Fractional Integral of which function is equal to Riemann's Zeta-Function?

[benorin]

TL;DR

I need help solving a fractional integral equation involving the Reimann Zeta Function

So the problem I’m attempting to solve is $\lim_{x\to a} I_{\alpha}f(x)=\zeta (\alpha )$ for f, and a, where $\zeta (\cdot )$ is the Riemann zeta function and $I_{\alpha}$ is the Riemann-Liouville left fractional integral operator, namely the integral equation

$$\lim_{x\to a}\frac{1}{\Gamma (\alpha )}\int_{t=0}^{x}(x-t)^{\alpha -1}f(t)\, dt = \zeta (\alpha )$$

for some value of $a$. I’ve only studied fractional calculus a little bit, but I’m attacking this problem from another angle as well in this thread. It’d be helpful to know how to define fractional derivatives if $\Re \left[ \alpha \right] > 0$ since the only definition I’ve come across involves the ceiling function of $\alpha$ which I don’t think is defined for complex values of $\alpha$. I think maybe the integral

$$\Gamma (z) \zeta (z)=\int_{u=0}^{\infty} \frac{u^{z-1}}{e^u-1}\, du , \, \Re\left[ z \right] >1$$

could be an alternative way to solve this problem if I can just make an appropriate substitution... will think on that.

Edit: Spelling corrected thanks @zinq

 

[benorin]

I figured it out by changing the integral operator definition to the Hadamard Fractional Integral Operator which Wikipedia defines as

$$_a D_t^{-\alpha}f(t):= \frac{1}{\Gamma (\alpha )}\int_a^t \log ^{\alpha -1} \left( \frac{t}{s}\right) f(s)\frac{ds}{s}$$

because if in the integral

$$\Gamma (\alpha )\zeta (\alpha ) =\int_{u=0}^{\infty} \frac{u^{\alpha -1}}{e^u-1}\, du , \, \Re\left[ z \right] >1$$

you substitute $u=\log \frac{t}{s}\Rightarrow du= -\frac{ds}{s}$ and use that negative to flip the bounds of integration you obtain

$$\zeta (\alpha ) =\frac{1}{ \Gamma (\alpha )} \int_{u=0}^{t} \log ^{\alpha -1}\left( \frac{t}{s}\right) \frac{1}{\frac{t}{s}-1}\frac{ds}{s}\,, \, \Re\left[ z \right] >1$$

and hence

$$\lim_{t\to 1} D_t^{-\alpha} \left(\frac{t}{t-1}\right) =\zeta (\alpha ) , \Re \left[ \alpha \right] >1$$

where that^ should read $_0 D_t^{-\alpha}$ but the combination of the _0 and the limit caused a “math processing error” so I omitted it.

Anybody think of a way to extend this result to include the critical line $\Re \left[ \alpha \right] = \frac{1}{2}$?

 

[zinq]

One way to define the order β fractional derivative is to take the (integer order) nth derivative that's more than you need, and then take the order α fractional integral of that, so that n-α = β as desired.

(If I may be obsessive about spelling: It's Riemann and Liouville.)