LAURENT EXPANSION RIEMANN ZETA FUNCTION

In this post we will derive the Laurent expansion for the Riemann Zeta Function

$\zeta(s)=\frac{1}{s-1}+\sum_{n=0}^\infty \frac{(-1)^n\gamma_n}{n!}(s-1)^n$

Recall the first order Euler Maclaurin expansion for the riemann zeta function shown here

$\zeta(s)=\frac{1}{s-1}+\frac12-s\int_1^\infty \frac{P_1(x)}{x^{s+1}}\,dx$ (1)

We can rewrite (1) as following:

$\begin{aligned} \zeta(s)&=\frac{1}{s-1}+\frac12-s\int_1^\infty \frac{P_1(x)}{x^{s+1}}\,dx\\ &=\frac{1}{s-1}+\frac12-s\int_1^\infty \frac{x-\lfloor x\rfloor-\frac12}{x^{s+1}}\,dx\\ &=\frac{1}{s-1}+\frac12-\frac{s}{2}\int_1^\infty \frac{dx}{x^{s+1}}-s\int_1^\infty \frac{x-\lfloor x\rfloor}{x^{s+1}}\,dx\\ &=\frac{1}{s-1}+\frac12-\frac{s}{2} \cdot \frac{1}{s}-s\int_1^\infty \frac{x-\lfloor x\rfloor}{x^{s+1}}\,dx\\ &=\frac{1}{s-1}+1-s\int_1^\infty \frac{x-\lfloor x\rfloor}{x^{s+1}}\,dx\\ \end{aligned}$

Now, define the function:

$\phi(s)=\zeta(s)-\frac{1}{s-1}=1-s\int_1^\infty \frac{x-\lfloor x\rfloor}{x^{s+1}}\,dx$ (2)

$\phi(s)$ is an analytic function, therefore, we can represent it by a series expansion around s=1 in the following form:

$\phi(s)=\sum_{n=0}^\infty \frac{\phi^{(n)}(s_0)}{n!}(s-s_0)^n\Big|_{s_0=1}$ (3)

Lets calculate the coefficients of (3) by calculating the derivatives of $\phi(s)$ using eq. (2) above.

For the first derivative we have:

$\begin{aligned} \phi^{\prime}(s)&=\frac{d}{ds}\left(1-s\int_1^\infty \frac{\{x\}}{x^{s+1}}\,dx\right)\\ &=\frac{d}{ds}\left(1-s\int_1^\infty \{x\}e^{-(s+1)\ln(x)}\,dx\right)\\ &=-\int_1^\infty \frac{\{x\}}{x^{s+1}}\,dx+s\int_1^\infty \frac{\{x\}\ln(x)}{x^{s+1}}\,dx\\ &=-\int_1^\infty \frac{\{x\}}{x^{s+1}}\left(1-s\ln(x) \right)\,dx \qquad \blacksquare \end{aligned}$

In the point s=1 we obtain

$\phi^{\prime}(1)=-\int_1^\infty \frac{\{x\}}{x^{2}}\left(1-\ln(x) \right)\,dx$

$\begin{aligned} \phi^{(2)}(s)&=\frac{d}{ds}\left(-\int_1^\infty \frac{\{x\}}{x^{s+1}}\,dx+s\int_1^\infty \frac{\{x\}\ln(x)}{x^{s+1}}\,dx\right)\\ &=\int_1^\infty \frac{\{x\}\ln(x)}{x^{s+1}}\,dx+\int_1^\infty \frac{\{x\}\ln(x)}{x^{s+1}}\,dx-s\int_1^\infty \frac{\{x\}\ln^2(x)}{x^{s+1}}\,dx\\ &=2\int_1^\infty \frac{\{x\}\ln(x)}{x^{s+1}}\,dx-s\int_1^\infty \frac{\{x\}\ln^2(x)}{x^{s+1}}\,dx\\ &=\int_1^\infty \frac{\{x\}}{x^{s+1}}\left(2-s\ln(x) \right)\ln(x)\,dx \qquad \blacksquare \end{aligned}$

for s=1 we obtain

$\phi^{(2)}(1)=\int_1^\infty \frac{\{x\}}{x^{2}}\left(2-\ln(x) \right)\ln(x)\,dx$

$\begin{aligned} \phi^{(3)}(s)&=\frac{d}{ds}\left(2\int_1^\infty \frac{\{x\}\ln(x)}{x^{s+1}}\,dx-s\int_1^\infty \frac{\{x\}\ln^2(x)}{x^{s+1}}\,dx\right)\\ &=-2\int_1^\infty \frac{\{x\}\ln^2(x)}{x^{s+1}}\,dx-\int_1^\infty \frac{\{x\}\ln^2(x)}{x^{s+1}}\,dx+s\int_1^\infty \frac{\{x\}\ln^3(x)}{x^{s+1}}\,dx\\ &=-3\int_1^\infty \frac{\{x\}\ln^2(x)}{x^{s+1}}\,dx+s\int_1^\infty \frac{\{x\}\ln^3(x)}{x^{s+1}}\,dx\\ &=-\int_1^\infty \frac{\{x\}}{x^{s+1}}\left(3-s\ln(x) \right)\ln^2(x)\,dx \qquad \blacksquare \end{aligned}$

for s=1 we obtain

$\phi^{(3)}(1)=-\int_1^\infty \frac{\{x\}}{x^{2}}\left(3-\ln(x) \right)\ln^2(x)\,dx$

$\begin{aligned} \phi^{(4)}(s)&=\frac{d}{ds}\left(-3\int_1^\infty \frac{\{x\}\ln^2(x)}{x^{s+1}}\,dx+s\int_1^\infty \frac{\{x\}\ln^3(x)}{x^{s+1}}\,dx\right)\\ &=3\int_1^\infty \frac{\{x\}\ln^3(x)}{x^{s+1}}\,dx+\int_1^\infty \frac{\{x\}\ln^3(x)}{x^{s+1}}\,dx-s\int_1^\infty \frac{\{x\}\ln^4(x)}{x^{s+1}}\,dx\\ &=4\int_1^\infty \frac{\{x\}\ln^3(x)}{x^{s+1}}\,dx-s\int_1^\infty \frac{\{x\}\ln^4(x)}{x^{s+1}}\,dx\\ &=\int_1^\infty \frac{\{x\}}{x^{s+1}}\left(4-s\ln(x) \right)\ln^3(x)\,dx \qquad \blacksquare \end{aligned}$

for s=1 we obtain

$\phi^{(4)}(1)=\int_1^\infty \frac{\{x\}}{x^{2}}\left(4-\ln(x) \right)\ln^3(x)\,dx$

If we keep this process further we may obtain the following general form :

$\phi^{(n)}(s)=\int_1^\infty \frac{\{x\}}{x^{s+1}}\left(n-s\ln(x) \right)\ln^{n-1}(x)\,dx$ (4)

For $n\geq 1$ . Evaluated at s=1 we obtain:

$\phi^{(n)}(1)=(-1)^n\int_1^\infty \frac{\{x\}}{x^{2}}\left(n-\ln(x) \right)\ln^{n-1}(x)\,dx$ (5)

But we have already proved in a previous post that the integral in (5) is precisely the Stiltjies constants!

$\gamma_{n}=\lim _{N \rightarrow \infty}\left\{\sum_{k=1}^{N} \frac{(\ln k)^{n}}{k}-\int_{1}^{N} \frac{(\ln x)^{n}}{x} d x\right\}=\lim _{N \rightarrow \infty}\left\{\sum_{k=1}^{N} \frac{(\ln k)^{n}}{k}-\frac{(\ln N)^{n+1}}{n+1}\right\}$ (6)