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Results tagged with riemann-zeta-function
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user 84768
The Riemann zeta function is the function of one complex variable $s$ defined by the series $\zeta(s) = \sum_{n \geq 1} \frac{1}{n^s}$ when $\operatorname{Re}(s)>1$. It admits a meromorphic continuation to $\mathbb{C}$ with only a simple pole at $1$. This function satisfies a functional equation relating the values at $s$ and $1-s$. This is the most simple example of an $L$-function and a central object of number theory.
3
votes
Xi Function on Critical Strip - Mellin Transform
Let $f(x) = 2\sum_{n = 1}^\infty e^{- \pi n^2 x^2}$ and $E(s) = \pi^{-s/2} \Gamma(s/2)\zeta(s)$.
For $Re(s) >0$ : $2\int_0^\infty x^{s-1} e^{-\pi n^2 x^2} dx = n^{-s} \pi^{-s/2}\Gamma(s/2) $ so we ha …
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1
vote
On a derivative involving the Riemann zeta function
Many basic complex-analyis misunderstandings here.
First of all, $\frac{d^n}{ds^n}$ means $n \in \mathbb{N}$ (otherwise you have to define it). Then you probably wanted to write the Laurent series …
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2
votes
Accepted
On Riemann zeta function and Dirac delta function/distribution
Let $$C_n = \int_{-1}^1 (1-\frac{x^2}{2})^n dx \sim \int_{-\pi/2}^{\pi/2} |\cos(x)|^n dx$$
If $f$ is continuous
$$\lim_{n \to \infty} \int_{-1}^{1} \frac{(1-\frac{x^2}{2})^{n} }{C_n} f(x)=\lim_{n \to …
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2
votes
$L_2$ bounds for $\zeta(1/2 + it)$ and a related integral
For $\Re(s) > 1$, $$\frac{\zeta(s)}{s} = \int_1^\infty (\sum_{1 \le n \le x} 1) x^{-s-1}dx, \qquad \frac{\zeta(s)}{s-1} = \int_1^\infty (x\sum_{1 \le n \le x} \frac1n) x^{-s-1}dx$$ Thus for $\Re(s) \ …
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3
votes
A generating function for non-trivial zeros of Riemann zeta function
There is no good reason to take the imaginary part of the zeros or to keep only those in $\Re(s)\ge 1/2$.
From the residue theorem we get for $\Re(x) >0$ the holomorphic function
$$F(x)=\sum_{\Im(\rho …
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9
votes
Is regularization of infinite sums by analytic continuation unique?
Let $$f_k(s) = k^{-s}+(s+1)k^{-s-2},\qquad f_k(-1)=k$$ then $$F(s)=\sum_k f_k(s) = \zeta(s)+(s+1)\zeta(s+2), \qquad F(-1)=-1/12+1$$
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8
votes
Accepted
$\psi(x)-x$ on average
Impossible as this would imply that $\frac{\zeta'(s)}{\zeta(s)}+\frac1{s-1}$ is analytic on $\Re(s)\ge 1/2$.
Your bound
$$\int_1^x |\psi(y)-y|^2 dy = O(x^a)$$ for some $a < 2$ implies (with Cauchy-Sch …
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3
votes
Accepted
Fourier transform of the von Mangoldt function?
Start with the explicit formula
$$\sum_{n \le x}\Lambda(n) =\frac1{2i\pi} \int_{2-i\infty}^{2+i\infty} \frac{-\zeta'(s)}{\zeta(s)}\frac{x^s}s ds=1_{x > 1}\sum Res(\frac{-\zeta'(s)}{\zeta(s)}\frac{x^s …
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1
vote
Continuing an analytic continuation of the Dirichlet $\eta$-function?
I abandoned making a precise answer but for a continuous branch of $(N+1/2 +ix)^{-s}$, $F_N(s)= \int_{-\infty}^{\infty} \frac{(N+1/2 +ix)^{-s}}{\cosh(\pi x)}\,dx$ converges and is analytic for $\Im(s …
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16
votes
A new way of approaching the pole of the Riemann zeta function - and a new conjectured formula
The first formula is trivial. $$f(s)= \frac1{s-1}+\gamma +O(s-1)$$ $$g(z)=1+2^{-z}+3^{-z}+4^{-z}+O(5^{-z})=1+2^{-z}(1+(3/2)^{-z}+(4/2)^{-z}+O(5/2)^{-z})$$
$$f(g(z)) = \frac1{2^{-z}(1+(3/2)^{-z}+(4/2) …
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2
votes
A Hadamard product representation for Keiper's $\tau$-function?
$$\frac{\xi'(s)}{\xi(s)}=\sum_\rho \frac1{s-\rho}+\frac1{\rho} = -\sum_{j\ge 1} s^j\rho^{-j-1}$$ the Taylor series is valid for $|s|<|1/2+14.13i|$
Thus the coefficients have exponential decay which i …
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3
votes
Regularisation of $\sum_{n=0}^\infty \frac{1}{(a+n^2)^p}$
As I said in the comment, $$F_a(s) = \sum_{n=0}^\infty (n^2+a)^{-s}, \qquad Re(s) > 1/2$$
Has an analytic continuation in term of the Riemann zeta function :
$$F_a(s) = \sum_{n= 0}^{A-1} (n^2+a)^{-s …
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4
votes
Accepted
Analytic extension of the Hurwitz ζ function
Summing by parts, using the binomial series and inverting the double sum works fine, obtaining
$$\zeta(s,a)-s \zeta(s,N)
= \sum_{n=1}^{N-1} n ((n+a-1)^{-s}-(n+a)^{-s})\\+ \sum_{k=0}^\infty {-s \cho …
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0
votes
Geometric / physical / probabilistic interpretations of Riemann zeta($n>1$)?
The sequence $a(2k)=-2(2i\pi)^{-2k}\zeta(2k),a(1)=\frac{-1}2$ is the convolutive inverse of the sequence $\frac1{(k+1)!},k\ge 0$
-2
votes
On infinite sum containing logarithmic derivative of Zeta function and Möbius function:
Your question doesn't make much sense. Let $a_d =\sum_{m|d} \mu(m) \frac{e^{-(d/m)/2}}{d/m}=O(1)$. As $\frac{\zeta'(s)}{\zeta(s)}=O(2^{-s})$ then for $\Re(s) >0$, $F(s) = \sum_{d\ge 1} a_d\frac{\zeta' …
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