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For analytic $f$, how can I represent the expression $f(z)\cdot\exp\left({s\,\log(z)}\right)$, i.e. $f(z)\cdot z^s$ in the form

$$\sum_{n}^\infty\left(\sum_{k}^\infty a_k s^k\right)z^n,$$

at least as a formal power series, where the index runs over the number needed?

I got to $$f(z)\cdot\exp\left({s\,\mathrm{log}(z)}\right)$$ $$=\left(\sum_{m}^\infty\frac{1}{m!}f^{(m)}(0)\,z^m\right) \left(\sum_{j=0}^\infty \frac{1}{j!} \left(s\,\log{(z)}\right)^j\right)$$ $$=\sum_{m}^\infty\sum_{j=0}^\infty \frac{1}{m!\,j!}f^{(m)}(0)\,s^j\cdot z^m\, \log{(z)}^j,$$

and I know

$$\log(z)=\sum_{l=1}^\infty (-1)^{l+1}\frac{1}{l}(z-1)^{l}.$$

I'm motivated by wanting to understand the Mellin transform (and the Laplace transform, for that matter) and here I approach this by looking at what it does to series components of a function

$$f(z)=\sum c_n z^n\mapsto \mathcal M(f)=\sum c_n^\mathcal{M} z^n$$

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Is $x$ a typo ? – Claude Leibovici Feb 25 '14 at 8:29
@ClaudeLeibovici: Yes, thx. – NikolajK Feb 25 '14 at 8:46

Unless $s$ is an integer and non-negative, your function has a singularity at 0, and no such series is possible.

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The particular case $f(z) = 1$ asks us to expand $z^s$ as a power series centered at $z=0$. – GEdgar Feb 25 '14 at 18:10
@GEdgar, that is the point: Unless $s$ is a non-negative integer, $z^s$ has no power series expansion around $z = 0$. – vonbrand Feb 25 '14 at 18:12

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