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February 7th, 2018, 11:52 PM   #1
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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*(n-1)*)/2=n^2-x 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?
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February 8th, 2018, 04:30 AM   #2
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Almost everything you've written is wrong. Especially the reference to integrals.

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February 11th, 2018, 11:27 PM   #3
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Quote:
Originally Posted by v8archie View Post
Almost everything you've written is wrong. Especially the reference to integrals
Sorry, I didn't type out the equation correctly; it should have read $\displaystyle \frac{1}{2}n^2+\frac{1}{2}n$.

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.
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Last edited by Sebastian Garth; February 11th, 2018 at 11:31 PM.
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February 12th, 2018, 03:35 AM   #4
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  1. The sum of the all integers is not $-\frac1{12}$;
  2. The partial sums of the integers form a function of a discrete variable whose integral is not $-\frac1{12}$;
  3. You have no justification for moving to a continuous variable, so you are going to struggle to explain any link.
The point about "resorting" to Analytic Continuation is that it is a rigorously proved technique with a firm logical basis.

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 03:38 AM.
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February 12th, 2018, 04:47 AM   #5
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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!
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February 12th, 2018, 01:55 PM   #6
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Originally Posted by Micrm@ss View Post
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!
Great work man! Your analysis of the general case really elucidates the crux of the matter quite well; if we take the quadratic roots for the $\displaystyle p=1$th power of $\displaystyle n$ and find the integral in between those two points we get $\displaystyle -\frac{1}{12}$ (thus making an extension from the integers to all of the real numbers).

Last edited by Sebastian Garth; February 12th, 2018 at 02:11 PM.
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February 14th, 2018, 09:19 PM   #7
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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.
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