Calculating half derivative of $x$ using Cauchy's repeated integral formula

Question

As far as I know, Cauchy's repeated integral formula is defined as: $$(I^nf)(t)=\frac{1}{\Gamma(n)}\int_a^x (x-t)^{n-1}\space f(t)\space\space dt,\space\space n\notin Z_{\le 0}$$

I know that Gamma function is not defined for negative integers and $0$. I also know that differentiation and integration are inverse operators so I thought of finding $\frac{-1}{2}$th antiderivative of $f(x)=x$ so that I can find the half derivative of $x$. However it seems that it does not work. What is the problem with this? Am I missing something or is it not possible? I know some other ways of finding half derivative of $x$, so what I am asking is the problem about what I tried.

My attempt: $$\frac{1}{\Gamma\left(-\frac{1}{2}\right)} \int^x_0(x-t)^{-\frac{3}{2}}\space t\space dt=\frac{x}{2\sqrt{\pi}}\int^0_x u^{-\frac{3}{2}}\space (x-u)\space du$$

This definite integral is divergent and so I cannot get to a result.

Answer 1

Admin related note - you shouldn't post a second question if your first question hasn't been answered, you should update the post.

You seem to have discovered that the formula doesn't make sense for $n\le 0$. Here's some things you could do instead.

1. In the case of $f(x) = x$, you already know that $If = x^2/2, I^2 f = x^3/3!$, and in general $$( I^n f) (x) = \frac{x^{n+1}}{(n+1)!} = \frac{x^{n+1}}{\Gamma(n+2)}$$ This algebraic formula makes perfect sense for $n=-1/2$. So you could set it as the definition of $I^{-1/2}f$: $$ I^{-1/2}f := \frac{x^{n+1}}{\Gamma(n+2)}\Big|_{n=-1/2} = \frac{\sqrt{x}}{\sqrt{\pi}/2}$$
2. The formula generalises for all $n\in [0,\infty)$ with convergent integrals. So if you want to differentiate halfway, you can differentiate once, then use the formula for $n=+1/2$. This leads to (one choice for) the Caputo fractional derivative of order 1/2: $$ I^{-1/2}f(x):=I^{1/2}(f')(x) = \frac1{\Gamma(1/2)}\int_0^x\frac 1{\sqrt{x-t}}f'(t)dt$$ in the case of $f(x)=x$, $$ I^{-1/2}f(x) = \frac1{\sqrt{\pi}}\int_0^x\frac 1{\sqrt{x-t}}dt =\frac1{\sqrt{\pi}}2\sqrt{x}$$
3. You could also do it in reverse order - "integrate $\frac12$ times", then differentiate. This is (one choice for) the Riemann-Liouville fractional derivative of order 1/2: $$ I^{-1/2}f(x):= \frac{d}{dx} \frac1{\Gamma(1/2)}\int_0^x\frac 1{\sqrt{x-t}}f(t)dt$$ In the case of $f(x)=x$, $$ I^{-1/2}f(x) = \frac{d}{dx} \frac1{\sqrt{\pi}}\int_0^x\frac 1{\sqrt{x-t}}tdt =\frac1{\sqrt{\pi}} \frac 43 \frac{d}{dx}x^{3/2} = \frac{1}{\sqrt{\pi}} 2\sqrt{x} $$ These formulas happen to match.

Answer 2

In your integral, it's supposed to be $f(t)=t$, not $f(x)=x$.

$$\int (x-t)^{-3/2}\cdot t\ dt = \cdots$$