Research proof index

Consecutive prime gaps are small relative to the primes themselves. If $p_n$ is the $n$th prime, then $p_{n+1}-p_n=o(p_n).$ Formally, there is a sequence $c_n\to0$ such that $p_{n+1}-p_n=c_np_n$ for every $n$.

For every $\varepsilon>0$, every sufficiently large real number $x$ has a prime $p$ in the short multiplicative interval $x<p<(1+\varepsilon)x$.

Let $R$ be the axis-parallel rectangle with opposite corners $z,w\in\mathbb C$. If a point $p$ lies strictly to the left, right, below, or above both corners, then $p\notin R$; equivalently, the rectangle and the singleton $\{p\}$ are disjoint.

An argument-principle identity on an axis-parallel rectangle. If $f$ and its logarithmic derivative $f'/f$ are meromorphic on the rectangle, every point has finite meromorphic order, and no zero or pole of $f$ lies on the boundary, then $\frac{1}{2\pi i}\oint_{\partial R}\frac{f'(z)}{f(z)}\,dz=\sum_{p\in R}\operatorname{ord}_p(f).$ Thus the normalized boundary integral counts zeros positively and poles negatively, with multiplicity.

The Residue Theorem on a rectangle for functions with simple poles.

A vertical splitting identity for rectangular contour integrals. Under integrability of $f$ along the two horizontal sides, splitting the rectangle with real coordinates from $x_0$ to $x_1$ at $a$ gives $\int_{\partial R(x_0,x_1;y_0,y_1)}f=\int_{\partial R(x_0,a;y_0,y_1)}f+\int_{\partial R(a,x_1;y_0,y_1)}f.$ The two integrals over the shared vertical side cancel.