I have run into a problem while computing the fractional derivatives of order $\alpha$ for the Riemann zeta function. My Theorem states
Let $s\in\mathbb{C}$, $\mathfrak{Re}(s)>1$, then the fractional derivative of order $\alpha$ of the Riemann zeta function is given by $$\zeta^{(\alpha)}(s)\equiv\sum_{k=2}^{\infty}\frac{e^{i \pi \alpha}\log^{\alpha}(k)}{k^s}\text{ .} $$
Research by Guariglia showed that this is indeed the desired expression for the frac. der. of zeta. I tried to prove it the following way:
Proof. We proceed by employing the fact that $| \hspace{0.5mm}.\hspace{0.5mm} |$ is a metric in $\mathbb{R}$. Let $\zeta_N(s)=\sum\limits_{k=1}^{N}\dfrac{1}{k^s}$, where $\lim\limits_{N\rightarrow\infty}\zeta_N(s)=\zeta(s)$, then
\begin{align*}
|\zeta^{(\alpha)}(s)-\mathfrak{D}_{s}^{\alpha}\zeta_N(s)|&=\Big| \zeta^{(\alpha)}(s) - \lim _{h \rightarrow 0^+} \frac{1}{h^\alpha}\sum_{m = 0}^{\infty}(-1)^{m}{\alpha\choose m} \zeta_N(s-m h) \Big|\\
&=\Big| \zeta^{(\alpha)}(s) - \lim _{h \rightarrow 0^+} \frac{1}{h^\alpha}\sum_{m = 0}^{\infty}(-1)^{m}{\alpha\choose m} \sum_{k=1}^{N}\frac{1}{k^{s-m h}} \Big|\\
&=\Big| \zeta^{(\alpha)}(s) -\sum_{k=1}^{N}\frac{1}{k^{s}}\left[\lim _{h \rightarrow 0^+}\frac{1}{h^\alpha}\sum_{m = 0}^{\infty}(-1)^{m}{\alpha\choose m}k^{mh} \right]\Big|\\
&=\Big| \zeta^{(\alpha)}(s) - \sum_{k=1}^{N}\frac{1}{k^{s}}\left[\lim _{h \rightarrow 0^+}\frac{1}{h^\alpha}\sum_{m = 0}^{\infty}{\alpha\choose m}(-k^{h})^m \right] \Big|\\
\end{align*}
Comparing the last series in the last line with the generalized binomial theorem one can notice, that it goes to $(1-k^h)^{\alpha}$. Overall $\zeta_N^{(\alpha)}(s)$ now equates to
$$\zeta^{(\alpha)}_N(s)=\sum_{k=1}^{N}\frac{1}{k^{s}}\lim _{h \rightarrow 0^+}\left(\frac{1-k^h}{h}\right)^\alpha = \sum_{k=1}^{N}(-1)^\alpha \frac{1}{k^{s}}\left(\lim _{h \rightarrow 0^+}\frac{k^h-1}{h}\right)^\alpha\text{ .}$$
The next step is to realize, that the expression $\lim _{h \rightarrow 0^+}\dfrac{k^h-1}{h}$ is simply the limit definition of the natural logarithm. Thus we overall arrive at
\begin{align*}
|\zeta^{(\alpha)}(s)-\mathfrak{D}_{s}^{\alpha}\zeta_N(s)|&=\Big|\sum_{k=1}^{\infty}e^{i\pi\alpha} \frac{\log^\alpha(k)}{k^{s}}-\sum_{k=1}^{N}(-1)^{\alpha} \frac{\log^\alpha(k)}{k^{s}}\Big|\\
&=\Big|\sum_{k=N+1}^{\infty}e^{i\pi\alpha} \frac{\log^\alpha(k)}{k^{s}}\Big|\\
&\le\sum_{k=N+1}^{\infty} \Big|e^{i\pi\alpha} \frac{\log^\alpha(k)}{k^{s}}\Big|\\
&=\sum_{k=N+1}^{\infty} \frac{\log^\alpha(k)}{k^{\mathfrak{Re}(s)}}\\
&<\sum_{k=N+1}^{\infty} \frac{k^\alpha}{k^{\mathfrak{Re}(s)}}\\
&=\zeta(\mathfrak{Re}(s)-\alpha)-\zeta_N(\mathfrak{Re}(s)-\alpha)\text{ ,}
\end{align*}
where the last inequality followed from the fact that $k>\log(k)$ for all $k\in\mathbb{N}$. Notice that when we let $N\rightarrow\infty$ our upper estimate also goes to zero, meaning\
$$\lim\limits_{N\rightarrow\infty}|\zeta^{(\alpha)}(s)-\mathfrak{D}_{s}^{\alpha}\zeta_N(s)|<\lim\limits_{N\rightarrow\infty}\zeta(\mathfrak{Re}(s)-\alpha)-\zeta_N(\mathfrak{Re}(s)-\alpha)=0\text{ .}$$
By the properties of a metric this implies that
$$\lim\limits_{N\rightarrow\infty}\mathfrak{D}_{s}^{\alpha}\zeta_N(s)=\lim\limits_{N\rightarrow\infty}\zeta^{(\alpha)}(s)=\zeta^{(\alpha)}(s)$$
and hence the claim follows.
Now here's the problem: The generalized binomial theorem only converges to $(1-k^h)^\alpha$ for $|-k^h|<1$ which is clearly not the case. But for some reason the limit of h and the corresponding 1/h keep the whole expression regularized and make it go to $(-1)^\alpha\log^\alpha(k)$.
Is there some rigorous way to justify what is happening here? I am trying to resolve this issue for several days now but nothing seems to be working. Help is highly appreciated, thank you!
Best regards,
--Jens