I've recently came across the notion of a fractional derivative of a function $f$ that is defined as
$$\big(D^{\frac{1}{2}} f\big)(x)= \frac{1}{\Gamma(\frac{1}{2})}\int_0^x (x-t)^{-\frac{1}{2}}f(t) dt$$
Now, I typically know of functions of differential operators to be defined in terms of the Fourier Transform. So in my case I'd think of the fractional derivative to be defined as
$$\big(D^{\frac{1}{2}}f\big)(x) = \frac{1}{2\pi}\int_{\mathbb{R}}e^{i \langle x, \xi \rangle} \sqrt{\xi} \cdot \hat{f}(\xi) d\xi$$
I imagine these should be the same but how would I go about showing this?
First of all, try understanding how the Fourier transform changes the homogenepus degree of a function. Try to show that if $f$ is a function (or more generaly a distribution) satisfying that $f(t x) = t^\alpha f(x)$, ten its Fourier transform would be something satisfying $\widehat{f}(t \xi) = t^{-n + \alpha} f(\xi)$. The same follows for the inverse Fourier transform.
After that, take inverse Fourier transforms in the second equation and uses that products become convolutions under the inverse Fourier. The only thing that rests is to adjust the constants.