I need a help with integral below,
$$\int_{0}^{1}x\sin (bx) J_0\left(a\sqrt{1-x^2}\right)\,\mathrm dx$$
where $a \geq 0$ and $b$ are constants and $J_0(x)$ is the zeroth-order of Bessel function of the first kind.
This integral has a closed form solution ?
I found some integrals similar to the integral above, but I don't have any idea on how to apply it.
I found this integral expression below on Gradshteyn and Ryzhik's book 7th edition, section 6.677, number 6:
$$\int_{0}^{a} \cos (cx)J_0\left(b\sqrt{a^2-x^2}\right)\,\mathrm dx = \frac{\sin (a\sqrt{b^2+c^2})}{\sqrt{b^2+c^2}} \quad [b\geq 0]$$
I try use the definition of zeroth-order of Bessel function of the first kind to solve this integral:
$$J_0(z) = \sum_{k=0}^\infty (-1)^k\frac{(\frac{1}{4}z^2)^k}{(k!)^2}$$
then i found:
$$J_0(a) \sum_{k=0}^\infty \int_{0}^{1}x\sin (bx)\left(1-x^2\right)^k\,\mathrm dx $$
On Wolfram Alpha i found that:
$$\int_{0}^{1}x\sin (bx)\left(1-x^2\right)^k\,\mathrm dx = \frac{1}{4}\sqrt{\pi}a\Gamma (k+1)\tilde{F}_1\Big(;k+\frac{5}{2};-\frac{a^2}{4}\Big)$$
where $\tilde{F}_1$ is the Regularized Hypergeometric Function, where:
$$\tilde{F}_1\Big(;k+\frac{5}{2};-\frac{a^2}{4}\Big) = \frac{J_{k+\frac{3}{2}}(a) }{(\frac{a}{2})^{k+\frac{3}{2}}}$$
therefore:
$$\int_{0}^{1}x\sin (bx) J_0\left(a\sqrt{1-x^2}\right)\,\mathrm dx = J_0(a)\frac{1}{4}\sqrt{\pi}a \sum_{k=0}^\infty \Gamma (k+1) J_{k+\frac{3}{2}}(a)\Big(\frac{a}{2}\Big)^{-(k+\frac{3}{2})} $$
The solution found above is right ?
Thanks in advance.
