Research proof index

A first-order Euler–Maclaurin formula. For $0\le a\le b$ and a differentiable function $f$ whose derivative is continuous on $[a,b]$, the sum over integers $\lfloor a\rfloor<k\le\lfloor b\rfloor$ equals $f(a)B_1(a)-f(b)B_1(b)+\int_a^b f(t)\,dt+\int_a^b f'(t)B_1(t)\,dt,$ where $B_1(t)=t-\lfloor t\rfloor-\tfrac12$.

Truncated contour shift through a simple pole. The left vertical side stays as the symmetric truncation; only the right vertical side is required to be Bochner integrable.

Set-integral form of tendsto_truncated_vertical_shift_with_simple_pole, normalized as (2π)⁻¹ ∫ f(σ + it) dt.

For two $n$-times continuously differentiable functions $f$ and $g$, the $n$th iterated derivative distributes over subtraction: $(f-g)^{(n)}=f^{(n)}-g^{(n)}.$

Let $f$ be a twice differentiable complex-valued function whose first two derivatives are integrable, and let $g$ be a twice differentiable compactly supported cutoff that equals $1$ on $[-1,1]$ and vanishes outside $(-2,2)$. Then $g(x/R)f(x)$ converges to $f$ as $R\to\infty$ in the project norm $\|h\|=\int_{\mathbb R}|h(x)|\,dx+\frac{1}{4\pi^2}\int_{\mathbb R}|h''(x)|\,dx.$

The Chebyshev function is asymptotic to the identity: $\psi(x)\sim x$ as $x\to\infty$. Equivalently, $\psi(x)/x\to1$.