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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.
7
votes
Accepted
Formula in common: How to search for same/similar equations in other knowledge domains?
If you can turn your formula into an integer sequence, put it into OEIS. This includes:
Plug in integer values for your parameters and enter the outputs.
If your construction produces a sequence of …
9
votes
Bounds for $\sum_{k=1}^\infty\frac{(-1)^{k+1}x^k}{(k-1)!\zeta(2k)}$
This answer is inspired by a now deleted answer by Abhinav Kumar:
A perhaps simpler form is
$$\sum_{n=1}^{\infty} \mu(n) \frac{x}{n^2} e^{- x/n^2}$$
where $\mu$ is the Möbius function.
Proof: Expand $ …
6
votes
How can one deduce an approximation for the density function of prime numbers from this Eule...
Let $d \pi$ be the measure on $\mathbb{R}_{>0}$ which is a delta function of size $1$ at the primes. In other words, $\int_{y}^x d \pi = \pi(x) - \pi(y)$. The prime number theorem is that $\pi(x)$ is …
87
votes
$\prod_{n=1}^{\infty} n^{\mu(n)}=\frac{1}{4 \pi ^2}$
We have
$$\frac{1}{\zeta(s)} = \sum_{n=1}^{\infty} \frac{\mu(n)}{n^s} \quad \mbox{for} \ Re(s)>1.$$
Taking the derivative with respect to $s$, we get the following
$$- \frac{\zeta'(s)}{\zeta(s)^2} = - …
3
votes
On the Dirichlet series for $1/\zeta(s)$ for real $s$ and the zeros of zeta
Let $s_0>0$. The right statement is that the following are equivalent:
The sum $\sum_{n=1} \tfrac{\mu(n)}{n^s}$ converges for $s>s_0$.
$\zeta(s)$ has no zeroes with real part $>s_0$.
$1/\zeta(s)$ h …