I got a bit twisted.

integral(x=1+e,2, of sigma(n=1,infinity, of floor(n^x)))

i.e. the numbers:

1^(1+e) through to 1^2

+

2^(1+e) through to 2^2

+

...

The point being that by Bertands, every prime is represented by the

integral, (and the sum is infinite, but I am not currently concerned with

this).

The second statement is the identity that the sigma and integral are

interchangeable:

sigma(n=1,infinity, of integral(x=1+e,2, of floor(n^x)))

Jon Perry

perry@...
http://www.users.globalnet.co.uk/~perry/maths
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-----Original Message-----

From: Phil Carmody [mailto:

thefatphil@...]

Sent: 02 May 2002 20:05

To: primenumbers

Subject: RE: [PrimeNumbers] Number crunch

--- Jon Perry <

perry@...> wrote:

> I came up with this:

>

> integral(x=1,infinity of sigma(n=1,2 of x^n)) contains all the

> primes, and

> further more:

>

> integral(x=1,infinity of sigma(n=1,2 of x^n))==sigma(n=1,2 of

> integral(x=1,infinity of x^n)

I'm fighting your notation somewhat. I assume sigma is summation?

Is sigma(n=1,2 of x^n) just x+x^2 = x(x+1) ?

The integral from 1..+oo of x(x+1) diverges.

What is "contains" supposed to mean in the context "+oo contains all

the primes"?

So I'm fighting your terminology as well as your notation.

Phil

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