What is the fractional integral of the second derivative? 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? :
$$
2\int\frac{d^2y}{dx^2}\frac{d^{\frac{1}{2}}y}{dx^{\frac{1}{2}}} =  ?
$$
 A: 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
$$
D_{a^{+}}^{\alpha }(I_{a^{+}}^{\alpha }f(x))=f(x).
$$
