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I have just seen that this is a duplicate but I don't follow how to arrive at the final power rule from where this question cuts off!

$$D_\alpha^nf(x)=\frac{1}{\Gamma(\lceil n\rceil-n)}\frac{d}{dx^{\lceil n\rceil}}\int_\alpha^xf(t)(x-t)^{\lceil n\rceil-n-1}\space dt$$ Where $\alpha$ is the base point for which $F(\alpha)=0$, $F\prime(x)=f(x)$ - I think, anyway; the video I saw that justified the repeated integration formula relied on this fact to make the formula work...

Anyway, aside from a small quibble about what $\alpha$ really is here, I think I understand this formula, but I do not understand where the following power rule comes from:

$$D_\alpha^n(x^k)=\frac{\Gamma(k+1)}{\Gamma(k+1-\alpha)}x^{k-\alpha}$$

I appreciate the arguments that take the ordinary power rule and just extend it to the reals with the use of $\Gamma$, but I would like to understand how to get to this formula through the integral definition.

I've tried integration by parts on the following expression but I can't make it work - also, since $x$ is a bound of the integral, does that affect how we differentiate the integral with respect to $x$ if we tried to move the derivative operator inside the integral?

$$\frac{1}{\Gamma(\lceil n\rceil-n)}\frac{d}{dx^{\lceil n\rceil}}\int_\alpha^xt^k(x-t)^{\lceil n\rceil-n-1}\space dt$$

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After some learning at home, and realising how to continue the work of the proof I linked at the beginning of my question, I have the following proof (without tediously detailed working) for any who might be curious: $$J_x^\kappa f(x)=\frac{1}{\Gamma(\kappa)}\int_\beta^x(x-\tau)^{\kappa-1}f(\tau)d\tau $$ Letting $\kappa=\lceil\varphi\rceil$: $$\begin{align*}&D_x^\varphi f(x)=\frac{d^\kappa}{dx^\kappa}D_x^{\varphi-\kappa}f(x)=\frac{d^\kappa}{dx^\kappa}J_x^{\kappa-\varphi}f(x) \\ \\ &D_x^\varphi f(x)=\frac{1}{\Gamma(\kappa-\varphi)}\frac{d^\kappa}{dx^\kappa}\int_\beta^x(x-\tau)^{\kappa-\varphi-1}f(\tau)d\tau \\ \\ &D_x^\varphi f(x)=\frac{1}{\Gamma(\kappa-\varphi)}\int_\beta^x\frac{\partial^\kappa}{\partial x^\kappa}(x-\tau)^{\kappa-\varphi-1}f(\tau)d\tau \\ \\ &D_x^\varphi f(x)=\frac{1}{\Gamma(\kappa-\varphi)}\frac{\Gamma(\kappa-\varphi)}{\Gamma(-\varphi)}\int_\beta^x(x-\tau)^{-\varphi-1}f(\tau)d\tau \\ \\ &D_x^\varphi f(x)=\frac{1}{\Gamma(-\varphi)}\int_\beta^x(1-\frac{\tau}{x})^{-\varphi-1}x^{-\varphi-1}f(\tau)d\tau \\ \end{align*} $$ Letting $\nu=\frac{\tau}{x}$ and $\beta=0$: $$ D_x^\varphi f(x)=\frac{1}{\Gamma(-\varphi)}\int_0^1(1-\nu)^{-\varphi-1}x^{-\varphi}f(x\nu)d\nu $$ Suppose $f(x)=x^\omega$, and let $B$ denote the Beta function: $$\begin{align*}&D_x^\varphi x^\omega=\frac{1}{\Gamma(-\varphi)}\int_0^1(1-\nu)^{-\varphi-1}x^{-\varphi}(x\nu)^\omega d\nu \\ \\ &D_x^\varphi x^\omega=\frac{1}{\Gamma(-\varphi)}x^{\omega-\varphi}\int_0^1(1-\nu)^{-\varphi-1}\nu^\omega d\nu \\ \\ &D_x^\varphi x^\omega=\frac{1}{\Gamma(-\varphi)}x^{\omega-\varphi}B(\omega+1,-\varphi) \\ \\ &D_x^\varphi x^\omega=\frac{\Gamma(\omega+1)}{\Gamma(\omega-\varphi+1)}x^{\omega-\varphi} \\ \end{align*} $$ And as such, we have one valid fractional derivative power law for $x^\omega$ as required.

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