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