Fourier analysis and fractional calculus Do Fourier transform properties still hold in the case of fractional derivatives ? 
i.e I have seen many times that some lectures define fractional derivative as : 
$$\frac{d^{\alpha}}{dx^{\alpha}}f=\mathscr{F}^{-1}\big[\mathscr{F}[f(x)](w)\cdot w^{\alpha}\big](x)$$
Indeed fractional derivatives of exponentials do not really look like exponentials ... 
Thanks
 A: Too long for a comment. Your spelling of the name of the mathematician Joseph Fourier is incorrect. Also your formula is almost impossible to read: your fractional derivative on the lhs is the Fourier multiplier $(i\xi)^\alpha$ and thus formally you find
$$
\left(\left(\frac{d}{dx}\right)^\alpha f\right)(x)=\int e^{i x\xi} (i\xi)^\alpha \hat f(\xi) d\xi/(2π).
$$
You may also say that the inverse Fourier transform of the homogeneous $(i\xi)^\alpha$, say for $\Re\alpha>-1$, is also an homogeneous distribution of degree $-1-\alpha$ and that the above fractional derivative is the convolution with
$$
c_+(\alpha) x_+^{-\alpha-1}+c_-(\alpha) x_-^{-\alpha-1}=w_\alpha(x),
$$
where $c_\pm(\alpha)$ are constants so that the formal formula is 
$$
\left(\left(\frac{d}{dx}\right)^\alpha f\right)(x)=\int f(y) w_\alpha (x-y) dy.
$$
Note that if $-\alpha-1$ happens to be a negative integer $-k$, then we have 
$
x_+^{-k}=c_k \delta_0^{(k-1)},
$
where $\delta_0^{(k-1)}$ stands for the $(k-1)$th derivative of the Dirac mass at 0.
