# Attempted formulation of the Riemann-Liouville (RL) Fractional Derivative.

I am a high school student studying fractional calculus. I recently came across a couple of issues regarding the formulation of the Riemann-Liouville (RL) Fractional Derivative from the Riemann-Liouville (RL) Fractional Integral. Consider the Riemann-Liouville (RL) Fraction Integral below.

$$I^{\alpha}_x f(x) =\frac{1}{\Gamma(\alpha)}\int_a^xf(t)(x-t)^{\alpha-1}~\mathrm dt$$

Since differentiation is the inverse operation to anti-differentiation, we attempt to formulate the Riemann-Liouville (RL) Fractional Derivative as such.

$$I^{-\alpha}_x f(x) = D^{\alpha}_x f(x) =\frac{1}{\Gamma(-\alpha)}\int_a^x\frac{f(t)}{(x-t)^{\alpha+1}} dt$$

However, this has a couple of apparent issues.

• Firstly, we cannot define the Gamma function for negative integer inputs through analytic continuation.
• Through Wolfram Alpha, I have found the integral does not converge for any $$f(t)$$ (so far).

Of course, the actual Riemann-Liouville (RL) Fractional Derivative is given by following in which $$\alpha > 0$$ and is such that $$\lceil\alpha\rceil = n$$.

$$D^{\alpha}_x f(x) = \frac{1}{\Gamma(n-\alpha)}\frac{d^{n}}{dx^{n}}\int_0^xf(t)(x-t)^{n-\alpha-1},dt.$$

• Does the integral, $$\int_a^x\frac{f(t)}{(x-t)^{\alpha+1}} dt$$ ever converge for a function $$f(t)$$?
• If it does not, how can we prove this?
• Are there any other issues in this attempted formulation of the Riemann-Liouville (RL) Fractional Derivative?

The issue is that the integral fails to converge in a neighborhood of $$t=x$$ unless $$f(t)$$ has a root of sufficiently large order at $$t=x$$ for the point of interest, but unless $$f$$ is zero over an entire interval this does not give any meaningful results.

Proving the divergence when $$f(x)\ne0$$ can easily be done by comparing to

$$\int_a^x\frac{f(x)}{(t-x)^{\alpha+1}}~\mathrm dt=-\frac{f(x)}{\alpha(t-x)^\alpha}\bigg|_{t=a}^{t=x}$$

which fails to exist when $$t=x$$ and $$\alpha>0$$.

Assuming by reformulation of the RL derivative you mean

$$D_x^\alpha f(x)=\frac1{\Gamma(1-\alpha)}\frac{\mathrm d^n}{\mathrm dx^n}\int_0^xf(t)(x-t)^{n-\alpha-1}~\mathrm dt$$

then yes, there is a minor error, as we should have

$$D_x^\alpha f(x)=\frac1{\Gamma(\color{red}n-\alpha)}\frac{\mathrm d^n}{\mathrm dx^n}\int_0^xf(t)(x-t)^{n-\alpha-1}~\mathrm dt$$

and further that we should use $$n=\lfloor\alpha\rfloor+1$$ to avoid $$n=\alpha$$ causing, once again, for the integral to diverge. Another formulation which does not require $$\alpha$$ to be real or for us to restrict $$\alpha>0$$ as a different case would be to use sufficiently large integer $$n$$ or in other words

$$D_x^\alpha f(x)=\lim_{n\to\infty}\frac1{\Gamma(n-\alpha)}\frac{\mathrm d^n}{\mathrm dx^n}\int_0^xf(t)(x-t)^{n-\alpha-1}~\mathrm dt$$

which becomes constant once the integral converges.