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 $\lceil\alpha\rceil = n$.

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

This leads me to ask the following questions:

  • 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?

1 Answer 1


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.


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