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
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
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));
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)
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)