My Math Forum  

Go Back   My Math Forum > College Math Forum > Number Theory

Number Theory Number Theory Math Forum


Reply
 
LinkBack Thread Tools Display Modes
August 24th, 2019, 12:01 PM   #1
Newbie
 
Joined: Aug 2019
From: spain

Posts: 2
Thanks: 1

Generalization of Riemann-weil formula for arithmetic functions

$\displaystyle \sum_{n=1}^{\infty} \frac{\mu(n)}{\sqrt{n}}g(\log n)=\sum_{\gamma}\frac{h( \gamma)}{\zeta '( \rho )}+\sum_{n=1}^{\infty} \frac{1}{\zeta ' (-2n)} \int_{-\infty}^{\infty}dxg(x)e^{-(2n+1/2)x} </math>$

$\displaystyle \sum_{n=1}^{\infty} \frac{\lambda(n)}{\sqrt{n}}g(\log n) = \sum_{\gamma}\frac{h( \gamma)\zeta(2 \rho )}{\zeta '( \rho)}+ \frac{1}{\zeta (1/2)}\int_{-\infty}^{\infty}dx g(x)$

$\displaystyle \sum_{n=1}^{\infty} \frac{\varphi (n)}{\sqrt{n}}g(\log n)= \frac{6}{\pi ^2} \int_{-\infty}^{\infty}dx g(x)e^{3x/2}+ \sum_{\gamma}\frac{h( \gamma)\zeta(\rho -1 )}{\zeta '( \rho)}+ \frac{1}{2}\sum_{n=1}^{\infty} \frac{\zeta (-2n-1)}{\zeta ' (-2n)} \int_{-\infty}^{\infty}dx g(x)e^{-x(2n+1/2)} $

think this could be interesting is based on a idea by myself could someone help me to get a Ph D thesis based on this ?
Rmath is offline  
 
Reply

  My Math Forum > College Math Forum > Number Theory

Tags
arithmetic, formula, functions, generalization, riemannweil



Thread Tools
Display Modes


Similar Threads
Thread Thread Starter Forum Replies Last Post
Riemann-von Mangolt formula 2 raul21 Number Theory 6 May 24th, 2014 09:50 AM
Riemann-Von Mangolt formula raul21 Number Theory 2 May 23rd, 2014 10:04 AM
formula of zeta riemann mhss12345 Number Theory 1 November 2nd, 2012 11:59 AM
How to compute Weil Pairing? miller Abstract Algebra 0 August 20th, 2010 12:11 AM
Solving Euler's Product Formula (Riemann-Zeta) Jalaska13 Number Theory 0 May 30th, 2010 03:30 PM





Copyright © 2019 My Math Forum. All rights reserved.