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?
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.