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Nontrivial zero's of the zeta function:

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Gerry

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Mar 3, 2011, 7:08:02 AM3/3/11
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Hi All,

just another integral for the zeta function:

zeta(n)=-I*I^n*(2*Pi)^(n-1)/(n-1)*Int_t=0^1(log(1-1/t)^(1-n))

defining the log integral as follows :

L(n)=Int_t=0^1(log(1-1/t)^n


I found the nice identity :

gamma(1+s*I)*zeta( s*I)*L(-s*I)
exp(s*Pi) = ---------------------------------
gamma(1-s*I)*zeta(-s*I)*L( s*I)

So all nontrivial zeta zero's must be zero's of L(n) and L(1-n).

Can anyone give a reference to this integral and its zero's?

Regards

Gerry

Gerry

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Mar 4, 2011, 3:41:38 AM3/4/11
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Forgot the closing bracket for L(n) so again the log integral is :

L(n) = Int_t=0^1(log(1-1/t)^n)

And the identity

gamma(1+s*I)*zeta( s*I)*L(-s*I)
exp(s*Pi) = ---------------------------------
gamma(1-s*I)*zeta(-s*I)*L( s*I)

gives a second reflection formula relating for instance zeta(1/2) to
zeta(-1/2)

1/2*I*sqrt(Pi)*zeta(1/2)*L(-1/2) = sqrt(Pi)*zeta(-1/2)*L(1/2)

giving numerically :

0.476874264894483676*(-1-I)

This is close to the Plouffe number :

1/log(Porter)*Golomb^2/zeta(1/2)^2
=
0.476874268970744

Gerry

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Mar 7, 2011, 2:53:23 AM3/7/11
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>  0.476874268970744- Hide quoted text -
>
> - Show quoted text -

So the identity gives a nice formula for the Integer Sequence:

A079484 a(n)=(2n-1)!!*(2n+1)!!, where the double factorial is
A000165.
1, 3, 45, 1575, 99225, 9823275,

a(s) = 2^(1+2*s)*exp(s*Pi*I)*gamma(3/2+s)/gamma(1/2-s);
a(s) = -I*2^(1+2*s)*zeta(-1/2-s)*L(1/2+s)/(zeta(s+1/2)*L(-1/2-s));

Gerry

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Mar 9, 2011, 5:11:35 AM3/9/11
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> a(s) = -I*2^(1+2*s)*zeta(-1/2-s)*L(1/2+s)/(zeta(s+1/2)*L(-1/2-s));- Hide quoted text -

>
> - Show quoted text -

So using the log integral :

L(n)= intnum(t=0,1,log(1-1/t)^n)

I get that the following function B(s) :

B(s)=-I*s*gamma(s)*exp(I*s*Pi)*L(1-s)/(Pi*(1-s))

for even s gives the bernoulli numbers and
for uneven it does not evaluate to zero!

for fractionals I found that :

zeta( 1/2) = 2*B(1/2)/(-I-1)
zeta(-1/2) = B(-1/2)/(2*Pi*(I-1)

Gerry

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Mar 17, 2011, 5:01:29 AM3/17/11
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>  zeta(-1/2) =   B(-1/2)/(2*Pi*(I-1)- Hide quoted text -

>
> - Show quoted text -

Here's another nice identity :

zeta(1/2-n) (2*n-2)!
------------------- = -------------------------
L(1/2-n)*((-1)^n-I) (n-1)!*2^(2*n-1)*Pi^(1/2)

with : L(n)= int_t=0^1(log(1-1/t)^n)

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