# Tag Info

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### Is the Riemann zeta function surjective?

The Riemann zeta function is surjective. First, $\zeta(1/z)$ is holomorphic in the punctured disk $0<|z|<1$. Looking at $z=(1/2+it)^{-1}$ with $t\to\infty$ reveals that $\zeta(1/z)$ has an ...

### A new way of approaching the pole of the Riemann zeta function - and a new conjectured formula

The paper entitled Euler constant as a renormalized value of Riemann zeta function at its pole by Andrei Vieru contains a derivation of the first formula in the OP (Benoȋt Cloitre's formula), and ...

### Could the Riemann zeta function be a solution for a known differential equation?

The Riemann zeta function is "hypertranscendental" in the sense shown HERE It is not the solution $y(x)$ of a differential equation of the form $$F(x,y,y',y'',\dots,y^{(n)})=0$$ where $F$ is a ...
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### values of $\zeta$ function are linearly independent?

People have been exerting steady effort to prove/disprove irrationality of the Riemann zeta values in your list. Of course, $\zeta(3)$ is known to be irrational due to Roger Apery. Such investigations,...

### Could the Riemann zeta function be a solution for a known differential equation?

The fact that $\zeta$ satisfy no algebraic differential equation is due to its famous relation with the Gamma function which was proved by Hölder not to satisfy such an equation. Detailed answer can ...

### Is there a closed form for $\int_0^\infty\frac{\tanh^3(x)}{x^2}dx$?

Rewrite the integrand and apply Taylor expansion to $\frac1{(1+e^{-2x})^3}$ so that $$\frac{\tanh^3x}{x^2}=\sum_{j\geq0}(-1)^j\binom{j+2}2 \frac{(1-e^{-2x})^3}{x^2}\binom{j+2}2e^{-2jx}.$$ Integrate ...
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### $P(s)=1-\sqrt{\frac{2}{\zeta(s)}-\sqrt{\frac{2}{\zeta(2s)}-\sqrt{\frac{2}{\zeta(4s)}-\sqrt{\frac{2}{\zeta(8s)}-...}}}}$

This has been answered in the comments by Lucia. The identity $$P(s)=1-\sqrt{\frac{2}{\zeta(s)}-\sqrt{\frac{2}{\zeta(2s)}-\sqrt{\frac{2}{\zeta(4s)}-\sqrt{\frac{2}{\zeta(8s)}-\cdots}}}}$$ is false. By ...

### Algebraic independence of shifts of the Riemann zeta function

Hmm, it was more difficult than I expected to leverage universality to establish the claim. But one can proceed by probabilistic reasoning instead, basically exploiting the phase transition in the ...