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Numerical calculations on the zeroes of the Riemann zeta function have reached a very high degree of refinement and sophistication and I think that the first $10^{20}$ (with positive imaginary part) o …

Consider $\rho$ be a $C^\infty$ function supported in $(-1/8,1/8)$ with integral 1 and set $ w=\chi_I\ast \rho, $ so that, for $n\ge 1$, we have $$ w^{(n)}(x)=\bigl(\chi_I\ast \rho^{(n)}\bigr)(x)= \bi …

Let me assume that $a=-\infty, b=+\infty, x_0=0$ and $f$ smooth and compactly supported near 0. Then after a suitable change of variable, you get that $ I(\lambda)=\int g(t) e^{i\lambda t^3/3} dt, $ …

The function $\mathbb R\ni t\mapsto\zeta(\frac12+it)$ is analytic and smaller in absolute value than $C(1+\vert t\vert)^{1/6}$ (the $1/6$ may be replaced by $9/56$ and even by a slightly smaller numb …

The trivial zeroes of the Riemann Zeta function are located on $-2\mathbb N^*$ and they are simple. It is not difficult to see that, but the proof I have in mind is using the fact that $\xi(-2k)=\xi(1 …

The function $\text{Li}$ (logarithmic integral) is defined for $x>0$ by $$ \text{Li}(x)=\int_2^{x}\frac{dt}{\ln t}. $$ The prime number theorem, proven by Hadamard and de la Vallée-Poussin in 1896 ass …