Notes on Whittaker & Watson, Chapter XIII, part 3

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

13.5 Inequalities satisfied by $\zeta(s,a)$ when $\sigma > 0$. $$ \zeta(s,a)=\sum_{n=0}^N(n+a)^{-s} - \frac1{(1-s)(N+a)^{s-1}} - \sum_{n=N}^\infty f_n(s), $$ where $$ \begin{align*} f_n(s) &= \frac1{1-s}\left(\frac{1}{(n+a+1)^{s-1}}-\frac{1}{(n+a)^{s-1}}\right) - \frac{1}{(n+a+1)^s} = s\int_{n}^{n+1}\frac{u-n}{(u+a)^{s+1}}du. \end{align*} $$ The series $\sum_{n=N}^\infty f_n(s)$ is absolutely and locally uniformly convergent since $|f_n(s)|\le|s|(n+a)^{-\sigma-1}$ for $\sigma>0$. Hence the series is convergent to an analytic function by Weierstrass’s theorem.

[The claim that the series $\sum_{n=N}^\infty f_n(s)$ is uniformly convergent need be qualified locally on compact sets.]

Setting $N=\lfloor|t|\rfloor$, we have $$ \begin{align*} |\zeta(s,a)|&<\sum_{n=0}^{\lfloor|t|\rfloor}(n+a)^{-\sigma}-\frac{(\lfloor|t|\rfloor+a)^{1-\sigma}}{|t|}+|s|\sum_{n=\lfloor|t|\rfloor}^\infty(n+a)^{-\sigma-1} \newline &<a^{-\sigma} +\int_0^{\lfloor|t|\rfloor}(x+a)^{-\sigma}dx-\frac{(\lfloor|t|\rfloor+a)^{1-\sigma}}{|t|}+|s|\int_{\lfloor|t|\rfloor-1}^{\infty}(x+a)^{-\sigma-1}dx. \end{align*} $$

Then we have the asymptotic behaviour of $\zeta(s,a)$ when $|t|\to\infty$ for $\sigma>0$ that $$ |\zeta(s,a)| = \begin{cases} O(|t|^{1-\sigma}) & \delta\le\sigma\le 1-\delta,\newline O(|t|^{1-\sigma}\log|t|) & 1-\delta\le\sigma\le 1+\delta,\newline O(1) &\sigma\ge1+\delta. \end{cases} $$ 13.51 Inequalities satisfied by $\zeta(s,a)$ when $\sigma \le 0$.

For the Riemann Zeta function, the asymptotic behaviour for $\sigma\le\delta$ can be obtained using the functional equation $$ \zeta(s)=2^s\pi^{s-1}\Gamma(1-s)\zeta(1-s)\sin\left(\frac{s\pi}2\right). $$ By Stirling’s series, we have $\Gamma(1-s)=O\left(\exp\left\{\left(\frac12-s\right)\log(1-s)-1+s\right\}\right)$, and thus $\zeta(s)=O(|t|^{\frac12-s})\zeta(1-s)$.

For the generalized $\zeta(s,a)$, consider $$ \zeta_a(1-s)=\sum_{n=1}^\infty\frac{e^{2n\pi ia}}{n^{1-s}} $$ which can be extended to $0\le\sigma\le1-\delta$ through $$ (1-e^{2\pi ia})\zeta_a(1-s)=e^{2\pi ia}+\sum_{n=2}^Ne^{2n\pi ia}\left(n^{s-1}-(n-1)^{s-1}\right) + (s-1)\sum_{n=N+1}^\infty e^{2n\pi ia}\int_{n-1}^nu^{s-2}du. $$ Thus $$ |\sin(\pi a)\zeta_a(1-s)|\le1+\sum_{n=2}^Nn^{\sigma-1}+(n-1)^{\sigma-1}+|s-1|\sum_{n=N+1} ^\infty\int_{n-1}^nu^{\sigma-2}du. $$ By setting $N=\lfloor|t|\rfloor$, we have $$ \zeta_a(1-s)=\begin{cases}O(|t|^\sigma) & \delta\le\sigma\le1-\delta\newline O(|t|^\sigma\log|t|) & -\delta\le\sigma<\delta \newline O(1) & \sigma<-\delta. \end{cases} $$ Since $\zeta(s,a)=\frac{2\Gamma(1-s)}{i(2\pi)^{1-s}}\left(e^{\frac12s\pi i}\zeta_a(1-s) - e^{-\frac12s\pi i}\zeta_{-a}(1-s)\right)$ (§13.15), we have consequently $$ \zeta(s,a)=\begin{cases}O(|t|^{\frac12-\sigma}) & \sigma\le-\delta\newline O(|t|^\frac12) &\delta\le\sigma\le1-\delta\newline O(|t|^\frac12\log|t|)& -\delta\le\sigma\le\delta. \end{cases} $$ 13.6 The asymptotic expansion of $\log\Gamma(z+a)$.

From the Weierstrass product, we have $$ \left(1+\frac{z}{a}\right)\prod_{n=1}^\infty\left(1+\frac{z}{a+n}\right)e^{-\frac zn}=\frac{e^{-\gamma z}\Gamma(a)}{\Gamma(z+a)}. $$ Taking the logarithm on both sides $$ \begin{align*} \log\frac{e^{-\gamma z}\Gamma(a)}{\Gamma(z+a)} & = \log\left(1+\frac{z}{a}\right)+\sum_{n=1}^{\infty}\log\left(1+\frac{z}{a+n}\right)-\frac{z}{n}\newline & = \sum_{m=1}^{\infty}\frac{(-1)^{m-1}}{m}\frac{z^m}{a^m} + \sum_{n=1}^{\infty}\left(\frac{-az}{n(n+a)}+\sum_{m=2}^{\infty}\frac{(-1)^{m-1}}{m}\frac{z^m}{(a+n)^m}\right) & \text{convergent}\ |z|<a\newline & = \frac{z}{a}-\sum_{n=1}^\infty\frac{az}{n(n+a)}+\sum_{m=2}^\infty\frac{(-1)^{m-1}}{m}z^m\zeta(m,a), \end{align*} $$ and use the contour integral on the Hankel contour $C$ with enclosing $[2, \infty)$ which is convergent for $|z|<1$, $$ \begin{align*} \log\frac{\Gamma(a)}{\Gamma(z+a)}&=-z\frac{\Gamma'(a)}{\Gamma(a)}-\frac{1}{2\pi i}\int_C\frac{\pi z^s}{s\sin\pi s}\zeta(s,a)ds\newline &= -z\frac{\Gamma'(a)}{\Gamma(a)}+\frac{1}{2\pi i}\int_{\frac32-\infty i}^{\frac32+\infty i}\frac{\pi z^s}{s\sin\pi s}\zeta(s,a)ds & \rho\to\infty:\left|\int_{|s-\frac32|=\rho,\sigma\ge\frac32}\right|\to0\newline &=-z\frac{\Gamma'(a)}{\Gamma(a)}+\frac{1}{2\pi i}\int_{-n-\frac12-\infty i}^{-n-\frac12+\infty i}\frac{\pi z^s}{s\sin\pi s}\zeta(s,a)ds + \sum_{m=-1}^nR_m, & R\to\infty:\left|\int_{-n-\frac12\pm Ri}^{\frac32\pm Ri}\right|\to0 \end{align*} $$ where the integral is analytic for $|\arg z|\le\pi-\delta$, and $R_m$ the residues of the integrand at $s=-m$. Since on the path from $-n-\frac12-\infty i$ to $-n-\frac12+\infty i$, $\left|\frac{\pi z^s}{s\sin\pi s}\right|<Kz^{-n-\frac12}e^{-\delta|t|}$ [there is a typo in the text, read $\left|\frac{\pi z^s}{s\sin\pi s}\zeta(s,a)\right|<Kz^{-n-\frac12}e^{-\delta|t|}|t|^{\tau(-n-\frac12)}$], $$ \log\frac{\Gamma(a)}{\Gamma(z+a)}=-z\frac{\Gamma'(a)}{\Gamma(a)} + \sum_{m=-1}^nR_m+O\left(z^{-n-\frac12}\right). $$ The residues $$ R_m=\begin{cases}\frac{(-1)^{m-1}z^{-m}\zeta(-m,a)}{m}=\frac{(-1)^mz^{-m}\phi'_{m+2}(a)}{m(m+1)(m+2)} & m>0\newline \left(\frac12-a\right)\log z+\zeta'(0, a)=\left(\frac12-a\right)\log z+\log\Gamma(a)-\frac12\log2\pi & m=0\newline -z\log z+z\frac{\Gamma'(a)}{\Gamma(a)}+z & m=-1 \end{cases} $$ Thus for $|\arg z|\le\pi-\delta$, $$ \log\Gamma(z+a)=\left(z+a-\frac12\right)\log z-z+\frac12\log2\pi+\sum_{m=1}^n\frac{(-1)^{m-1}\phi'_{m+2}(a)}{m(m+1)(m+2)z^m} + O\left(z^{-n-\frac12}\right). $$ Note that this result is obtained for $0<a\le1$. We write the asymptotic expansion for $a=1$, then substitute $z\to z+a$, with the condition that $|\arg(z+a)|\le\pi-\delta$ and $|\arg(z)|\le\pi-\delta$, and cancel the term $\log(z+a)$ from both sides, thus extending the result beyond $0<a\le1$ $$ \log\Gamma(z+a)=\left(z+a-\frac12\right)\log z-z+\frac12\log2\pi+o(1). $$