Bessel and cosine function identity formula

by expanding into series ( sorry i have tried but get no answer) how could i prove that

$$ \sqrt \pi\frac{d^{1/2}}{dx^{1/2}}J_{0} (a\sqrt x) = \frac{\cos(a\sqrt x)}{\sqrt x}$$

Solution 1:

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 $$ \sqrt{\pi} \frac{d^{\frac{1}{2}}}{dx^{\frac{1}{2}}}J_{0} (a\sqrt x) =\frac{1}{\sqrt{x}}\,\sum _{{\it k}=0}^{\infty }{\frac { \left( -1 \right)^{{\it k}} \,(a\sqrt{x})^{2k}}{ \Gamma(2k+1)}}=\frac{\cos(a\sqrt x)}{\sqrt x} \,. $$

Solution 2:

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 $x\to a x$.