Ha, interestingly enough, Milnor has a problem in his book on complex dynamics about this very problem; in the section dedicated to parabolic fixed points and Abel functions. Although he doesn't provide a proof, he asks that one prove that,
\(
f(z)= z - 1/z\\
\)
has a neutral fixed point at \( \infty \) and to prove the Julia set is the Real-line including infinity, and the parabolic basins are the upper half plane and the lower half plane. This would imply we can define two Abel functions \( \alpha_+, \alpha_- \) holomorphic on \( \Im(z) > 0 \) and \( \Im(z) < 0 \) (respectively), in which,
\(
\alpha(f(z)) = \alpha(z) +1\\
\)
As, \( \alpha(z) \sim z \) as \( z \to \infty \) in either half-plane we know that an inverse exists in a neighborhood of infinity, call it \( \alpha^{-1} \), and we can define,
\(
f^{\circ s}(z) = \alpha^{-1}(\alpha(z) + s)\\
\)
which is holomorphic for \( \Re(z) > R, \Im(z)>\delta \) (choosing \( \alpha_+ \) appropriately) and \( \Re(s) > R \) at least. We can expect that this is asymptotic to \( s \) as \( s \to \infty \) and therefore,
\(
\Gamma(1-s)f^{\circ s}(z) = \sum_{n=0}^\infty f^{\circ n+1}(z) \frac{(-1)^n}{n!(n+1-s)} + \int_1^\infty (\sum_{n=0}^\infty f^{\circ n+1}(z) \frac{(-x)^n}{n!})x^{-s}\,dx\\
\)
Converges for \( \Re(z) > R, \Im(z) >\delta \) and \( \Re(s) > R \). Which would imply the Ramanujan method does work in this case (though I'm not sure how to specify \( R \) or \( \delta \), but it shouldn't be hard).
Edit:
We could also modify the domains (which I guess is practical) to \( |z| > R \) and \( \Im(z) > \delta \) and make \( \Re(s) > R' \) (I simply chose to amalgamate \( R \) and \( R' \) into their maximum). That would work too, and is a tad more general. Recalling we are thinking of the point at infinity in the Riemann Sphere sense. I'm curious to wonder if we could pull this back to its maximal domain--I'm not sure what it would be. I'm guessing it may be something of the form \( \Im(z) > 0 \) and \( s \in \mathcal{S} \) where \( \mathcal{S} \) is either a half-plane (where the Ramanujan theorem would hold) or \( \mathbb{C} \) minus various branch-cuts including a halfplane (though Ramanujan's form will not converge for all of \( \mathcal{S} \), it necessarily can only converge in a half-plane by the nature of the Mellin Transform).
\(
f(z)= z - 1/z\\
\)
has a neutral fixed point at \( \infty \) and to prove the Julia set is the Real-line including infinity, and the parabolic basins are the upper half plane and the lower half plane. This would imply we can define two Abel functions \( \alpha_+, \alpha_- \) holomorphic on \( \Im(z) > 0 \) and \( \Im(z) < 0 \) (respectively), in which,
\(
\alpha(f(z)) = \alpha(z) +1\\
\)
As, \( \alpha(z) \sim z \) as \( z \to \infty \) in either half-plane we know that an inverse exists in a neighborhood of infinity, call it \( \alpha^{-1} \), and we can define,
\(
f^{\circ s}(z) = \alpha^{-1}(\alpha(z) + s)\\
\)
which is holomorphic for \( \Re(z) > R, \Im(z)>\delta \) (choosing \( \alpha_+ \) appropriately) and \( \Re(s) > R \) at least. We can expect that this is asymptotic to \( s \) as \( s \to \infty \) and therefore,
\(
\Gamma(1-s)f^{\circ s}(z) = \sum_{n=0}^\infty f^{\circ n+1}(z) \frac{(-1)^n}{n!(n+1-s)} + \int_1^\infty (\sum_{n=0}^\infty f^{\circ n+1}(z) \frac{(-x)^n}{n!})x^{-s}\,dx\\
\)
Converges for \( \Re(z) > R, \Im(z) >\delta \) and \( \Re(s) > R \). Which would imply the Ramanujan method does work in this case (though I'm not sure how to specify \( R \) or \( \delta \), but it shouldn't be hard).
Edit:
We could also modify the domains (which I guess is practical) to \( |z| > R \) and \( \Im(z) > \delta \) and make \( \Re(s) > R' \) (I simply chose to amalgamate \( R \) and \( R' \) into their maximum). That would work too, and is a tad more general. Recalling we are thinking of the point at infinity in the Riemann Sphere sense. I'm curious to wonder if we could pull this back to its maximal domain--I'm not sure what it would be. I'm guessing it may be something of the form \( \Im(z) > 0 \) and \( s \in \mathcal{S} \) where \( \mathcal{S} \) is either a half-plane (where the Ramanujan theorem would hold) or \( \mathbb{C} \) minus various branch-cuts including a halfplane (though Ramanujan's form will not converge for all of \( \mathcal{S} \), it necessarily can only converge in a half-plane by the nature of the Mellin Transform).