
Number Theory Number Theory Math Forum 
 LinkBack  Thread Tools  Display Modes 
February 7th, 2018, 10:52 PM  #1 
Member Joined: Jul 2010 Posts: 64 Thanks: 0  Integral of the sum of all integers: 1/12
The sum of all integers over the complex plane by analytic continuation is typically given as the Riemann zeta function evaluated at z({1,0})=1/12. But the partial sum at index 'n'' of the series is (n*(n1)*)/2=n^2x and looking at the definite integral S there we find that it's precisely z(1). Perhaps we should be calling z(1) an integral instead of a sum?

February 8th, 2018, 03:30 AM  #2 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,268 Thanks: 2434 Math Focus: Mainly analysis and algebra 
Almost everything you've written is wrong. Especially the reference to integrals. 
February 11th, 2018, 10:27 PM  #3  
Member Joined: Jul 2010 Posts: 64 Thanks: 0  Quote:
Anyhow if you look at the definite integral (see plot at bottom) you find that it's actually $\displaystyle zeta(1) = \frac{1}{12}$. And so instead of resorting to the peculiarities of analytic continuation required by the Reimman zeta function, we can simply point to the generating polynomial for the sum of integers up to $\displaystyle n$ and examine it's integral to gain a little better insight into this odd discovery of Euler and Ramanujan. Last edited by Sebastian Garth; February 11th, 2018 at 10:31 PM.  
February 12th, 2018, 02:35 AM  #4 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,268 Thanks: 2434 Math Focus: Mainly analysis and algebra 
I expect that their is a link between the zeta function and the result. And the result probably expresses some truth about the sum of the positive integers, but that doesn't give us equality. Last edited by v8archie; February 12th, 2018 at 02:38 AM. 
February 12th, 2018, 03:47 AM  #5 
Senior Member Joined: Oct 2009 Posts: 364 Thanks: 120 
In general, the sum of the $p$th power of the first few integers is $$\sum_{k=0}^N k^p = \frac{B_{p+1}(N+1)  B_{p+1}(0)}{p+1}$$ where $B_{p+1}$ is the $p+1$th Bernouilli polynomial. Your idea is now to see $N$ as continuous and integrate this. Notice that $$\int_a^0 B_{p+1}(x+1)dx = \frac{B_{p+2}(1)  B_{p+2}(a+1)}{p+2}$$ Thus the integral of the above is $$\frac{B_{p+2}(1)  B_{p+2}(a+1) + aB_{p+1}(0)(p+2)}{(p+2)(p+1)}$$ Keeping in mind that for $p\geq 0$, we have $B_{p+2}(1) = B_{p+2}(0)$, evaluating the above in $1$ yields $$\frac{B_{p+1}(0)}{p+1}$$ which is exactly $\zeta(p)$ for $p>0$. Very well noticed link, OP! 
February 12th, 2018, 12:55 PM  #6  
Member Joined: Jul 2010 Posts: 64 Thanks: 0  Quote:
Last edited by Sebastian Garth; February 12th, 2018 at 01:11 PM.  
February 14th, 2018, 08:19 PM  #7 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,268 Thanks: 2434 Math Focus: Mainly analysis and algebra 
Except that there is still no justification for why we should consider $N$ to be continuous. It stops the sum from being the sum of the $p$th power of the integers and the summation stops making any sense.


Tags 
1 or 12, integers, integral, sum 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Integers  yo79  Math Events  4  March 17th, 2013 12:50 AM 
Sum of Odd Integers  sulonski  Number Theory  11  August 31st, 2012 03:49 PM 
Integers mod 4  DanielThrice  Abstract Algebra  1  November 7th, 2010 01:17 AM 
integers  Julie13  Calculus  0  August 18th, 2010 10:08 PM 
Integers:  cafegurl  Elementary Math  3  January 7th, 2009 07:21 AM 