What is the fractional integral of the second derivative?

Question

The following example illustrates the issue. The derivative of derivative is: $$ \frac{d}{dx}(\frac{dy}{dx}) = \frac{d^2y}{dx^2} $$ The derivative of square of derivative is: $$ \frac{d}{dx}(\frac{dy}{dx})^2 = 2 \frac{dy}{dx}\frac{d^2y}{dx^2} $$ Therefore the integral of the double derivative is : $$ 2\int\frac{d^2y}{dx^2}dy = (\frac{dy}{dx})^2 $$ or $$ 2\int\frac{d^2y}{dx^2}\frac{dy}{dx} = (\frac{dy}{dx})^2 $$ For fractional integral (ex: 1/2) : $$ 2\int\frac{dy}{dx}\frac{d^{\frac{1}{2}}y}{dx^{\frac{1}{2}}} = (\frac{d^{\frac{1}{2}}y}{dx^{\frac{1}{2}}})^2 $$ Now, what is the fractional integral of the second derivative? : $$ 2\int\frac{d^2y}{dx^2}\frac{d^{\frac{1}{2}}y}{dx^{\frac{1}{2}}} = ? $$

Answer 1

Riemann- Liouville fractional Derivative definition is: $$ D_{a^{+}}^{\alpha }f(x)=\frac{1}{\Gamma (1-\alpha )}\frac{d}{dx}% \int\limits_{a}^{x}(x-t)^{-\alpha }f(t)dt $$ and Fractional Integral definition is: $$ I_{a^{+}}^{\alpha }f(x)=\frac{1}{\Gamma (\alpha )}\int\limits_{a}^{x}(x-t)^{% \alpha -1}f(t)dt. $$ Fractional derivative of the fractional integral; $$ D_{a^{+}}^{\alpha }(I_{a^{+}}^{\alpha }f(x))=f(x)??. $$

Let's calculate; \begin{eqnarray*} D_{a^{+}}^{\alpha }(I_{a^{+}}^{\alpha }f(x)) &=&\frac{1}{\Gamma (1-\alpha )}% \frac{d}{dx}\int\limits_{a}^{x}(x-t)^{-\alpha }(I_{a^{+}}^{\alpha }f(t))dt \\ && \\ &=&\frac{1}{\Gamma (1-\alpha )}\frac{d}{dx}\int\limits_{a}^{x}(x-t)^{-\alpha }\frac{1}{\Gamma (\alpha )}\int\limits_{a}^{t}(t{-}s)^{\alpha -1}f(s)dsdt \end{eqnarray*} Let's change the order and boundary of the integration.

\begin{array}{c} a<t<x \\ a<s<t% \end{array}

\begin{array}{c} a<s<t<x \end{array}%

$$ D_{a^{+}}^{\alpha }(I_{a^{+}}^{\alpha }f(x))=\frac{1}{\Gamma (1-\alpha )}% \frac{1}{\Gamma (\alpha )}\frac{d}{dx}\int\limits_{a}^{x}f(s)\left( \int\limits_{s}^{x}(x-t)^{-\alpha }(t-s)^{\alpha -1}dt\right) ds $$ Let's calculate the last integral. $$ t=s+\left( x-s\right) u, $$ \begin{eqnarray*} &&\int\limits_{s}^{x}(x-t)^{-\alpha }(t-s)^{\alpha -1}dt=\int\limits_{0}^{1}u^{-\alpha }(1-u)^{\alpha -1}du \\ && \\ &=&\Gamma (\alpha )\Gamma (1-\alpha ). \end{eqnarray*} So, \begin{eqnarray*} &&D_{a^{+}}^{\alpha }(I_{a^{+}}^{\alpha }f(x)) \\ &=&\frac{1}{\Gamma (1-\alpha )}\frac{1}{\Gamma (\alpha )}\frac{d}{dx}% \int\limits_{a}^{x}f(s)\left( \Gamma (\alpha )\Gamma (1-\alpha )\right) ds \\ &=&\frac{1}{\Gamma (1-\alpha )\Gamma (\alpha )}\left( \Gamma (\alpha )\Gamma (1-\alpha )\right) \frac{d}{dx}\int\limits_{a}^{x}f(s)ds \end{eqnarray*} and finally, we get

$$ D_{a^{+}}^{\alpha }(I_{a^{+}}^{\alpha }f(x))=f(x). $$