Bessel and cosine function identity formula by expanding into series ( sorry i have tried but get no answer) how could i prove that
$$  \sqrt \pi\frac{d^{1/2}}{dx^{1/2}}J_{0} (a\sqrt x) = \frac{\cos(a\sqrt x)}{\sqrt x}$$
 A: A related problem. The power series of $J_{0} (a\sqrt x)$ is
$$J_{0} (a\sqrt x)= \sum _{{\it k}=0}^{\infty }{\frac { \left( -1 \right) ^{{\it k}}a^{2k}
x^{{\it k}}}{  2^{2k}\Gamma  \left( 1+
{\it k}   \right) ^{2}}} $$
Applying the formula for fractional derivative of a monomial
$$ \frac{d^q}{dx^q} x^m = \frac{\Gamma(m+1)}{\Gamma(m-q+1 )} x^{m-q}\,, $$
to the above series yields
$$   \frac{d^{\frac{1}{2}}}{dx^{\frac{1}{2}}}J_{0} (a\sqrt x) =\sum _{{\it k}=0}^{\infty }{\frac { \left( -1 \right) ^{{\it k}}a^{2k}
\Gamma(k+1)\,x^{{\it k-\frac{1}{2}}}}{  2^{2k}\Gamma(k+\frac{1}{2})\Gamma\left( 1+{\it k}\right) ^{2}}} \,. $$
Simplifying the above series and using the identity $ \Gamma(2k+1)=\frac{1}{\sqrt{\pi}}2^{2k}\Gamma(k+1)\Gamma(k+\frac{1}{2}) $, we get 
$$   \frac{d^{\frac{1}{2}}}{dx^{\frac{1}{2}}}J_{0} (a\sqrt x) =\frac{1}{\sqrt{\pi}\sqrt{x}}\,\sum _{{\it k}=0}^{\infty }{\frac { \left( -1 \right)^{{\it k}}
\,(a\sqrt{x})^{2k}}{ \Gamma(2k+1)}} \,. $$
Multiplying both sides of the above equation by $ \sqrt{\pi} $, we reach the desired result follow
$$  \sqrt{\pi} \frac{d^{\frac{1}{2}}}{dx^{\frac{1}{2}}}J_{0} (a\sqrt x) =\frac{1}{\sqrt{x}}\,\sum _{{\it k}=0}^{\infty }{\frac { \left( -1 \right)^{{\it k}}
\,(a\sqrt{x})^{2k}}{ \Gamma(2k+1)}}=\frac{\cos(a\sqrt x)}{\sqrt x} \,. $$
A: For simplicity let $a=1$. 
Using the definition of the fractional derivative 
we find that if we can show that
$$\int_0^x \frac{J_0(\sqrt{t})}{(x-t)^{1/2}} dt
= 2\sin\sqrt{x},$$
then we will have the quoted result. 
(The derivative of this result is the fractional derivative of interest.)
A straightforward approach is to expand $J_0$ in small $t$ and integrate term by term.
The resulting integrals are related to the beta function, and after some manipulation we arrive at the series expansion for $2\sin\sqrt{x}$. 
Finally, we have 
$$\begin{eqnarray*}
\sqrt \pi\frac{d^{1/2}}{dx^{1/2}}J_{0} (\sqrt x)
&=& \frac{d}{dx}  2\sin\sqrt{x} \\
&=& \frac{\cos{\sqrt{x}}}{\sqrt{x}}.
\end{eqnarray*}$$
To recover $a$, let $x\to a x$.
