I found this interesting stuff with Riemann's ZF

[rpg16]

RZF --> Zeta Function
While working with integrals of fractional parts, I came out with
this... which I haven't quite seen before (on the internet, or books
that I own)...

Namely, the equation
RZF(n) = 1/(n-1) - integral _1 ^oo 1/r^(1/n) dr
0 < n < 1, n belongs to N

I myself am doubtful about the correctness of this equation... But I
have rechecked the proof many times... and still dont find any
errors...

Do you think its correct???

[Tonico]

In what book/paper did you find this? Let us know, please.

Tonio

[Jon Slaughter]

No, it looks totally wrong. First, 0 < n < 1 and n in N?

Second, for positive n > 1, RZF(n) >= 1 yet your expression is always < 1.

So unless I'm missing something or you made a mistake in notation, I would
say it's just nonsense.

If you mean 0 < n < 1 then this still does not make much sense but could
potentially be fixed if there is a notational error.

What I can say is that numerically the formula is wrong.

Zeta(1/2) = -1.46

while your RZF(1/2) = -3 (exactly)

[rpg16]

yea... i think you are right...
for the typo errors... i am sorry i didnt notice i wrote these two at
the same time, 0 < n < 1 and n belongs to N... shit!!!

i actually meant n != 1 n > 0...

in fact i also found some errors... I corrected those and am getting
the following formula...

RZF(n) = n/(n-1) - lim a -> oo ( (1/a) integral _1 ^oo (frac(a/x^(1/
n))) dx )
where, n > 1, n is positive real and frac(x) = x - floor(x)

Now, i think its a correct formulaton...

[Raymond Manzoni]

rpg16 a �crit :

> On Nov 22, 2:52 am, "Jon Slaughter" <Jon_Slaugh...@Hotmail.com> wrote:
>> rpg16 wrote:
>>> RZF --> Zeta Function
>>> While working with integrals of fractional parts, I came out with
>>> this... which I haven't quite seen before (on the internet, or books
>>> that I own)...
>>> Namely, the equation

(snip)

>
> RZF(n) = n/(n-1) - lim a -> oo ( (1/a) integral _1 ^oo (frac(a/x^(1/
> n))) dx )
> where, n > 1, n is positive real and frac(x) = x - floor(x)
>
> Now, i think its a correct formulaton...
>

The "Perron's formula" applied to Riemann zeta gives :

zeta(s)= s int_1^oo {x}/x^(s+1) dx
( with {x} the fractional part of x )

See here : <http://en.wikipedia.org/wiki/Perron%27s_formula>

A proof of Perron's formula may be found in Apostol's book :
<http://books.google.com/books?id=Il64dZELHEIC&pg=PA243>
with application to zeta as an exercise...

Hoping it helped,
Raymond

[rpg16]

On Nov 22, 6:54 pm, Raymond Manzoni <raym...@free.fr> wrote:
> rpg16 a écrit :

Thanks Raymond :)

[Raymond Manzoni]

Raymond Manzoni a �crit :

> rpg16 a �crit :
>> On Nov 22, 2:52 am, "Jon Slaughter" <Jon_Slaugh...@Hotmail.com> wrote:
>>> rpg16 wrote:
>>>> RZF --> Zeta Function
>>>> While working with integrals of fractional parts, I came out with
>>>> this... which I haven't quite seen before (on the internet, or books
>>>> that I own)...
>>>> Namely, the equation
> (snip)
>>
>> RZF(n) = n/(n-1) - lim a -> oo ( (1/a) integral _1 ^oo (frac(a/x^(1/
>> n))) dx )
>> where, n > 1, n is positive real and frac(x) = x - floor(x)
>>
>> Now, i think its a correct formulaton...
>>
>
> The "Perron's formula" applied to Riemann zeta gives :
>
> zeta(s)= s int_1^oo {x}/x^(s+1) dx
> ( with {x} the fractional part of x )

Oops...
zeta(s)= s int_1^oo (x-{x})/x^(s+1) dx
= s int_1^oo 1/x^s dx - s int_1^oo {x}/x^(s+1) dx
= s/(s-1) - s int_1^oo {x}/x^(s+1) dx
of course...

Note that the improper integral converges when Re(s)>0 (instead of
Re(s)>1 at the start) and is analytic in the Re(s)>0 domain while
s/(s-1) provides the simple pole at s=1 of zeta(s).

Merry Christmas to all!
Raymond