The Grünwald–Letnikov fractional derivative of the function $x^{-1/2}$

Question

Let $f(x) = 1/\sqrt{x}$, then how can we find the Grünwald–Letnikov fractional derivative for this function. By definition, $$\operatorname{D}_{0}^{p}f(x) = \lim_{h \to 0,\, n \cdot h = x - a}\left( h^{-p} \cdot \sum_{k=0}^{n}\left( -1 \right)^{k} \cdot \binom{p}{k} \cdot f\left( x - k \cdot h \right) \right). $$ I know for smooth functions, we can have a nice integral form for the Grünwald–Letnikov fractional derivative, but for these types of discontinuous function, how we can the derivative?

Answer 1

For $s\in\mathbb{C}$ and $x\in\mathbb{R}$, denote $$s!:=\Gamma(s+1),\qquad x_+^s:=\begin{cases}x^s,&x>0\\0,&x\leqslant 0\end{cases}.$$

Then we have $\color{blue}{D^\beta x_+^\alpha=\frac{\alpha!\,x_+^{\alpha-\beta}}{(\alpha-\beta)!}}$ for $\alpha,\beta\in\mathbb{C}$ with $\Re\alpha>-1$.

This amounts to showing that $$\lim_{h\to 0^+}S(\alpha,\beta,h)=\frac{\alpha!}{(\alpha-\beta)!},\qquad S(\alpha,\beta,h):=h^{-\beta}\sum_{n=0}^\infty\binom\beta{n}(-1)^n(1-nh)_+^\alpha.$$

I'm following my approach from here (somewhat extended).

The point of departure is Hankel's integral representation of the reciprocal gamma function $$\frac1{\Gamma(s)}=\frac1{2\pi i}\int_\lambda z^{-s}e^z\,dz\qquad(s\in\mathbb{C})$$ where $\lambda$ is a simple contour encircling the negative real axis. If $\Re s>1$, the contour can be deformed to give $\int_\lambda=\int_{c-i\infty}^{c+i\infty}$ for any $c>0$; then, for any $x\in\mathbb{R}$ and $\alpha\in\mathbb{C}$ with $\Re\alpha>0$, we have $$x_+^\alpha=\frac{\alpha!}{2\pi i}\int_{c-i\infty}^{c+i\infty}z^{-\alpha-1}e^{xz}\,dz.$$

Indeed, the case of positive $x$ is reduced to the above after an obvious substitution, and for $x\leqslant 0$ the integral equals $\lim_{r\to\infty}\int_{\lambda_r}$ with $\lambda_r$ the bounary of $\{z\in\mathbb{C} : |z|<r \land \Re z>c\}$, and the last integral equals $0$ by Cauchy.

Taking this at $x=1-nh$, and using the binomial series, we get $$S(\alpha,\beta,h)=\frac{\alpha!}{2\pi i}\int_{c-i\infty}^{c+i\infty}z^{-\alpha-1}e^z\left(\frac{1-e^{-hz}}h\right)^\beta dz$$ (for $\alpha,\beta\in\mathbb{C}$ with $\Re\alpha>0$).

It's tempting to take $h\to 0^+$ right now, but this would work only for $\Re\beta<\Re\alpha$. The idea to overcome this issue is to deform the contour back, but, in turn, this meets two obstacles. The easier one is that the integrand may grow exponentially with $|z|$, and this is handled by considering small enough values of $h$, such that $h\Re\beta<1$.

The harder one is that the integrand has branch points at $z=2n\pi i/h$, so that, instead of a single integral along $\lambda$, we get an infinite series of such integrals: $$S(\alpha,\beta,h)=\frac{\alpha!}{2\pi i}\sum_{n\in\mathbb{Z}}\int_\lambda\left(z+\frac{2n\pi i}h\right)^{-\alpha-1}e^z\left(\frac{1-e^{-hz}}h\right)^\beta dz.$$

Now taking $h\to 0^+$ is not a problem anymore, and we get $$\lim_{h\to 0^+}S(a,b,h)=\frac{\alpha!}{2\pi i}\int_\lambda z^{\beta-\alpha-1}e^z\,dz=\frac{\alpha!}{(\alpha-\beta)!}$$ as expected, but we're still under $\Re\alpha>0$.

To reach $\Re\alpha>-1$, one might rewrite the above as $$S(\alpha,\beta,h)=\frac{h^{\alpha+1}}{2\pi i}\int_\lambda\operatorname{Li}_{-\alpha}(e^{-hz})e^z\left(\frac{1-e^{-hz}}h\right)^\beta dz$$ using the polylogarithm function, which is analytically continued onto the whole of $\alpha\in\mathbb{C}$, and take the limit, which now exists if $\Re\alpha>-1$. I'm still looking for a more elementary approach, perhaps even an algebraic one.