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

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The discussion focuses on finding a function f such that the limit of its Riemann-Liouville left fractional integral equals the Riemann zeta function, specifically for some value of a. The integral equation is presented, emphasizing the need for a suitable definition of fractional derivatives for complex values of alpha. An alternative approach using the Hadamard Fractional Integral Operator is proposed, which leads to a relationship between the zeta function and the integral. The conversation also touches on extending results to the critical line where the real part of alpha equals 0.5. Overall, the thread explores complex connections between fractional calculus and the Riemann zeta function.
benorin
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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
 
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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}##?
 
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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.)
 

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