11/15/2007, 10:45 PM

Gottfried Wrote:Well in this case, for the result s being real only, alpha is a unique function of beta. If you take +beta you need one alpha, and if you take -beta, you need another alpha adapted from this; in this case, the two alphas are even equal (the graph looks symmetrical w.r.t. y-axis).

So no ambiguity is here, just use the formula of my paper, by which the alpha is determined by beta to get a real s.

Gottfried

I do not take +- beta, I take +- I . beta is still 0<beta<pi. In that case, IMAG(t) at beta=pi/2 would be -I, so graph will be on the right lower quadrant. However, at beta - pi/2, if we take +i, we get IMAG(t) = -i -> in the left lower quadrant, while at -I it will be where it is now-left upper quadrant.

The question is, can we even use beta = - pi/2 as that would imply h(e^-pi/2) diverges which it does not, and s in that case is not 4,81 = e^pi/2......... but 0,20........=i^i.

Best regards,

Ivars