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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.

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    The answer is just identity $(59)$ here: http://mathworld.wolfram.com/BesselFunctionoftheFirstKind.html – 2017-02-05
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    But isn't that for integrals of products of the form $x^{m}J_{m-1}(x)$? – 2017-02-05
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    Sure, but there is a magic thing called *integration by parts* ;) – 2017-02-05
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    And there is no wonder in $\int_{0}^{+\infty}x^{1/2}J_{\nu}(x)\,dx$ being non-convergent, since the asymptotic behaviour of $J_\nu(x)$ for large $x$ is $\sqrt{\frac{\pi}{2x}}\cos(x-\theta_\nu)$. – 2017-02-05

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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 $$\int_0^\infty x^m J_n(x)\,dx=2^m\frac{ \Gamma \left(\frac{1}{2} (n+m+1)\right)}{\Gamma \left(\frac{1}{2} (n-m+1)\right)}\qquad \text{if}\qquad \Re(m+n)>-1\land \Re(m)<\frac{1}{2}$$