Notes on Whittaker & Watson, Chapter XIII, part 2

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

13.2 Hermite’s formula for $\zeta(s,a)$.

Applying the Abel-Plana formula to the summation $\zeta(s,a)=\sum_{n=0}^{\infty}(n+a)^{-s}$, with the summand function $\phi(z)=(a+z)^{-s}$, and $z=x+iy$, we have for $\sigma>1$ $$ \zeta(s,a)=\lim_{x_2\to\infty}\frac{1}{2}\alpha^{-s} + \int_{x_1=0}^{x_2}(a+x)^{-s}dx + \frac1i\int_{0}^{\infty}\frac{\phi(x_2+iy)-\phi(x_1+iy) -\phi(x_2+iy)+\phi(x_1+iy)}{e^{2\pi y}-1}dy. $$ With $\arg(a+z)=\arctan\frac{y}{a+x}$, consider the function $$ q(z) = \frac{\phi(z)-\phi(\bar z)}{2i}= - \left((a+x)^2+y^2\right)^{-\frac12s}\sin\left(s\arctan\frac{y}{a+x}\right). $$ By the inequalities that $|\sin z|=|\sinh iz|\le \big|\sinh|z|\big|$, $\arctan\frac{y}{a+x}\le \frac{|y|}{a+x}$, and the fact that $\arctan \le \frac{\pi}{2}$, we have $$ \begin{align*} |q(z)|&\le \left((a+x)^2+y^2\right)^{\frac12-\frac12\sigma}\left|y^{-1}\right|\sinh \frac{\pi|s|}2, & \frac{|y|}{\left((a+x)^2+y^2\right)^{\frac12}}\le1\newline |q(z)|&\le \left((a+x)^2+y^2\right)^{-\frac12\sigma}\sinh \frac{y|s|}{a+x}. \end{align*} $$ From the first bound on $|q(z)|$ we have that the integrability of $\frac{q(x+iy)}{e^{2\pi y}-1}$ in $y$ over $(a, \infty)$, and from the second bound that $\frac{q(x+iy)}{e^{2\pi y}-1}$ can be integrated over $(0,a)$ since the limit $\lim_{y\to0}\frac{\sinh|s|y/(a+x)}{e^{2\pi y}-1}$ exists. We have that the integral $\int_{0}^\infty\frac{q(x+iy)}{e^{2\pi y}-1}dy$ is of $O(x^{-\sigma})$ as $x\to\infty$. Hence Hermite’s formula for $\zeta(s,a)$ $$ \zeta(s,a)=\frac{1}{2}a^{-s} + \frac{a^{1-s}}{s-1}+ 2\int_{0}^{\infty}\left(a^2+y^2\right)^{-\frac12s}\sin\left(s\arctan\frac{y}{a}\right)\frac{dy}{e^{2\pi y}-1}. $$ The integral is convergent $\forall s\in\mathbb{C}$. The analycity follows from the convergence of the integral with differentiation under the integration sign. The differentiation of the integrand has one term involving $\sin s\arctan\frac{y}{a}$ to which the bound on $|q(z)|$ developed above applies, and the other term involving $\cos s\arctan\frac{y}{a}$, which can be bounded similarily by the relation that $|\cos z|\le |\cosh \Im z|\le \big|\cosh|z|\big|$. Therefore, Hermite’s formula provides an analytic continuation for $\zeta(s,a)$.

13.21 Deductions from Hermite’s formula.

When $s=0$, $\zeta(0,a)=\frac12-a$.

In the limit of $s\to1$, applying Binet’s second expression of $\log\Gamma$, we have $$ \lim_{s\to1}\zeta(s,a)-\frac{1}{s-1} = -\frac{\Gamma'(a)}{\Gamma(a)}. $$ By the same expression of $\log\Gamma$, $$ \left.\frac{d}{ds}\zeta(s,a)\right|_{s=0} = \log\Gamma(a)-\frac{1}{2}\log2\pi. $$ With $a=1$ $$ \begin{gather} \lim_{s\to1}\zeta(s)-\frac{1}{s-1} = \gamma\newline \zeta'(0) = -\frac{1}{2}\log2\pi. \end{gather} $$ 13.3 Euler’s product for $\zeta(s)$.

For $\sigma\ge 1+\delta$ $$ \prod_p\left(1-\frac1{p^s}\right)=\frac1{\zeta(s)}. $$ 13.31 Riemann’s hypothesis concerning the zeros of $\zeta(s)$.

From Riemann’s functional equation $$ \zeta(s)=2^{s-1}\frac1{\Gamma(s)\cos(\frac{s\pi}{2})}\zeta(1-s) $$ the zeros of $\zeta(s)$ can either be due to $\Gamma(s)^{-1}\cos^{-1}(\frac{s\pi}2)$ if $\sigma<0$, since $\zeta(s)$ has no zero for $\sigma\ge 1+\delta$ (by Euler’s product), i.e. at $s=-2,-4,\cdots$, or lie in the critical strip $0\le \sigma\le 1$. Riemann conjectured that all zeros in the strip $0\le \sigma\le 1$ lie on the line $\sigma=\frac12$.

13.4 From the Eulerian integral of the second kind $n^{-s}\Gamma\left(\frac12s\right)\pi^{-\frac12s}=\int_0^\infty e^{-n^2\pi x}x^{\frac12s-1}dx$, we have $$ \zeta(s)\Gamma\left(\frac12s\right)\pi^{-\frac12s}=\lim_{N\to\infty}\int_0^{\infty}\sum_{n=1}^Ne^{-n^2\pi x}x^{\frac12s-1}dx $$ Let $\varpi(x)=\sum_{n=1}^\infty e^{-n^2\pi x}$ ($\psi$ in Riemann’s notation, see H.M. Edwards, _Riemann’s Zeta Function_, cf. Jacobi theta function), the integral $\int_0^\infty \varpi(x)x^{\frac12s-1}dx$ converges since $\lim_{x\to 0^+}x^{\frac12}\varpi(x)=\frac12$. The limit can be found from $1+2\varpi(x)=x^{-\frac12}(1+2\varpi(1/x))$ which is a result of applying the Poisson summation formula to $\varpi(x)$.

[The text has the limit $\lim_{x\to 0}x^{\frac12}\varpi(x)=1$ which seems to be an error.]

The remainder can be bounded $$ \begin{align*} \left|\int_0^\infty\sum_{n=N+1}^\infty e^{-n^2\pi x}x^{\frac12s-1}dx\right| &\le \int_0^\infty\sum_{n=N+1}^\infty e^{-n(N+1)\pi x}x^{\frac12\sigma-1}dx \newline &=\int_0^\infty\frac{e^{-(N+1)^2\pi x}x^{\frac12\sigma-1}}{1-e^{(N+1)\pi x}}\newline & \le \frac{1}{(N+1)\pi} \int_0^\infty e^{-(N^2+2N)\pi x}x^{\frac12\sigma-2}dx & \frac{x e^{-x}}{1-e^{-x}} < 1 ,\forall x>0\newline & = \frac{\big((N^2+2N)\pi\big)^{1-\frac12\sigma}}{(N+1)\pi}\Gamma\left(\frac12\sigma-1\right)\to0 & N\to\infty,\sigma>2 \end{align*} $$ and vanishes when $\sigma>2$.

Making the change of variable $x\to\frac1x$ for $\varpi(x)$ on the interval $(0,1)$ and rearranging the terms, we have $$ \zeta(s)\Gamma(s)-\frac{1}{s(s-1)}=\int_1^\infty \left(x^{\frac12(1-s)x}+x^{\frac12s}\right)x^{-1}\varpi(x)dx. $$ The integral on the RHS is analytic for all $s\in\mathbb{C}$.

Riemann’s Xi-function $\xi(t)$ ($\Xi$ in modern notation) on the line $s=\frac12+it$ $$ \begin{align*} \xi(t) &=\frac12s(s-1)\zeta(s)\Gamma\left(\frac12s\right)\pi^{-\frac12s}\newline &= \frac12-\left(t^2+\frac14\right)\int_1^\infty x^{-\frac34}\varpi(x)\cos\left(\frac12t\log x\right)dx. \end{align*} $$