A question on the use of fractional derivatives in Riemann Hypothesis We already know that Riemann-zeta function on the critical band is defined as follows:
$$(1-2^{1-\alpha})\zeta(\alpha) = \sum_{k=1}^{\infty} (-1)^{k+1}k^{-\alpha},\quad \Re(\alpha) \in ]0, 1[  $$
Is it possible to say that
$$ (1-2^{1-\alpha})\zeta(\alpha) = \left[\frac{d^{-\alpha}}{dw^{-\alpha}} \sum_{k=1}^{\infty}(-1)^{k+1}e^{iwk}\right]_{w=0} = \left.
\frac{d^{-\alpha}}{dw^{-\alpha}}\frac{e^{iw}}{1+e^{iw}}\right|_{w=0} \;?
$$
how can this fractional derivative
$$\left.
\frac{d^{-\alpha}}{dw^{-\alpha}}\frac{e^{iw}}{1+e^{iw}}\right|_{w=0}  = \left.
\frac{d^{\alpha-1}}{dw^{\alpha-1}}\frac{e^{iw}}{1+e^{iw}}\right|_{w=0} \ $$
be evaluated or numerically estimated using the fractional derivative or antiderivative definition that has exponentials as eigenfunctions?
Thanks
Note. I know this method does not fully comply with absolute convergence and I am not sure if using such fractional derivatives makes sense.
 A: (Disclaimer I need to verify if your integral of $\frac{1}{1+e^{iw}}$ trick works but here's an alternative framework which should meet your needs)
So we wish to evaluate
$$(1-2^{1-a} ) \zeta(a) =  \sum_{k=1}^{\infty} (-1)^{k+1} k^{-a} $$
Our first trick is to consider the function
$$ f = \frac{e^x}{1+e^x} = e^x - e^{2x} + e^{3x} - ... $$
So if we take integrals of this we end up with your function namely:
$$ \frac{d^{-a}}{dx^{-a}} \left[ f(x) \right] |_{x = 0} = (1-2^{1-a}) \zeta(a)$$
Now if you pick your integration bounds carefully there probably is way to make that work. If we want to use the fractional "Derivative" as opposed to the integral our strategy changes a bit.
We want to fractionally differentiate the following function
$$ g= e^{x} - e^{\frac{1}{2}x} + e^{\frac{1}{3}x} - ...  $$
But this series does not converge the way its given, so we can expand all the exponentials to yield
$$ g = \left( 1 - 1 + 1 -  ...\right) + \left( 1 - \frac{1}{2} + \frac{1}{3} - ... \right) x + \left( 1 - \frac{1}{4} + \frac{1}{9} - .... \right) $$
We can sum these
$$ g = \frac{1}{2} + \ln(2)x + \frac{\pi^2}{12}x^2 ... $$
So now we have that for this magic $g$
$$ \frac{d^{a}}{dx^a} [g] |_{x=0} = (1 - 2^{1-a})  \zeta(a) $$
But this is not as profound as it may seem initially...
Really given any function $f(a)$ if we consider the taylor series
$$ h(x) = \frac{1}{0!}f(0) + \frac{1}{1!}f(1)x + \frac{1}{2!}f(2)x^2 + ... $$
Then:
$$ \frac{d^{a}}{dx^{a}}[h(x)]|_{x=0} = f(a) $$
And what we have done above is a restatement of that... (Quite literally this is how you get the definition of the gamma function).
Now I'm actually pretty poor with complex integration so I need to review how to evaluate your integral for the first definition (although I suspect i'll end up refinding the functions you found after using a suitable contour).
