entire function close to sexp ??

[JmsNxn]

Im a bit puzzled by the " -s ".

e^[3] = e^(e^e)

But e^[-3] = ln(ln(ln(1))) = ln(ln(0)) = ??

So I do not know what the "-s" means.

However your last formula seems to end that problem so it seems

\( \frac{1}{(^s e)} \approx \psi(s) = \frac{1}{\Gamma(-s)} (\sum_{n=0}^\infty \frac{(-1)^n}{n!(^n e)(n-s)} + \int_1^\infty e^{-\lambda w}\vartheta(-w)w^{-s-1}\,dw) \)

Is the kind of formula we look for.

However, Im more concerned about wheither or not this is analytic !?
Convergeance is an intresting subject, but if it is analytic somewhere then by analytic continuation we can extend it.

But why should that be analytic ? Assuming it is analytic then its derivative must equal the term by term derivative. Can you show it - the term by term derivative formulation - converges ?

Actually I bet that this function is not analytic and I even bet sheldon agrees with me.

Intresting though. But I suspect you have mentioned this function before, not ?

regards

tommy1729

very easy to see this is analytic. No idea what you're talking about. lol. Mellin transform of an entire function is analytic. Multiply it by the inverse gamma which is entire and voila, analytic and entire.
I'd explain why but it's literally an excercise in analysis 101. These are all uniform limits of analytic functions, so it's neessarily analytic.