Re: integral zeta

David C. Ullrich wrote:



On Thu, 16 Aug 2007 17:51:34 EDT, tommy1729
<tommy1729@xxxxxxxxx>
wrote:

David C.Ullrich
wrote:

On Thu, 16 Aug 2007 07:16:32 EDT, tommy1729
<tommy1729@xxxxxxxxx>
wrote:

tommy1729 a écrit :
Denis Feldmann wrote:
tommy1729 a écrit :
where do the zero's lie of integral zeta ?
Not to mention the problem of Riemann
surfaces,
what do you take as
integration constant?


Always getinf your questions out of your
hat,
do
you?

I though that he got them from somewhere
else.......


haha very funny guys

*rolleyes*

how about actually answering a question and
doing
or explaining math , apart from just making fun
of
the OP ??

tommy1729


But this was an answer : what about that
constant?
(and Riemann surfaces)

a primitive integral ...

If you'd think about what you're saying instead
of trying so hard to be a smartass you'd
understand
when people explain why the question makes no
sense.

we understood that by "integral zeta" you meant a
primitive integral.

yet at the end of this post you seem to have
forgotten it hmm ?


There are several problems with
that. First, the fact that zeta has a pole with
non-vanishing residue at 1 means that it does not
_have_ a primitive integral in various sets;
the integral is "multivalued", just like the
integral of 1/z. (That's a simple-minded way
of expressing what Denis is getting at when he
mentions Riemann surfaces).


you dont completely understand what i mean


but i take almost full responsability for it

i will specify below *



Second, even if we ignore that problem, the
question
still makes no sense, because a given function
has many different integrals. What are the zeroes
of integral (x) on the line? Well, integral(x)
could be x^2 + 1, in which case it has no zeroes.
Or it could be x^2 - 4, which has two zeroes. Etc.

like i said you seem to have forgotten here what a
primitive integral is


the primitive integral of x is x^2 / 2 and only
that.

Well, then I didn't understand what you meant by
"primitive integral".

Unfortunately you need to give a _definition_ of the
concept instead of just an example. What does
"primitive integral" mean, exactly?

not x^2 - 4 or similar

so primitive integral of zeta(z) = 0*





************************

David C. Ullrich



*the primitive integral of zeta the way i meant it :

Ah. So you expected us to know what you _meant_,
even though you didn't _say_ what you meant, and
then you complain when people try to explain
why what you actually _said_ is meaningless.

Think about that next time you feel like
shooting your mouth off about how stupid
people are.

z + sum n=2 -> infinity [elog(n)^-1]*n^(-z)

I imagine that "elog" just means the logarithm?
Inventing your own private notation is not the
best way to communicate.

Just curious, still: What _definition_ of
"primitive integral" allows us to conclude
that _that_ is the primitive integral of zeta?

and the question is where do the zero's lie of


primitiveintegralzeta(z)=0


you might worry about continuation and riemann
surfaces ,

but that is not so important as it seems ...


as an example of that


the zero's of sum n=1 -> infinity n^(-z)

lie in the critical line

so it also converges there to zero.

even if its neighbourhood is not analiticly
continued;

not even finite !

and not caring about riemann surfaces or similar ...

Your comments here are just dripping with ignorance.

You cannot talk about the zeroes of that sum except
for Re(z) > 1 (where there are none) because
elsewhere
the sum simply does not converge.

You _can_ talk about the zeroes of the zeta function
elsewhere in the plane. The _reason_ you can talk
about that is that the function defined by that
sum _has_ a _unique_ analytic continuation to the
plane
minus {0}. That is not true for the "primitive
integral" of zeta, as defined by the sum above.

This is not an example of why we don't need to
worry about things, it's an illustration of
exactly why we _do_.

that might ammaze you , since it is not often found
in the books ...

No, the fact that the sum defining zeta(z) for Re(z)
1
converges elsewhere is indeed "not often found in
books".
There'a a reason for that - it's not true.

or it might ammaze others ...


but if you consider that any function must have all
of its zero's or at least an unempty subset of its
zero's

then we still have these zero's despite no analytic
continuation or even continuation at all ...

tommy1729


************************

David C. Ullrich

you are terribly wrong David

thats the most polite way i can say it ...

terribly wrong


apparantly you dont even know the integral of x

it is x^2 / 2 + C

and a primitive integral means C = 0

this is very basic math ...

( and in the books )

and you are even more wrong about zeta and your other comments

( outside of the books )

tommy1729

**********************************************************
" we should be gratefull and feel honored that David C. Ullrich is here on sci.math " David C. Ullrich 's lawyer in a threath that was a super ode to David ...
.

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