Lets say that I have the following function $$f(x,\gamma)=\frac{x^{\gamma-2}}{(\gamma-2)!}+\frac{x^{\gamma-4}}{(\gamma-4)!}+...$$ $$f(x,\gamma)=x^\gamma\sum_{k=1}^\infty \frac{x^{-2k}}{(\gamma-2k)!}$$ When gamma is an integer: $$f(x,1)=0$$ $$f(x,2)=1$$ $$f(x,3)=x$$ $$f(x,4)=1 + \frac{x^2}{2!} $$ $$f(x,n)=1+\frac{x^2}{2!}+ \frac{x^4}{4!}+...$$ where n is an even number. Ive tried the obvious extention $$f(x,\gamma)=x^\gamma\sum_{k=1}^\infty \frac{x^{-2k}}{\Gamma(\gamma-2k + 1)}$$ However for non integer values the series still diverges, so how can I extend this for all possible inputs.
-----EDIT-------
I was thinking about using Ramanujan summation to look for an extension of this series that is convergent.
$$f(x,\gamma)=-x^\gamma\sum_{k=1}^\infty \frac{B_k}{k!} \frac{d^k}{dn^k}\left[\frac{x^{-2n}}{\Gamma(\gamma-2n+1)}\right]_{n=0}$$
Ive tried expressing it in the following form $$f(x,\gamma)=-x^\gamma\sum_{k=1}^\infty \frac{B_k}{k!} \frac{d^k}{dn^k}\left[e^{-2n\ln x - \ln \Gamma(\gamma-2n+1)}\right]_{n=0}$$
$\ln \Gamma(\gamma-2n+1)$ is $\psi^{(-1)}(\gamma-2n+1)$
$$f(x,\gamma)=-x^\gamma\sum_{k=1}^\infty \frac{B_k}{k!} \frac{d^k}{dn^k}\left[e^{-2n\ln x - \psi^{(-1)}(\gamma-2n+1)}\right]_{n=0}$$
However Im struggling to find an expression for $\frac{d^k}{dn^k}\left[e^{-2n\ln x - \psi^{(-1)}(\gamma-2n+1)}\right]_{n=0}$. If anyone can find this expression that would be helpful. Many Thanks Josh.
