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In a book I read about Riemann-Liouville fractional derivative, it says, $$_0D_t^\alpha 1=\frac{t^{-\alpha}}{\Gamma(1-\alpha)},\alpha\geq0,t\geq0$$ which identically vanishes for $\alpha\in\mathbb{N}$, due to the poles of Gamma function.(????)

I have two questions, the first is, why will the equation be applicable for $\alpha\geq0$? Isnt it that gamma function is only defined for positive arguments? Because i was thinking why the restriction is not $0<\alpha<1$.

My second question is what does that phrase "which identically vanishes for $\alpha\in\mathbb{N}$, due to the poles of Gamma function" mean.

Thank you.

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2 Answers 2

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  1. The $\Gamma$ function is defined in the whole complex plane except for its simple poles at non positive integers.
    Here is a representation of the absolute value of $\Gamma$ :

Gamma

  1. The reciprocal of $\Gamma$ is an entire function with zeros for non positive integers.

reciprocal

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We have two somehow strange funtions. The negative power and the gamma funtion. The negative power assumed to be prolonged to negative t's with zeroes - we can write t^(-a).u(t) - has a pole at t=0 when a is a positive integer, a=N. The gamma function is defined for all t, but it has poles at t<=0. the function 1/gamma(t) is analytic with zeros at t<=0. This means that the function you wrote above has problems when a=N. What happens in this case? we obtain the Dirac delta and its derivatives. This is why we should be carefull in using the Riemann-Liouville or Caputo derivatives. Avoid them.

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