# Fractional derivative of a constant (Riemann-Liouville Derivative)

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.

## 2 Answers

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