Dirichlet series

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

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.

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 $T\ge 2$.