On the wikipage about fractional calculus, there's a general formula for the fractional derivative:

$D^\alpha$ is the derivative operator.

$$D^\alpha f(x)=\frac1{\Gamma(1-\alpha)}\cdot\frac{d}{dx}\int_0^x\dfrac{f(t)}{(x-t)^\alpha}dt,\quad0 < \alpha < 1$$

Since it had a derivative of an integral and I wanted to remove it I put it into wolfram alpha, however the solution has something I'm unsure how I should deal with. (I omitted the factor of $1/\Gamma(1-\alpha)$ in WA)

$$\frac1{\Gamma(1-\alpha)}\cdot\left(\int_0^x\dfrac{\alpha f(t)(x-t)^{-\alpha}}{t-x}dt+0^{-\alpha}f(x)\right)$$

As you can see this equation contains $0^{-\alpha}$, however $\alpha$ is always between $0$ and $1$ so this is division by zero.

What happened here, and what is a nicer equation?

  • $\begingroup$ the problem occurs because the integrand has a singularity at $t=x$ (although the function is integrable for $0 \lt \alpha \lt 1$). however the usual formula for $\frac{d}{dx} \int_0^x g(x,t)dt$ is $g(x,x)+ \int_0^x \frac{\partial}{\partial x}g(x,t)dt$, but here $g(x,x)$ blows up $\endgroup$ Aug 28, 2014 at 13:28

1 Answer 1


David Holden posted this answer in the comments.

The problem occours because of a singularity at $t=x$, however the function is still integrateable in $0<\alpha<1$. However the usual formula for derivative of definite integral dosen't work because of the singularity.


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