Skip to content

Commit

Permalink
MPNT
Browse files Browse the repository at this point in the history
  • Loading branch information
AlexKontorovich committed Jan 28, 2025
1 parent 96393ff commit f6c75bc
Showing 1 changed file with 2 additions and 3 deletions.
5 changes: 2 additions & 3 deletions PrimeNumberTheoremAnd/MediumPNT.lean
Original file line number Diff line number Diff line change
Expand Up @@ -59,8 +59,7 @@ noncomputable def SmoothedChebyshev (SmoothingF : ℝ → ℝ) (ε : ℝ) (X :
VerticalIntegral' (SmoothedChebyshevIntegrand SmoothingF ε X) 2

/-%%
\begin{lemma}[integrable_x_mul_Smooth1]\label{integrable_x_mul_Smooth1}
\lean{integrable_x_mul_Smooth1}\leanok
\begin{lemma}[integrable_x_mul_Smooth1]\label{integrable_x_mul_Smooth1}\lean{integrable_x_mul_Smooth1}\leanok
Fix a nonnegative, continuously differentiable function $F$ on $\mathbb{R}$
with support in $[1/2,2]$, and total mass one, $\int_{(0,\infty)} F(x)/x dx = 1$. Then for any $\epsilon>0$, the function
$x \mapsto x \cdot \widetilde{1_{\epsilon}}(x)$ is integrable on $(0,\infty)$.
Expand Down Expand Up @@ -137,7 +136,7 @@ lemma integrable_x_mul_Smooth1 {SmoothingF : ℝ → ℝ} (diffSmoothingF : Cont
\uses{Smooth1Properties_above}
We have
from Lemma \ref{Smooth1Properties_above}
that $\widetilde{1_{\epsilon}}(x) = 0$ for all $x \geq 1+c\epsilon$.
that $\widetilde{1_{\epsilon}}(x) = 0$ for all $x \leq 1+c\epsilon$.
So the claimed function is integrable on $(0,\infty)$.
\end{proof}
%%-/
Expand Down

0 comments on commit f6c75bc

Please sign in to comment.