Consider the following integral:
$$\int x^{\nu - 1}J_{\nu}(x) \ \mathrm{d}x,$$
where $\nu > 0$ and $J_{\nu}$ denotes the Bessel function of the first kind. For which values $\nu$ (either integers or half-integers) do we have a closed-form expression for this integral? For instance, in the case where $\nu = 1$, then we have the well known identity
$$\displaystyle \int J_{1}(x) \ \mathrm{d}x = -J_{0}(x).$$
If $\nu = 2$ then we also have
$$\displaystyle \int xJ_{2}(x) \ \mathrm{d}x = -2J_0(x) - xJ_1(x),$$
and it seems that for positive integers $\nu$ we can obtain an expression involving Bessel functions. Is there a general formula?
If $\nu$ is a half-integer then I am not sure a closed form expression exists. Wolfram Alpha says that
$$\displaystyle \int x^{1/2}J_{3/2}(x) \ \mathrm{d}x = \sqrt{\frac{2}{\pi}}\left(\mathrm{Si}(x) - \sin(x)\right),$$
where $\mathrm{Si}(x)$ denotes the sine integral. The presence of the $\sin$ also seems to suggest that the integral wouldn't converge on an infinite interval such as $[0, \infty)$, say.
If you accept hypergeometric functions, there is $$\int x^m J_n(x)\,dx=\frac{ x^{m+n+1}}{2^n\,(m+n+1)\, \Gamma (n+1)} \,\, _1F_2\left(\frac{m+n+1}{2};\frac{m+n+3}{2},n+1;-\frac{x^2}{4}\right)$$ and $$\int_0^\infty x^m J_n(x)\,dx=2^m\frac{ \Gamma \left(\frac{1}{2} (n+m+1)\right)}{\Gamma \left(\frac{1}{2} (n-m+1)\right)}\qquad \text{if}\qquad \Re(m+n)>-1\land \Re(m)<\frac{1}{2}$$