\documentclass[12pt]{article} \usepackage{graphicx} \setlength{\textwidth}{6.5in} \setlength{\textheight}{9.5in} \setlength{\topmargin}{-0.5in} \setlength{\oddsidemargin}{-0.3cm} \setlength{\evensidemargin}{-0.3cm} \setlength{\parindent}{1cm} \setlength{\parskip}{2ex} \usepackage{amssymb} \newcounter{ctr} \newenvironment{lista}{\begin{list}{(\alph{ctr})}% {\usecounter{ctr} \setlength{\itemsep}{0cm} \setlength{\parskip}{0cm} \setlength{\leftmargin}{3em}}}{\end{list}} \newcounter{ctr1} \newenvironment{listr}{\begin{list}{(\roman{ctr1})}% {\usecounter{ctr1} \setlength{\itemsep}{0cm} \setlength{\parskip}{0cm} \setlength{\leftmargin}{3em}}}{\end{list}} \newcommand{\half}{\frac{1}{2}} \newcommand{\dis}{\displaystyle} \newcommand{\varep}{\varepsilon} \begin{document} \addtolength{\baselineskip}{1.5mm} \begin{center} {\bf 1.7. Dirichlet series and the zeta function.} \end{center} \begin{flushleft} {\it The zeta function of a real variable} \end{flushleft} A {\it Dirichlet series} is a series of the form \[ \sum_{n=1}^\infty \frac{a(n)}{n^s}. \] (By contrast, a {\it power} series is, of course, one of the form $\sum_{n=0}^\infty a(n)s^n$.) Any arithmetic function $a(n)$ has a corresponding Dirichlet series defined in this way. The $s$ is to be regarded as a complex variable, and we return shortly to the meaning and properties of $n^s$ when $s$ is complex. As mentioned previously, the (Riemann) {\it zeta function} is the case $a(n) = 1$: \[ \zeta (s) = \sum_{n=1}^\infty \frac{1}{n^s}. \] We summarize first the elementary properties of the zeta function as a function of a real variable (which we denote by $\sigma $). {\bf Proposition 1.7.1} {\it The series defining $\zeta (\sigma )$ converges for all $\sigma > 1$. For such $\sigma $, ~$\zeta (\sigma ) > 1$ and is decreasing. Also: \[ \frac{1}{\sigma -1} \leq \zeta (\sigma ) \leq \frac{1}{\sigma -1} + 1. \] Hence:\\[2mm] \begin{tabular}{rl} (i) & $\zeta (\sigma ) \to 1$ as $\sigma \to \infty $,\\ [2mm] (ii) & $\zeta (\sigma) \to \infty $ as $\sigma \to 1^+$, \\ [2mm] (iii) & $(\sigma - 1) \zeta (\sigma ) \to 1$ as $\sigma \to 1^+$. \end{tabular}\\[3mm] Further, if $\zeta_n (\sigma) = \sum_{r=1}^n (1/r^\sigma )$, then \[ \frac{1}{(\sigma -1)(n+1)^{\sigma -1}} \leq \zeta (\sigma) - \zeta_n(\sigma ) \leq \frac{1}{(\sigma -1)n^{\sigma -1}}. \] Proof}. Firstly, $\zeta (\sigma )$ is decreasing since this is true of each term $1/n^\sigma $. Now apply 1.4.4 with $f(x) = 1/x^\sigma $. The stated inequalities, and statements (i),(ii),(iii), all follow, since \[ \int_1^\infty \frac{1}{x^\sigma }\; dx = \frac{1}{\sigma -1}, \qquad \int_n^\infty \frac{1}{x^\sigma }\; dx = \frac{1}{(\sigma -1)n^{\sigma -1}}.\quad \Box \] {\it Note}. The function $f(x) = 1/x^\sigma $ is {\it convex}, hence $\half f(n-1) + \half f(n) \geq \int_{n-1}^n f(x)\; dx$ (i.e. the trapezium estimate is greater than the integral; cf. 1.4.12). By adding these inequalities, we can strengthen the left-hand inequality in DSZ1 to $\zeta (\sigma ) \geq 1/(\sigma -1) + \half $. The value of $\zeta (2)$ is $\pi ^2/6$ ($= 1.6449\ldots $). This can be derived from the Fourier series for $x^2$. Explicit values of this type for $\zeta (\sigma )$ are only known for even integers. The relevance of the zeta function to prime numbers will emerge in chapter 2. \begin{flushleft} {\it The function $x^s$ for complex s} \end{flushleft} We assume that the reader is familiar with the complex exponential function, which is defined by the usual series and satisfies~ $e^s e^w = e^{s+w}$ ~and~ $\frac{d}{ds}e^s = e^s$. For {\it positive, real x} and {\it complex s}, we define $x^s$ (without ambiguity) to be $e^{s \log x}$, where $\log x$ means the usual real-valued logarithm of $x$. Clearly, $(xy)^s = x^s y^s$ ~and~ $x^s x^w = x^{s+w}$. We follow the rather strange traditional notation of analytic number theory, in which a complex variable is written as $s = \sigma + it$. With this notation, we have \[ x^s = e^{\sigma \log x} e^{it \log x} = x^\sigma [\cos (t\log x) + i\sin (t\log x)], \] \[ |x^s| = x^\sigma , \] \[ x^{\overline{s}} = \overline{x^s} \quad \mbox{(where $\overline{z}$ denotes the complex conjugate of $z$)}. \] Next, we consider derivatives and integrals of $x^s$, firstly as a function of $s$. Since $\frac{d}{ds} e^{as} = ae^{as}$, we have \[ \frac{d}{ds} x^s = x^s \log x. \] Now regard $x^s$ as a function of $x$. This is a complex function of a real variable; for the reader who needs it, the basic principles of the calculus of such functions are set out in appendix A. In particular, the appropriate form of the chain rule applies to give (exactly as when $s$ is real) \[ \frac{d}{dx} x^s = \frac{d}{ds} e^{s \log x} = \frac{s}{x} e^{s \log x} = s x^{s-1}. \] The ``fundamental theorem of calculus" now applies to give (again as in the real case) \[ \int_a^b x^s \; dx = \frac{1}{s+1} (b^{s+1} - a^{s+1}) \] and for Re $s > 1$, \[ \int_a^\infty \frac{1}{x^s}\; dx = \frac{1}{(s-1)a^{s-1}}. \] \begin{flushleft} {\it Basic properties of Dirichlet series} \end{flushleft} First, we consider the region of {\it absolute} convergence. As the next result shows, it is either the whole plane, empty or a half-plane. {\bf Proposition 1.7.2} {\it Let $a(n)$ be a sequence. Suppose that $\dis{\sum_{n=1}^\infty \frac{|a(n)|}{n^\alpha }}$ is convergent (with sum $M$) for some real number $\alpha $. Then $\dis{\sum_{n=1}^\infty \left| \frac{a(n)}{n^s} \right|}$ is convergent (with sum not greater than M) for all $s = \sigma +it$ with $\sigma \geq \alpha $. Consequently, there exists $\sigma_{a}$ (possibly $\infty $ or $-\infty $) such that $\sum_{n=1}^\infty a(n)/n^s$ ~is absolutely convergent when $\sigma > \sigma_{a}$, not absolutely convergent when $\sigma < \sigma_{a}$. Proof}. The first statement follows from the comparison test for series, together with the inequality $|\sum_{n=1}^\infty c_n| \leq \sum_{n=1}^\infty |c_n|$, since for $\sigma \geq \alpha $, \[ \left| \frac{a(n)}{n^s} \right| = \frac{|a(n)|}{n^\sigma} \leq \frac{|a(n)|}{n^\alpha }. \] Now let $E$ be the set of real numbers $\alpha $ such that $\sum_{n=1}^\infty (|a(n)|/n^\alpha )$ is convergent, and let $\sigma_{a}$ be the infimum of $E$. Let $s = \sigma +it$. If $\sigma > \sigma_{a} $, then (by the meaning of infimum) there exists $\alpha $ in $E$ such that $\alpha < \sigma $. By the first statement, it follows that $\sigma $ is in $E$ (as required). If $\sigma < \sigma_{a}$, then $\sigma $ is not in $E$. Since $|a(n)/n^s| = |a(n)|/n^\sigma $, this means that $\sum_{n=1}^\infty |a(n)/n^s| $ is not absolutely convergent. $\Box $ {\bf Proposition 1.7.3} {\it If $|a(n)| \leq 1$ for all $n$, then $\sum_{n=1}^\infty a(n)/n^s$ is absolutely convergent when {\rm Re} $s > 1$ (hence $\sigma_a \leq 1$). The same is true if $|a(n)| \leq \log n$ for all $n$. Proof}. The first statement follows from convergence of $\sum_{n=1}^\infty (1/n^\sigma )$. The second statement amounts to saying that $\sum_{n=1}^\infty (\log n/n^\sigma )$ is convergent. To show this, let $\sigma = 1+ 2\delta $. For large enough $n$, we have $\log n \leq n^\delta $, hence $\log n/n^\sigma \leq 1/n^{1+\delta }$. This implies the fact stated. $\Box $ For the zeta function, we have $\sigma_{a} = 1$, since the series diverges when $\sigma = 1$. Also, we can give the following description of the location of its values. {\bf Proposition 1.7.4} {\it We have $|\zeta (\sigma +it)| \leq \zeta (\sigma )$ when $\sigma > 1$. More precisely, $|\zeta (\sigma +it) - 1| \leq \zeta (\sigma ) -1$. Proof}. The first statement is immediate, and the second one is derived from $\zeta (s) -1 = \sum_{n=2}^\infty 1/n^s$. $\Box $ \noindent \begin{minipage}{10cm} \addtolength{\baselineskip}{1.5mm} \hspace*{1cm} This means that the values assumed by $\zeta (s)$ on a vertical line are within the circle shown. \vspace*{2ex} \hspace*{1cm} Since $\zeta (\sigma ) < 2$ for $\sigma \geq 2$, it also follows that $\zeta (s) \neq 0$ (in fact, Re $\zeta (s) > 0$) for $\sigma \geq 2$. (We shall improve on this statement in section 2.1). \end{minipage} % \begin{minipage}{6cm} % \begin{flushright} % \includegraphics[height=3cm]{dszsc.eps} % \end{flushright} % \end{minipage} Note also that since $n^{\overline {s}} = \overline {n^s}$, we have $\zeta (\overline{s}) = \overline{\zeta (s)}$. We now consider ordinary convergence (not absolute). The key is Abel summation. Restating 1.3.6 and 1.3.8 for the case $f(x) = 1/x^s$, we have: {\bf Proposition 1.7.5} {\it Let $a(n)$ be a sequence, and let $A(x) = \sum_{n\leq x} a(n)$. Then for any $X \geq 1$, \[ \sum_{n \leq X} \frac{a(n)}{n^s} = \frac{A(X)}{X^s} + s \int_1^X \frac{A(x)}{x^{s+1}}\; dx. \] Now suppose that $s \neq 0$ and $A(x)/x^s \to 0$ as $x \to \infty $. Then if one of \[ \sum_{n=1}^\infty \frac{a(n)}{n^s} \qquad \mbox{and } \qquad s \int_1^\infty \frac{A(x)}{x^{s+1}}\; dx \] converges, then so does the other, to the same value. For any $X \geq 1$, we then have} \[ \sum_{n > X} \frac{a(n)}{n^s} = - \frac{A(X)}{X^s} + s \int_X^\infty \frac{A(x)}{x^{s+1}}\; dx .\; \Box \] % Hence, for example, % \[ \zeta (s) = s \int_1^\infty \frac{[x]}{x^{s+1}}\; dx \quad \mbox{for Re } s > 1. \] We shall call an expression of the form \[ I(s) = \int_1^\infty \frac {f(x)}{x^{s+1}}\; dx \] a {\it Dirichlet integral}. The last result shows that every Dirichlet series is essentially a case of a Dirichlet integral. We now formulate a result on convergence for Dirichlet integrals, and derive one for Dirichlet series. Of course, some integrability condition is needed when discussing Dirichlet integrals. We shall make no attempt at maximum generality here. We shall simply assume the following condition, which is certainly satisfied by summation functions $A(x)$: (int) $f$ is continuous except at integers, and has left and right limits at each integer. This ensures that $f(x)/x^{s+1}$ is Riemann integrable on each interval $[n, n+1]$, and integrals on longer intervals are then simply obtained by combining these. % \pagebreak {\bf Proposition 1.7.6} {\it Suppose that f satisfies (int) and for some $\alpha $, we have $|f(x)| \leq Mx^\alpha $ for all $x \geq 1$. Then \[ \int_1^\infty \frac {f(x)}{x^{s+1}}\; dx \] is convergent for all $s = \sigma + it$ with $\sigma > \alpha $. Denote its value by $I(s)$, and let \[ \int_1^X \frac{f(x)}{x^{s+1}}\; dx = I_X(s). \] Then for $\sigma > \alpha $, \[ |I(s)| \leq \frac{M}{\sigma - \alpha } \qquad \mbox{and } \qquad |I(s) - I_X(s)| \leq \frac{M}{(\sigma - \alpha )X^{\sigma - \alpha }}. \] Proof}. We have \[ \left| \frac {f(x)}{x^{s+1}} \right| \leq \frac{M}{x^{\sigma - \alpha +1}}. \] Now for $X \geq 1$, \[ \int_X^\infty \frac{1}{x^{\sigma - \alpha +1}}\; dx = \frac{1}{(\sigma - \alpha )X^{\sigma - \alpha }}. \] The statements follow, by the inequality $|\int_a^b f(x)\: dx| \leq \int_a^b |f(x)|\: dx$ (for a proof of this inequality when $f(x)$ is complex, see appendix A). $\Box $ {\bf Proposition 1.7.7} {\it Let $a(n)$ be a sequence, and let $A(x) = \sum_{n \leq x} a(n)$. Suppose that $|A(x)| \leq Mx^\alpha $ for all $x \geq 1$, where $\alpha \geq 0$. Then $\sum_{n=1}^\infty a(n)/n^s $ is convergent for all $s = \sigma + it$ with $\sigma > \alpha $. Denote its sum by $F(s)$, and let $\sum_{n\leq X} a(n)/n^s = F_X(s)$. Then \[ |F(s)| \leq M \frac{|s|}{\sigma - \alpha } \qquad \mbox{and } \qquad |F(s) - F_X(s)| \leq \frac{M}{X^{\sigma - \alpha} }\left( \frac{|s|}{\sigma - \alpha }+ 1 \right). \] Proof}. This follows at once by combining 1.7.5 and 1.7.6. Note that $|A(x)/x^s| \leq M/x^{\sigma - \alpha }$, which tends to 0 as $x \to \infty $. $\Box $ {\it Remark}. In particular, if $|A(x)| \leq M$ for all $x$, then $|F(s)| \leq M|s|/\sigma $. The ratio $|s|/\sigma $ has a simple geometrical meaning: it equals $\sec \theta $ when $s$ is expressed as $re^{i\theta }$. We can now contrast absolute and non-absolute convergence. Recall that a {\it power} series converges absolutely within the circle where it converges at all. The situation is different for Dirichlet series. {\bf Proposition 1.7.8} {\it Suppose that~ $\sum_{n=1}^\infty a(n)/n^\alpha $ ~is convergent for a certain real $\alpha $. Then $\sum_{n=1}^\infty a(n)/n^s $ is convergent for all $s = \sigma + it$ with $\sigma > \alpha $. Consequently, there exists $\sigma_c$ (possibly $\infty $ or $-\infty $), such that the series is convergent for all $s = \sigma +it$ with $\sigma > \sigma_c$ and not convergent for $\sigma < \sigma_c$. Further, $\sigma_{a} \leq \sigma_c +1$. Proof}. Let $b(n) = a(n)/n^\alpha $ and $B(x) = \sum_{n\leq x}b(n)$. Then $\sum_{n=1}^\infty b(n)$ is convergent, so there exists $M$ such that $|B(x)| \leq M$ for all $x \geq 1$. By the case $\alpha = 0$ in 1.7.7, $\sum_{n=1}^\infty b(n)/n^s$ ~is convergent when Re $s > 0$. But \[ \frac{a(n)}{n^s} = \frac {b(n)}{n^{s-\alpha }}, \] so $\sum_{n=1}^\infty a(n)/n^s$ is convergent when $\sigma > \alpha $. The existence of $\sigma_c$ now follows in the same way as for $\sigma_{a}$. With $\alpha $ as above, we show that $\sum_{n=1}^\infty a(n)/n^s$ is absolutely convergent when $\sigma > \alpha + 1$: it then follows that $\sigma_{a} \leq \sigma_c + 1$. Now $|b(n)|$ is bounded, say by $K$, so \[ \left| \frac {a(n)}{n^s} \right| = \frac {|b(n)|}{|n^{s- \alpha }|} \leq \frac {K}{n^{\sigma - \alpha }}.\] Since $\sigma - \alpha > 1$, the series~ $\sum_{n=1}^\infty |a(n)/n^s|$ ~is convergent, as required. $\Box $ A further easy application of Abel summation shows that when $\sigma_c$ is positive, it is actually characterized by the condition in 1.7.7. We use the $O$ notation (see appendix E). {\bf Proposition 1.7.9} {\it Suppose that $\sum_{n=1}^\infty a(n)/n^\sigma $ converges for a certain $\sigma > 0$. Then $A(x) = O(x^\sigma )$ for $x \geq 1$. Let $\alpha $ be the infimum of the numbers $\sigma $ such that $A(x) = O(x^\sigma )$. If $\alpha > 0$, then $\sigma_c = \alpha $. Proof}. Let $b(n) = a(n)/n^\sigma $, so that $\sum_{n=1}^\infty b(n)$ is convergent. In particular, $|B(x)|$ is bounded, say by $M$. By Abel's summation formula, \[ A(x) = \sum_{n \leq x} n^\sigma b(n) = x^\sigma B(x) - \sigma \int_1^x t^{\sigma -1} B(t)\: dt. \] Hence \[ |A(x)| \leq Mx^\sigma + M\int_1^x \sigma t^{\sigma -1}\: dt \leq 2Mx^\sigma. \] With $\alpha $ as stated, let $\sigma > \alpha $. Take $\sigma _1$ such that $\alpha < \sigma_1 < \sigma$. Then $A(x) = O(x^{\sigma_1})$ for $x \geq 1$, so by 1.7.7, $\sum_{=1}^\infty a(n)/n^\sigma $ is convergent. So $\sigma_c \leq \alpha $. Now let $0 < \sigma < \alpha $. Then $A(x)$ is not $O(x^\sigma )$, so by the above, $\sum_{n=1}^\infty a(n)/n^\sigma $ is divergent. Hence $\sigma_c \geq \alpha $. \hfill $\Box $ Clearly, if $\alpha = 0$, then $\sigma_c \leq 0$. When $\sigma_c < 0$, it can be characterized using the {\it tail} of the series $\sum_{n=1}^\infty a(n)$: see exercise 5. If $a(n) \geq 0$ for all $n$, then absolute convergence of $\sum_{n=1}^\infty a(n)/n^\sigma $ (for real $\sigma $) coincides with convergence. Since $\sigma_c$ is the infimum of {\it real} $\sigma $ for which the series converges (and similarly for $\sigma_{a} $), it follows that $\sigma_{a} = \sigma_c$ in this case. In particular, for the zeta function, we have $\sigma_{a} = \sigma_c = 1$, since the series diverges when $\sigma = 1$. EXAMPLE. Let $a(n) = (-1)^{n-1}$. Then $A(n)$ is bounded, since it is alternately 1 and 0. So by 1.7.7, the series converges whenever $\sigma > 0$. It does not converge when $\sigma = 0$, so $\sigma_c = 0$. Since $|a(n)| = 1$ for all $n$, we have $\sigma_{a} = 1$. It is easy to express the sum $F(s)$ of this series in terms of $\zeta (s)$ for Re $s > 1$: \[ \zeta (s) - F(s) = 2 \sum_{n=1}^\infty \frac{1}{(2n)^s} = \frac{2}{2^s} \zeta (s), \] so~ $F(s) = (1 - 2^{1-s}) \zeta (s)$. $\Box $ {\it Note}. Let $f(x)$ be a function defined on $(0, \infty)$. The {\it Mellin transform} of $f$ is the function defined by \[ M(s) = \int_0^\infty x^{s-1} f(x)\; dx \] where this converges. For functions that are zero on $(0,1)$, the Mellin transform coincides with our ``Dirichlet integral", with $-s$ substituted for $s$. But a general study of Mellin transforms requires consideration of convergence of the integral on $(0,1)$ as well as $[1, \infty )$. % \pagebreak \begin{flushleft} {\it Differentiability} \end{flushleft} It remains to establish a very important property of functions defined by Dirichlet series, namely that they are {\it holomorphic}, i.e. differentiable in the sense of complex functions. We have seen that the derivative of $a(n)n^{-s}$ is $-a(n)n^{-s} \log n$. The point is to show that termwise differentiation of the series is valid. The key notion here is {\it uniform convergence}. Recall that if $f_n \; (n \geq 1)$ and $f$ are (complex) functions and $E$ is a set in the complex plane, then $(f_n)$ is said to converge to $f$ {\it uniformly on E} if for any $\varep > 0$, there exists $n_0$ such that for all $n \geq n_0$, one has~ $|f_n(s) - f(s)| \leq \varep $ for all $s \in E$. If $E$ is the closed rectangle consisting of $s = \sigma +it$ with $\sigma_1 \leq \sigma \leq \sigma_2$, ~$t_1 \leq t \leq t_2$, its {\it interior}, int$(E)$, is the set given by $\sigma_1 < \sigma < \sigma_2$, ~$t_1 < t < t_2$. We assume the following standard theorem from complex analysis: {\it If $(f_n)$ is a sequence of holomorphic functions that converges to a function f uniformly on some rectangle E, then f is holomorphic on} int$(E)$, {\it and $f'(s) = \lim_{n\to \infty } f'_n(s)$ for all s in} int$(E)$. (We remind the reader that a similar statement for {\it real} functions is not true!) {\bf Proposition 1.7.10} {\it Suppose that~ $\sum_{n=1}^\infty a(n)/n^s$ ~converges to $F(s)$ for {\rm Re} $s > \sigma_c$. Then $F(s)$ is holomorphic for such s, with derivative given by \[ F'(s) = -\sum_{n=1}^\infty \frac{a(n) \log n}{n^s}. \] Proof}. Choose $\alpha > \sigma_c\:$. As in 1.7.8, let $b(n) = a(n)/n^\alpha $. Then $|B(N)| \leq M$ (say) for all $N \geq 1$, and $F(s) = G(s-\alpha )$, where~ $G(s) = \sum_{n=1}^\infty b(n)/n^s$. We will show that $G(s)$ is holomorphic, with derivative as stated, for Re $s > 0$. It then follows that \[ F'(s) = G'(s- \alpha ) = -\sum_{n=1}^\infty \frac{b(n)\log n}{n^{s - \alpha }} = \sum_{n=1}^\infty \frac{a(n)\log n}{n^s} \] whenever Re $s > \alpha$, and hence in fact whenever Re $s > \sigma_c\: $. Write $G_N(s) =\sum_{n=1}^N b(n)/n^s$. Since this is a finite sum, we have \[ G'_N(s) = -\sum_{n=1}^N \frac {b(n) \log n}{n^s}. \] Take $\delta > 0$, $R > 0$ and let $E_{\delta ,R}$ be the semi-infinite rectangle consisting of $s = \sigma + it$ with $\sigma \geq \delta $ and $|t| \leq R$. We will show that $G_N(s) \to G(s)$ uniformly on $E_{\delta ,R}$. The statement then follows for all $s$ in int$(E_{\delta ,R})$. But every $s$ with $\sigma > 0 $ belongs to such a set for a suitable choice of $\delta $ and $R$, so the result holds for all such $s$. By 1.7.7 (the case $\alpha = 0$), we have \[ |G(s) - G_N(s)| \leq \frac{M}{N^\sigma } \left( \frac{|s|}{\sigma } + 1 \right). \] Now for all $s$ in $E_{\delta,R}$, \[ \frac{|s|}{\sigma } \leq \frac{\sigma + |t|}{\sigma } = 1 + \frac{|t|}{\sigma } \leq 1 + \frac{R}{\delta }, \] and hence \[ |G(s) - G_N(s)| \leq \frac {M}{N^\delta }\left( 2 + \frac{R}{\delta } \right), \] which tends to 0 as $N \to \infty $. This proves the required uniform convergence. $\Box $ {\it Remark}. Uniform convergence is rather easier if we are only interested in $\sigma > \sigma_{a}$. For if $\alpha > \sigma_{a}$, then the inequality appearing in the proof of 1.7.2, together with the ``M-test", shows that the series converges uniformly on the set of $s$ such that $\sigma \geq \alpha$. We will also need the fact that functions defined by Dirichlet {\it integrals}, under the conditions of 1.7.6, are holomorphic (though of course this is now clear when the Dirichlet integral is obtained from a Dirichlet series as in 1.7.5). The uniform convergence step is easy, by the estimate in 1.7.6. However, we now need information about the derivative (with respect to $s$) of the integral \[ I_X(s) = \int_1^X \frac{f(x)}{x^{s+1}}\; dx \] instead of a finite sum $\sum_{n=1}^N a(n)/n^s$. In place of differentiation of a finite sum, we need to know that ``differentiation under the integral sign" is valid, giving \[ I_X'(s) = - \int_1^X \frac {f(x) \log x}{x^{s+1}}\; dx. \] There is indeed a general theorem ensuring that this is valid. It is not very hard to give a proof specialized to the case we want: this is done in appendix D. Assuming this result, we deduce: {\bf Proposition 1.7.11} {\it Suppose that $f(x)$ satisfies (int) and $|f(x)| \leq Mx^\alpha $ for all $x \geq 1$. Let \[ I(s) = \int_1^\infty \frac{f(x)}{x^{s+1}}\; dx. \] Then $I(s)$ is holomorphic for $s = \sigma + it$ with $\sigma > \alpha $, and \[ I'(s) = - \int_1^\infty \frac{f(x) \log x}{x^{s+1}}\; dx. \] Proof}. We only need to check uniform convergence. Let $E_\delta = \{ s : \sigma \geq \alpha + \delta \}$. By 1.7.6, for $s \in E_\delta $ and integers $N$, \[ |I(s) - I_N(s)| \leq \frac{M}{(\sigma - \alpha )N^{\sigma - \alpha }} \leq \frac{M}{\delta N^\delta }. \] Hence~ $I_N(s) \to I(s)$ ~as $N \to \infty $, uniformly on $E_\delta $. $\Box $ In particular, for the zeta function, we have \[ \zeta'(s) = -\sum_{n=1}^\infty \frac{\log n}{n^s} = -\sum_{n=2}^\infty \frac{\log n}{n^s}. \] when Re $s > 1$. We finish this section with a rough estimation of this function. Clearly, $|\zeta'(s)| \leq -\zeta'(\sigma )$, where $s = \sigma + it$. {\bf Proposition 1.7.12} {\it For all $\sigma > 1$, we have \[ - \zeta'(\sigma ) \leq \frac{1}{(\sigma -1)^2} + \frac{3}{4}. \] Proof}. We assume the integral \[ \int_1^\infty \frac{\log x}{x^\sigma }\; dx = \frac{1}{(\sigma -1)^2}, \] which is readily established by intrgrating by parts on $[1,X]$ and considering the limit as $X \to \infty $. We use the integral to estimate the series, but this is not quite as simple as before because~ $\log x/x^\sigma $ ~does not decrease throughout $[1, \infty)$. In fact, differentiation shows that it decreases when~ $\log x \geq 1/\sigma $. Hence (for any $\sigma > 1$) it certainly decreases for all $x \geq 3$, so that \[ \sum_{n=4}^\infty \frac{\log n}{n^\sigma} \leq \int_3^\infty \frac{\log x}{x^\sigma }\; dx < \frac{1}{(\sigma 1)^2}. \] The terms for $n = 2,3$ contribute less than~ $\half \log 2 + \frac{1}{3} \log 3 \approx 0.7128 < \frac{3}{4}$. (If~ $\sigma \geq \frac{3}{2}$, then $\log x/x^\sigma $ decreases for all $x \geq 2$, so $\frac{3}{4}$ can be replaced by the term $\log 2/2^\sigma $. Actually, one can show that the $\frac{3}{4}$ is not needed at all.) \hfill $\Box $ \begin{center} {\it Exercises} \end{center} \noindent 1. By reversing the order of summation, prove that~ $\sum_{m=2}^\infty [\zeta (m) -1] = 1$, and obtain a series expression for~ $\sum_{m=2}^\infty \zeta'(m)$. \noindent 2. By using integral estimation for~ $\sum_{n=3}^\infty 1/n^\sigma $, show that \[ \zeta (\sigma) \leq 1 + \frac{\sigma +1}{\sigma -1} \frac{1}{2^\sigma }. \] \noindent 3. Let~ $A(x) = \sum_{n \leq x} a(n)$, and suppose that $|A(x)| \leq Mx^\alpha $ for all $x \geq 1$, where $\alpha > 0$. Write $b(n) = a(n) \log n$. Use Abel summation to show that $ |B(x)| \leq Mx^\alpha (\log x + \alpha ^{-1})$, and deduce that~ $\sum_{n=1}^\infty b(n)/n^s$ ~is convergent when $\sigma > \alpha $. Adapt this for the case $\alpha = 0$. \noindent 4. ({\it An estimate in terms of t}) Suppose that $|a(n)| \leq m$ and $|A(n)| \leq M$ for all $n$. Let $F(s) = \sum_{n=1}^\infty a(n)/n^s$ where it converges. Let $t \geq 0$ be given. By taking $X = t+2$ in 1.7.7, show that \[ |F(1 + it)| \leq m \log (t+2) + M + m. \] \noindent 5. ({\it Companion to 1.7.7 in terms of the tail of $\sum a(n)$})~ Suppose that $\sum_{n=1}^\infty a(n)$ is convergent (say to $L$), and write $R(x) = L - A(x) = \sum_{n > x} a(n)$. Suppose that for some $M$ and some $\alpha > 0$, we have $|R(x)| \leq Mx^{-\alpha }$ for all $x \geq 1$. By substituting $L - R(x)$ for $A(x)$ in Abel's summation formula, show that if Re $s < \alpha $, then $\sum_{n=1}^\infty a(n)n^s$ converges to \[ L + s\int_1^\infty x^{s-1}R(x)\: dx. \] Deduce that $\sigma_c \leq -\alpha $. \noindent 6. Suppose that $a(n) \geq 0$ for all $n$. Suppose also that~ $\sum_{n=1}^\infty a(n)/n^s$ ~converges to $F(s)$ for all real $s > 0$, and that $F(s) \to L$ as $s \to 0^+$. Prove that $\sum_{n=1}^\infty a(n)$ converges to $L$. Deduce the following variant: if $F(s) \to L_1$ as $s \to 1^+$, then~ $\sum_{n=1}^\infty a(n)/n = L_1$. \noindent 7. Suppose that $B(x)$ is bounded and tends to $L$ as $x \to \infty $. Let \[ G(s) = s\int_1^\infty \frac{B(x)}{x^{s+1}}\: dx. \] In this integral, replace $B(x)$ by $B(x)-L$ and split $[1, \infty )$ into $[1,X]$ and $[X,\infty )$ to show that $G(s) \to L$ as $s \to 0^+$ (with $s$ real). Now let $F(s) = \sum_{n=1}^\infty a(n)/n^s$, and suppose that $A(x) - \log x \to L$ as $x \to \infty $. Using the identity $s\int_1^\infty x^{-s-1}\log x\: dx = 1/s$, deduce that \[ F(s) - \frac{1}{s} \to L \quad \mbox{as } s \to 0^+. \] Deduce in particular that $\zeta (s) - 1/(s-1) \to \gamma $ as $s \to 1^+$. %and suppose that $A(x)/x \to L$ as $x \to \infty $. %Prove that $(s-1)F(s) \to L$ as $s \to 1^+$. \noindent 8. The {\it gamma function} is defined, for Re $s > 0$, by \[ \Gamma (s) = \int_0^\infty x^{s-1} e^{-x}\; dx. \] By substituting $x = ny$ in this expression and assuming validity of termwise integration of the relevant series, show that \[ \Gamma (s) \zeta (s) = \int_0^\infty \frac{x^{s-1}}{e^x -1}\; dx \] for Re $s > 1$. % (Termwise integration can be justified without the % dominated convergence theorem, by considering % the finite geometric series~ $\sum_{j=1}^n e^{-jx}$.) \noindent 9. ({\it Uniqueness and zeros}) Suppose that $\sum_{n=1}^\infty a(n)/n^s$ converges to $F(s)$ when Re $s > \sigma_c$, and that there is a sequence $s_k = \sigma_k + it_k$, with $\sigma_k \to \infty $ as $k \to \infty $, such that $F(s_k) = 0$ for all $k$. Show by the following steps that $a(n) = 0$ for all $n$. Supposing the opposite, let $N$ be the first $n$ such that $a(n) \neq 0$. Fix $c > \sigma_a$. Show that there is a constant $B$ such that, for $\sigma > c$, \[ \sum_{n=N+1}^\infty \frac{|a(n)|}{n^\sigma } \leq \frac{B}{(N+1)^{\sigma - c}}, \] and hence that $a(N) = 0$. Deduce further: (i) if $a(n) \neq 0$ for some $n$, then there exists $\sigma_1$ such that $F(s) \neq 0$ whenever $\sigma > \sigma_1$, (ii) if $\sum_{n=1}^\infty a(n)/n^\sigma $ = $\sum_{n=1}^\infty b(n)/n^\sigma $ for all large enough $\sigma $, then $a(n) = b(n)$ for all $n$. \end{document}