

If $\text{Re}(s) >-3$, the remainder goes to zero as $m$ goes to $\infty$. This is due to the oscillatory nature of $B_{3}(x - \lfloor x \rfloor)$. But at the very least it goes to zero for $\text{Re} (s) > -2$.

So

$$ \lim_{m \to \infty} \zeta(s) = \zeta(s) = \lim_{m \to \infty} \Big( \sum_{k=1}^{m} k^{-s} - \frac{m^{1-s}}{1-s} - \frac{m^{-s}}{2} + \frac{sm^{-s-1}}{12} \Big)$$Then assuming it is OK to differentiate inside of the limit,

$$\zeta'(s) = \lim_{m \to \infty} \Bigg(- \sum_{k=1}^{m} k^{-s} \log k - \frac{-m^{1-s} (1-s) \log m +m^{1-s}}{(1-s)^{2}} + \frac{m^{-s} \log m}{2} $$

$$ + \frac{1}{12} \left(m^{-s-1}- sm^{-s-1} \log m \right) \Bigg) $$

$$ \implies \zeta'(-1) = \lim_{m \to \infty} \Bigg( - \sum_{k=1}^{m} k \log k - \frac{-2m^{2} \log m + m^{2}}{4} + \frac{m \log m}{2} + \frac{1}{12} + \frac{ \log m}{12} \Bigg)$$

$$ = \lim_{m \to \infty} \Bigg( - \sum_{k=1}^{m} k \log k + \Big(\frac{m^{2}}{2} + \frac{m}{2} + \frac{1}{12} \Big) \log m -\frac{m^{2}}{4} + \frac{1}{12} \Bigg)$$

$$ = - \lim_{m \to \infty} \Bigg( \sum_{k=1}^{m} k \log k - \Big(\frac{m^{2}}{2}+\frac{m}{2} + \frac{1}{12} \Big) \log m + \frac{m^{2}}{4} \Bigg) + \frac{1}{12} = - \log A + \frac{1}{12}$$