Dirichlet series

A Dirichlet series is a series of the formMultiplication of absolutely convergent Dirichlet series corresponds to Dirichlet convolution of their coefficients.

Let $S(x)={\sum }_{n\le x}{a}_n$ satisfy $|S(x)|\le C{x}^{\alpha }$ for $x\ge 1$. The Abel summation formula gives $A(s)=s{\int }_1^{\infty }S(t)t^{-s-1}dt$ when $\mathrm{Re}s>\alpha$, since its finite-sum boundary term tends to zero. Split this integral into blocks $I_m(s)={\int }_m^{m+1}S(t)t^{-s-1}dt$. For every $\delta >0$, the estimate $|I_m(s)|\le C{m}^{-1-\delta }$ holds uniformly on $\mathrm{Re}s>\alpha +\delta$. Each block is an entire function, by differentiation under the finite-range integral on compact sets. The Weierstrass M-test makes their sum locally uniformly convergent and therefore holomorphic; multiplication by $s$ preserves holomorphy.

This proof distinguishes uniform convergence of the block sum on a whole half-plane from local uniform convergence after multiplication by $s$. The latter is all that holomorphy needs. In particular, a bound ${\sum }_{n\le x}\mu (n)=O(x^{\alpha })$ gives a holomorphic continuation of the reciprocal Riemann zeta function to this half-plane and excludes zeta zeros there, with the pole at one treated by cancellation.

absolute convergence of a Dirichlet series depends only on the real part of its variable. The set of real parts giving convergence is upward closed by the comparison test. If that set is nonempty and not all of the real line, its finite infimum is the abscissa: every larger real part gives convergence, and every smaller real part gives divergence. Boundary convergence is a separate question.

Writing $B(x)={\sum }_{n\le x}{b}_n$, the Abel summation formula gives $\sum b_n{n}^{-s}=s{\int }_1^{\infty }B(x)x^{-s-1}dx$. If $B$ is bounded, the integral and its differentiated integrands converge uniformly on compact subsets of $\mathrm{Re}s>0$, giving a holomorphic function. A nonprincipal Dirichlet character has bounded partial sums because it is periodic and its sum over a period is zero.

// Target: algebra.bigb

For a full-rank Euclidean lattice in ${\mathbb{R}}^n$, its Epstein zeta function is $E_{\Lambda }(s)={\sum }_{0\ne x\in \Lambda }|x{|}^{-2s}$, initially convergent for $\mathrm{Re}s>n/2$. A Mellin transform of its theta function gives its completed Epstein zeta function and a meromorphic continuation. Its sole pole has residue ${\pi }^{n/2}/[m(\Lambda )\Gamma (n/2)]$ at $s=n/2$.

The completion ${\mathcal{E}}_{\Lambda }(s)={\pi }^{-s}\Gamma (s)E_{\Lambda }(s)$ satisfies ${\mathcal{E}}_{\Lambda }(s)=m(\Lambda {)}^{-1}{\mathcal{E}}_{{\Lambda }^{\vee }}(n/2-s)$. Its simple poles at zero and $n/2$ have residues $-1$ and $m(\Lambda {)}^{-1}$. The pole-subtracted theta integral for an Epstein zeta function gives both the continuation and this dual lattice symmetry.

Put $a=n/2$, $m=m(\Lambda )$ and $A_{\Lambda }(s)={\int }_1^{\infty }({\Theta }_{\Lambda }(it)-1)t^{s-1}dt$, which is entire. Splitting the Mellin transform at one and using the lattice theta functional equation gives ${\mathcal{E}}_{\Lambda }(s)=A_{\Lambda }(s)+m^{-1}{A}_{{\Lambda }^{\vee }}(a-s)+m^{-1}/(s-a)-1/s$. The two rational terms explicitly retain the contributions of the zero lattice vector.

A Dirichlet polynomial is a finite Dirichlet series, $A(s)={\sum }_{n\le X}{a}_n{n}^{-s}$. Vertical integration separates equal product indices from oscillatory unequal ones.

For finite Dirichlet polynomials, ${\int }_T^{2T}A(\sigma +it)\overline{B(\sigma +it)}dt$ has diagonal term $T{\sum }_n{a}_n\overline{b_n}/n^{2\sigma }$. Its off-diagonal error is bounded by ${\sum }_{m\ne n}|a_n{b}_m|/(n^{\sigma }{m}^{\sigma }|\log (m/n)|)$. A weighted row-sum estimate further bounds this by $\log (2X){\sum }_{n\le X}{n}^{1-2\sigma }(|a_n{|}^2+|b_n{|}^2)$.

For odd $j$, expanding $({\sum }_{p\le X}{p}^{-\sigma }\cos (t\log p){)}^j$ gives products with different numbers of prime factors on the two sides. Unique prime factorization rules out every diagonal equality. The integral over $[T,2T]$ is therefore $O_j(X^j({\sum }_{n\le X^j}{n}^{-\sigma }{)}^2)$.

An Euler product factors a Dirichlet series into local factors indexed by prime numbers. For a multiplicative arithmetic function $f$ and in a half-plane of absolute convergence,

For $\sigma >1$, logarithmic expansion of the Euler product reduces positivity to $3+4\cos \theta +\cos 2\theta =2(1+\cos \theta {)}^2$. At primes dividing the modulus the zeta contribution is positive by itself. If the last factor is bounded at the line-one point, a zero of the middle factor would outweigh the zeta pole and force this product to zero. This proves the relevant nonvanishing of Dirichlet L-functions on the line one.

For a finite set of primes choose $0<g(p)<1$, put $k(p)=g(p)/(1-g(p))$, and let $W={\prod }_p(1+k(p))$. Give each squarefree integer $d$ on these primes probability $k(d)/W$. Inclusion of each prime is represented by a Bernoulli random variable with parameter $g(p)$; these are independent random variables, so $\mathbb{E}[\log d]={\sum }_{p}g(p)\log p$. If this is at most $\theta \log L$ with $0<\theta <1$, the Markov inequality givesThis turns a full Euler product into a lower bound for the truncated normalizing sum of a Selberg sieve.

For a finite set of primes,expands as the sum of $1/n$ over positive integers whose prime factors are at most $x$. It contains every term with $n\le x$, so these partial products dominate the harmonic series and diverge. Sinceuniformly for $0\le t\le 1/2$, convergence of ${\sum }_{p}1/p$ would force convergence of the logarithms of these products, a contradiction.

For $x\ge 2$, the harmonic number $H_{\lfloor x\rfloor }$ exceeds $\log x$. The finite Euler product over the primes at most $x$ contains every summand of this harmonic number. Taking logarithms and applying the geometric-series bound for the logarithmic remainder bounds the total remainder by $\frac{1}{2}{\sum }_{n=2}^{\infty }[n(n-1){]}^{-1}=1/2$, proving the displayed bound.

For every complex number $z$ with $|z|\le 1$,Applying this to each prime-power term in logarithms of Euler products proves thatfor $\sigma >1$ and every completely multiplicative $f$ bounded by one.

Perron's formula recovers a summatory arithmetic function from its Dirichlet series by the inverse Mellin integralwith the usual convergence and endpoint conventions.

The Perron's formula kernel $y^{s+1}/(s(s+1))$ integrates to $(y-1{)}_{+}$, giving the smoothed sum as an integral of $x^{s+1}/(\zeta (s)s(s+1))$. For each fixed height $T$, a zero-free rectangle permits shifting the central contour to a real part below one; its vertical contribution is $o(x^2)$, and its horizontal contributions are $O_T(x^2/\log x)$ when the original real part is $1+1/\log x$. A bound $1/\zeta (\sigma +it)=O({\log }^7(2+|t|))$ for $\sigma \ge 1$ bounds the remaining tails by $C{x}^2{\int }_T^{\infty }{\log }^7(2+t)t^{-2}dt$. Letting $x$ grow first and $T$ afterward proves the displayed cancellation, without uniform estimates to the left of one.

If the Dirichlet series $F(s)={\sum }_n{a}_n{n}^{-s}$ is absolutely convergent at $\sigma >0$, thenThe scalar kernel is $(2\pi {)}^{-1}{\int }_{\mathbb{R}}{e}^{v(\sigma +it)}(\sigma +it{)}^{-2}dt=v_{+}$. This follows by Fourier inversion applied to $u{e}^{-\sigma u}{1}_{u\ge 0}$, whose Fourier transform is $(\sigma +it{)}^{-2}$. Absolute convergence of the series and of $\int |\sigma +it{|}^{-2}dt$ justifies interchange. The factor $ds=idt$ is essential.

Suppose $|a_n|\le (\log (2n){)}^k$ and $A(x)={\sum }_{n\le x}{a}_n\log (x/n)=O(x(\log x{)}^{\beta })$, with $0\le \beta \le k$. For $0<h\le 1$, direct subtraction givesThe last term has absolute value $O((x{h}^2+h)(\log x{)}^k)$. Choose $h=(\log x{)}^{(\beta -k)/2}$ to obtain the displayed bound, with an additional $O((\log x{)}^k)$ term harmless as $x\to \infty$ for fixed $k$. This elementary finite-difference argument converts a logarithmically smoothed Perron formula bound to cancellation in its original partial sum.

A truncated Perron integral over $c-iT$ to $c+iT$ recovers a summatory function with an error controlled by

For $c=1+1/\log x$, $1\le y\le x$ and $2\le T\le x$, apply the truncated Perron formula with Von Mangoldt function coefficients at both endpoints on the same vertical line. The difference kernel is $((x+y{)}^s-x^s)/s={\int }_x^{x+y}{u}^{s-1}du$, of modulus $O(y)$. The usual near-integer error bound contributes $O(x{\log }^{2}x/T)$. The logarithmic derivative on this line is finite because of the Euler product.

For $v\ne 1$, $K_T(v)$ differs from $1_{v>1}$ by $O(v^c\min (1,(T|\log v|{)}^{-1}))$. Close the contour to the left for $v>1$ and to the right for $v<1$; horizontal sides give the logarithmic denominator and the nearby transition is bounded. At $v=1$, direct integration gives $K_T(1)={\pi }^{-1}\arctan (T/c)=1/2+O(c/T)$. This explains the half-weight endpoint in the truncated Perron formula. The constants may be taken uniform for $1\le c\le 2$ and $T\ge 2$.