Notes Whittaker & Watson, Chapter XIII, part 1

Whittaker & Watson, A Course of Modern Analysis. Chapter XIII, The Zeta Function of Riemann.

13.1 The definition of Zeta-function.

The series definition $$ \zeta(s)=\sum_{n=1}^\infty\frac{1}{n^s} $$ is uniformly convergent and analytic for $s=\sigma+it$ when $\sigma\ge 1+\delta$ with $\delta>0$.

13.11 The generalized Zeta-function (Hurwitz) $$ \zeta(s,a)=\sum_{n=0}^\infty\frac{1}{(a+n)^s} $$ 13.12 The expression of $\zeta(s,a)$ as an infinite integral.

By the integral $(n+a)^{-s}\Gamma(s) = \int_0^\infty x^{s-1}e^{-(n+a)x}dx$, for $\sigma>0$ and $\arg x=0$, we have $$ \begin{align*} \Gamma(s)\zeta(s,a) &=\lim_{N\to\infty}\sum_{n=0}^N \int_0^{\infty}x^{s-1}e^{-(n+a)x}dx\newline &=\lim_{N\to\infty}\int_0^{\infty}\frac{x^{s-1}e^{-ax}}{1-e^{-x}}dx - \int_0^{\infty}\frac{x^{s-1}e^{-(N+a+1)x}}{1-e^{-x}}dx\newline &=\lim_{N\to\infty}\int_0^{\infty}\frac{x^{s-1}e^{-ax}}{1-e^{-x}}dx -(N+a)^{1-\sigma}\Gamma(\sigma-1) \end{align*} $$ in which the last term tends to 0 when $\sigma\ge 1+\delta$. Hence $$ \zeta(s,a)=\frac{1}{\Gamma(s)}\int_0^\infty\frac{x^{s-1}e^{-ax}}{1-e^{-x}}dx. $$ 13.13 The expression of $\zeta(s,a)$ as a contour integral.

Consider the integral on the Hankel contour $$ \int_\infty^{(+0)}\frac{(-z)^{s-1}e^{-az}}{1-e^{-z}}dz $$ in which $(0+)$ indicates a counterclockwise contour around 0. In the similar way as in §12.22, $(-z)$ evaluates to $e^{-i\pi}x$, $\Delta e^{i\theta}$, $e^{i\pi}x$ on segments of the contour $\rho\to\Delta$, $|z|=\Delta$, $\Delta\to\rho$, such that $$ \begin{align*} \int_D\frac{(-z)^{s-1}e^{-az}}{1-e^{-z}}dz &= -2i\sin\pi s\int_\delta^{\rho}\frac{x^{s-1}e^{-ax}}{1-e^{-x}}dx-i\int_{-\pi}^{\pi}(-z)^{s-1}\frac{ze^{-az}}{1-e^{-z}}d\theta. \end{align*} $$ Since $\lim_{z\to 0}\frac{ze^{-az}}{1-e^{-z}}=1$, the last term tends to zero as $\Delta\to 0$ for fixed $s$ with $\sigma\ge 1+\delta$. We have $$ \int_\infty^{(+0)}\frac{(-z)^{s-1}e^{-az}}{1-e^{-z}}dz =\frac{ -2\pi i}{\Gamma(1-s)}\zeta(s,a) $$ or $$ \zeta(s,a)=-\frac{\Gamma(1-s)}{2\pi i}\int_\infty^{(+0)}\frac{(-z)^{s-1}e^{-az}}{1-e^{-z}}dz. $$ Since the contour integral is analytic for all $s$, the singularities of $\zeta(s,a)$ can only locate at the poles of $\Gamma(1-s)$: $s=1,2,\cdots$. Given that $\zeta(s,a)$ is analytic for $\sigma\ge 1+\delta$, the only singularity of $\zeta(s,a)$ is at $s=1$.

With $s=1$, the residue of the contour integral at $z=0$ is 1. Therefore the residue of $\zeta(s,a)$ at the pole $s=1$ is +1, since its residue for $\Gamma(1-s)$ is -1.

13.14 Values of $\zeta(s, a)$ for special values of $s=0,-1,-2\cdots$.

Differentiating the generating function of the Bernoulli polynomial $-z\frac{e^{-az}-1}{e^{-z}-1} = \sum_{n=1}^\infty\frac{(-1)^nz^n}{n!}\phi_n(a)$ (§7.2) with regard to $a$, we have $$ \frac{z^2e^{-az}}{e^{-z}-1}=\sum_{n=1}^\infty\frac{(-1)^nz^n}{n!}\phi'_n(a) $$ which gives the series expansion $$ \begin{align*} \frac{e^{-az}}{e^{-z}-1} & =\sum_{n=1}^\infty\frac{(-1)^nz^{n-2}}{n!}\phi'_n(a)\newline &=\frac{1}{z} + \sum_{n=0}^\infty\frac{(-1)^nz^n}{(n+2)!}\phi'_{n+2}(a) \end{align*} $$ in which the coefficient for $z^m$, ($m=0,1,2\cdots$) corresponds to the residue of the Hankel integral $\int_\infty^{(+0)}\frac{z^{-m-1}e^{-az}}{1-e^{-z}}dz$. Thus $$ \begin{align*} \zeta(-m,a) =-\frac{\Gamma(1+m)}{2\pi i}\int_\infty^{(+0)}\frac{(-z)^{-m-1}e^{-az}}{1-e^{-z}}dz = -\frac{\phi'_{m+2}(a)}{(m+2)(m+1)}, \quad -m=0,-1,-2\cdots. \end{align*} $$ With $a=1$, $$ \zeta(-m)=\zeta(-m, 1)=\begin{cases}-\frac{1}{2} & m=0,\newline 0 & m\ \text{even},\newline (-1)^{\frac{m+1}{2}}\frac{B_{\frac{m+1}{2}}}{m+1} & m\ \text{odd}.\end{cases} $$ 13.15 The formula of Hurwitz for $\zeta(s,a)$ when $\sigma<0$.

Consider the contour integral $\int\frac{(-z)^{s-1}e^{-az}}{1-e^{-z}}dz$ between the Hankel contour and $|z|=(2N+1)\pi$ $$ \frac{1}{2\pi i}\left(\int_{|z|=(2N+1)\pi} - \int_{(2N+1)\pi}^{(0+)}\right)\frac{(-z)^{s-1}e^{-az}}{1-e^{-z}}dz = \sum_{n=1}^N(R_n+R_n') $$ in which the residues $R_n, R'_n$ at simple poles $2n\pi i$ and $-2n\pi i$ can be calculated such that $R_n+R'_n= 2\cdot(2n\pi)^{s-1}\sin\left(\frac{s\pi}{2}+2\pi an\right)$.

Since for $0<a\le 1$, $\frac{e^{-az}}{1-e^{-z}}$ is uniformly bounded on the contour $|z|=(2N+1)\pi, \forall N$, the integral on the outer circle tends 0 as $N\to\infty$ for $\sigma<0$.

Thus the integral on the Hankel contour evaluating to the sum of residues, we have $$ \zeta(s,a)=\frac{2\Gamma(1-s)}{(2\pi)^{1-s}}\left(\sin\frac{s\pi}{2}\sum_{n=1}^\infty\frac{\cos 2\pi an}{n^{1-s}}+\cos\frac{s\pi}{2}\sum_{n=1}^\infty\frac{\sin 2\pi an}{n^{1-s}}\right). $$ 13.151 With $a=1$, the Zeta function $\zeta(s)=\frac{2\Gamma(1-s)}{(2\pi)^{1-s}}\sin\left(\frac{s\pi}{2}\right)\zeta(1-s)$. By the reflection formula of the Gamma function, we have Riemann’s functional equation $$ 2^{1-s}\Gamma(s)\zeta(s)\cos\left(\frac{s\pi}{2}\right)=\pi^s\zeta(1-s). $$