11/09/2007, 08:30 AM

bo198214 Wrote:Quote:So the problem is now turned to the properties of continuous application of logarithms with base e^pi/2 on -1?Yes, yes

Which in turn is calculated in each step via normal logarithms by formula:

log b (z) = (ln (z) + 2pik)/ ln (b)?

So now we need analytic expressions for that infinite taking of logarithms with base e^pi/2 of (-1) as a function of k.

Any ideas? I am searching as well.

if we put e^(pi/2) = e^((sqrt(pi)/sqrt(2))^(sgrt(pi)/sqrt(2) ) then

we can use Gamma(1/2) = sqrt(pi), so

e^(pi/2) = e^(Gamma^2 (1/2)/2)

and ln (e^pi/2) = Gamma^2 (1/2) /2 +- 2pik

next ln would be 2 lnGamma(1/2) - ln 2

next ln2+ ln*lnGamma(1/2) - lnln2 etc. ln2 and lnln2 will cancel out, so

in the end we will have lnln............ln (Gamma(1/2) when n-> infinity.