(05/11/2014, 04:30 PM)JmsNxn Wrote:(05/11/2014, 04:26 PM)tommy1729 Wrote: I totally agree. Although I added a minus sign to it, which I assume was a typo.
So lets think about \( \vartheta(-x) \).
Since the Taylor coefficients of \( \vartheta(-x) \) decay EXTREMELY FAST , I consider this as a function that is well approximated by a polynomial for a long time.
( many remainder theorems for Taylor series imply this )
This means the main behaviour of this \( \vartheta(-x) \) is like \( (-1)^n a_n x^n \) where n increases slowly with x.
This implies that \( \vartheta(-x) \) is not bounded by a polynomial and also that \( \vartheta(-x) \) = 0 infinitely often.
Therefore the integral diverges.
Even if we consider taking the limit of x going to +oo as the limit of the sequence x_i with \( \vartheta(-x_i) \) = 0.
regards
tommy1729
yes yes, I'm quite aware it diverges. That's only another trick we need to come up with to handle that. I have a few but I need to look deeper into the laplace transform.
I assumed you were aware of it. But some readers might not have been convinced.
With that in the back of my mind, I felt the neccessity to reply.
Its pretty hard to combine the properties of convergeance and the functional equation with fractional calculus and integral transforms ... or so it seems.
Maybe a bit of topic but finding an approximation to \( \exp^{[1/2]}(x) \) in terms of an integral transform seems to be closer to a solution due to the recent work and talk of myself and sheldon.
\( \exp^{[1/2]}(x) = \theta(x) \int_{\0}^{\infty} \((z (\2sinh^{[-1]}(z) - 1) )!)^{-1} x^{z}\,dz \)
where \( \theta(x) \) is bounded by a constant above and a constant below , and x > 27.
(And the factorial is computed with the gamma function ofcourse )
with approximation I mean that they have the same " growth rate ".
regards
tommy1729
