MHB The Euler Maclaurin summation formula and the Riemann zeta function

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The Euler-Maclaurin summation formula provides a method to approximate the sum of a function's values at discrete points using integrals and derivatives. It is expressed in terms of Bernoulli numbers and polynomials, allowing for the evaluation of sums like the Riemann zeta function. By applying this formula to the series sum of k raised to the power of -s, it establishes a relationship between the zeta function and its derivatives. Specifically, it shows that the derivative of the zeta function at -1 can be expressed in terms of the Glaisher-Kinkelin constant. This connection highlights the interplay between summation techniques and analytic number theory.
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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

$$A = \lim_{n \to \infty} \frac{\prod_{k=1}^{n} k^{k}}{n^{n^{2}+n/2+1/12} e^{-n^{2}/4}} . $$
 
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$$ \sum_{k=m}^{\infty} k^{-s} = \zeta(s) - \sum_{k=1}^{m-1} k^{-s} = \zeta(s) - \sum_{k=0}^{m} k^{-s} + m^{-s} $$

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

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

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

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,

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

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

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

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

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