(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?