(06/05/2014, 01:17 AM)JmsNxn Wrote:

(06/05/2014, 12:45 AM)mike3 Wrote: I decided to test your integral plugging the \( a_n \) in from the Kneser super-logarithm as calculated via other methods. If it works you should get that super-logarithm back out.

However, I wasn't able to quite get it to converge on a numerical test for \( z = 0.1 \). The infinite-sum-defined function exhibits what is probably a very slow and very rapidly (tetrationally, I bet! )-increasing period oscillation, and so I'm not sure how well the negative-power factor in the integrand damps it out, i.e. if it damps it out enough to converge.

I want to point out that the Kneser super-logarithm has singularities at the fixed points of the logarithm \( L \) and \( \bar{L} \). Therefore, on the boundary of the half-plane \( \Re(z) > \Re(L) \), there are two singularities. These are logarithmic singularities and so the function is exponentially unbounded on that half-plane. If you need a tight (and not just asymptotic) bound, try a half-plane \( \Re(z) > R > \Re(L) \) for some \( R \). Then there are no singularities on the boundary and the function is exponentially-bounded on the whole half-plane.

Yep that should fix it, makes a lot more sense.

And as to tommy's qblunt statement. lol, you're wrong.


\( \int_{\sigma - i\infty}^{\sigma + i \infty} \G(z) f(R-z) w^{-z} \,dz = 2 \pi i\sum_{n=0}^\infty f(R+n) \frac{(-w)^n}{n!} \)

and as well:

\( \int_{\sigma - i\infty}^{\sigma + i \infty} \G(z) f(e^{R-z}) w^{-z} \,dz = 2 \pi i \sum_{n=0}^\infty f(e^{R+n}) \frac{(-w)^n}{n!} = 2 \pi i \sum_{n=0}^\infty (f(R+n)+1) \frac{(-w)^n}{n!}= p(w) \)

Similarly:
\( \int_{\sigma - i\infty}^{\sigma + i \infty} \G(z) (f(R-z)+1) w^{-z} \,dz = p(w) \)
where \( R \) is an int


Now are you going to doubt the one to one nature of the fourier transform?

(06/05/2014, 12:25 PM)tommy1729 Wrote: It seems you are once again trying to use Ramanujan's master theorem.

As for the remark 3) :

Let N,M,i,j be integers.

we have slog(z+ 2 pi j i) = slog(z).

if tet (a_i) = b_i and tet ' (a_i) = 0

then slog(z) is not analytic at b_i.

it follows slog(z) is also not analytic at b_i + N 2pi i.

also tet ' (a_i + M) = 0

tet(a_i + M) = exp^[M](b_i) which is chaotic most of the time !!

SO your halfplane cannot contain all those points nor the fixpoints L.
Also none of the complex conjugates of those !!!

hence your half-plane is parallel to Re(z) > Q if it exists at all.

That was the reason for my comment about 3).

Maybe a link to H trapmann's paper is usefull.

I hope I have not offended you and you understand my viewpoint (now).

regards

tommy1729