Fractional derivative (differintegral) can be defined via Fourier or Laplace transform:

${\displaystyle D ^{q}f(t)={\mathcal {F}}^{-1}\left\{(i\omega )^{q}{\mathcal {F}}[f(t)]\right\}},$

${\displaystyle D ^{q}f(t)={\mathcal {L}}^{-1}\left\{s^{q}{\mathcal {L}}[f(t)]\right\}.}$

This way, one can similarly define the fractional discrete derivative (because $\Delta=e^D-1$):

${\displaystyle D ^{q}f(t)={\mathcal {F}}^{-1}\left\{(e^{i\omega}-1 )^{q}{\mathcal {F}}[f(t)]\right\}},$

${\displaystyle D ^{q}f(t)={\mathcal {L}}^{-1}\left\{(e^s-1)^{q}{\mathcal {L}}[f(t)]\right\}.}$

Discrete derivative also can be defined via Newton series:

${\displaystyle D ^{q}f(t)=\sum _{m=0}^{\infty }{\binom {q}{m}}\sum _{k=0}^{m}{\binom {m}{k}}(-1)^{m-k}f^{(k)}(x).}$

One can replace the derivative with difference operator to get fractional difference:

${\displaystyle D ^{q}f(t)=\sum _{m=0}^{\infty }{\binom {q}{m}}\sum _{k=0}^{m}{\binom {m}{k}}(-1)^{m-k}\Delta^k f(x).}$

The Fourier transform method can be realized in Mathematica as follows:

FourierTransform[InverseFourierTransform[f[t], t, w] (E^(I w) - 1)^q, 
  w, x] // FullSimplify

(Mathematica realizes a different definition of Fourier transform, so there is a slight difference in formula)

And the Laplace transform method is

LaplaceTransform[InverseLaplaceTransform[f[t], t, w] (1 - E^w)^q, w, 
   x] // FullSimplify

(again, due to specifics of Mathematica, the formula is a bit different to make it more powerful).

The 1/2-th difference of the function $1/x$ is thus $\frac{i \sqrt{\pi } \Gamma \left(x-\frac{1}{2}\right)}{2 \Gamma (x+1)}$

Plot of function $1/x$ (blue), imaginary part of its 1/2-th (yellow) and 1-st (green) finite differences.

For function $\frac{(-1)^s \Gamma (s+1) \Gamma (x-s)}{\Gamma (x+1)}$

Fractional derivative (differintegral) can be defined via Fourier or Laplace transform:

${\displaystyle D ^{q}f(t)={\mathcal {F}}^{-1}\left\{(i\omega )^{q}{\mathcal {F}}[f(t)]\right\}},$

${\displaystyle D ^{q}f(t)={\mathcal {L}}^{-1}\left\{s^{q}{\mathcal {L}}[f(t)]\right\}.}$

This way, one can similarly define the fractional discrete derivative (because $\Delta=e^D-1$):

${\displaystyle D ^{q}f(t)={\mathcal {F}}^{-1}\left\{(e^{i\omega}-1 )^{q}{\mathcal {F}}[f(t)]\right\}},$

${\displaystyle D ^{q}f(t)={\mathcal {L}}^{-1}\left\{(e^s-1)^{q}{\mathcal {L}}[f(t)]\right\}.}$

Discrete derivative also can be defined via Newton series:

${\displaystyle D ^{q}f(t)=\sum _{m=0}^{\infty }{\binom {q}{m}}\sum _{k=0}^{m}{\binom {m}{k}}(-1)^{m-k}f^{(k)}(x).}$

One can replace the derivative with difference operator to get fractional difference:

${\displaystyle D ^{q}f(t)=\sum _{m=0}^{\infty }{\binom {q}{m}}\sum _{k=0}^{m}{\binom {m}{k}}(-1)^{m-k}\Delta^k f(x).}$

The Fourier transform method can be realized in Mathematica as follows:

FourierTransform[InverseFourierTransform[f[t], t, w] (E^(I w) - 1)^q, 
  w, x] // FullSimplify

(Mathematica realizes a different definition of Fourier transform, so there is a slight difference in formula)

And the Laplace transform method is

LaplaceTransform[InverseLaplaceTransform[f[t], t, w] (1 - E^w)^q, w, 
   x] // FullSimplify

(again, due to specifics of Mathematica, the formula is a bit different to make it more powerful).

The 1/2-th difference of the function $1/x$ is thus $\frac{i \sqrt{\pi } \Gamma \left(x-\frac{1}{2}\right)}{2 \Gamma (x+1)}$

Plot of function $1/x$ (blue), imaginary part of its 1/2-th (yellow) and 1-st (green) finite differences.

For function $f(x)=1/x$ the general expression holds: $\Delta^q [1/x]= \frac{(-1)^q \Gamma (q+1) \Gamma (x-q)}{\Gamma (x+1)}$

Fractional derivative (differintegral) can be defined via Fourier or Laplace transform:

${\displaystyle D ^{q}f(t)={\mathcal {F}}^{-1}\left\{(i\omega )^{q}{\mathcal {F}}[f(t)]\right\}},$

${\displaystyle D ^{q}f(t)={\mathcal {L}}^{-1}\left\{s^{q}{\mathcal {L}}[f(t)]\right\}.}$

This way, one can similarly define the fractional discrete derivative (because $\Delta=e^D-1$):

${\displaystyle D ^{q}f(t)={\mathcal {F}}^{-1}\left\{(e^{i\omega}-1 )^{q}{\mathcal {F}}[f(t)]\right\}},$

${\displaystyle D ^{q}f(t)={\mathcal {L}}^{-1}\left\{(e^s-1)^{q}{\mathcal {L}}[f(t)]\right\}.}$

Discrete derivative also can be defined via Newton series:

${\displaystyle D ^{q}f(t)=\sum _{m=0}^{\infty }{\binom {q}{m}}\sum _{k=0}^{m}{\binom {m}{k}}(-1)^{m-k}f^{(k)}(x).}$

One can replace the derivative with difference operator to get fractional difference:

${\displaystyle D ^{q}f(t)=\sum _{m=0}^{\infty }{\binom {q}{m}}\sum _{k=0}^{m}{\binom {m}{k}}(-1)^{m-k}\Delta^k f(x).}$

The Fourier transform method can be realized in Mathematica as follows:

FourierTransform[InverseFourierTransform[f[t], t, w] (E^(I w) - 1)^q, 
  w, x] // FullSimplify

(Mathematica realizes a different definition of Fourier transform, so there is a slight difference in formula)

And the Laplace transform method is

-LaplaceTransform[InverseLaplaceTransform[f[t], t, w] (E^w - 1)^q, w, 
   x] // FullSimplify

(again, due to specifics of Mathematica, the formula is a bit different to make it more powerful).

The 1/2-th difference of the function $1/x$ is thus $-\frac{\sqrt{\pi } \Gamma \left(x-\frac{1}{2}\right)}{2 \Gamma (x+1)}$

Plot of function $1/x$ (blue) and its 1/2-th (green) and 1-st (yellow) finite differences.

For function $f(x)=1/x$ the general expression holds: $\Delta^q [1/x]=-\frac{\Gamma (q+1) \Gamma (x-q)}{\Gamma (x+1)}$

Fractional derivative (differintegral) can be defined via Fourier or Laplace transform:

${\displaystyle D ^{q}f(t)={\mathcal {F}}^{-1}\left\{(i\omega )^{q}{\mathcal {F}}[f(t)]\right\}},$

${\displaystyle D ^{q}f(t)={\mathcal {L}}^{-1}\left\{s^{q}{\mathcal {L}}[f(t)]\right\}.}$

This way, one can similarly define the fractional discrete derivative (because $\Delta=e^D-1$):

${\displaystyle D ^{q}f(t)={\mathcal {F}}^{-1}\left\{(e^{i\omega}-1 )^{q}{\mathcal {F}}[f(t)]\right\}},$

${\displaystyle D ^{q}f(t)={\mathcal {L}}^{-1}\left\{(e^s-1)^{q}{\mathcal {L}}[f(t)]\right\}.}$

Discrete derivative also can be defined via Newton series:

${\displaystyle D ^{q}f(t)=\sum _{m=0}^{\infty }{\binom {q}{m}}\sum _{k=0}^{m}{\binom {m}{k}}(-1)^{m-k}f^{(k)}(x).}$

One can replace the derivative with difference operator to get fractional difference:

${\displaystyle D ^{q}f(t)=\sum _{m=0}^{\infty }{\binom {q}{m}}\sum _{k=0}^{m}{\binom {m}{k}}(-1)^{m-k}\Delta^k f(x).}$

The Fourier transform method can be realized in Mathematica as follows:

FourierTransform[InverseFourierTransform[f[t], t, w] (E^(I w) - 1)^q, 
  w, x] // FullSimplify

(Mathematica realizes a different definition of Fourier transform, so there is a slight difference in formula)

And the Laplace transform method is

LaplaceTransform[InverseLaplaceTransform[f[t], t, w] (1 - E^w)^q, w, 
   x] // FullSimplify

(again, due to specifics of Mathematica, the formula is a bit different to make it more powerful).

The 1/2-th difference of the function $1/x$ is thus $\frac{i \sqrt{\pi } \Gamma \left(x-\frac{1}{2}\right)}{2 \Gamma (x+1)}$

Plot of function $1/x$ (blue), imaginary part of its 1/2-th (yellow) and 1-st (green) finite differences.

For function $\frac{(-1)^s \Gamma (s+1) \Gamma (x-s)}{\Gamma (x+1)}$