Prove the following identities: $$a) \int J_3(z) dz = -2J_2-J_0 + C$$ $$b) \int z^3 J_1(z)dz = z^3J_1-2z^2J_2+C$$
I tried using the recursive formulae, of course, namely $J_{\nu-1}-J_{\nu+1}=2J^{'}_{\nu}$ for the first one and for the second one I am a bit stuck. Can you give me a hint?
Here a solution for a). Your formula $J_{\nu-1}-J_{\nu+1}=2J'_{\nu}\;$ gives with $J_{-1} = -J_1$ the equations $$J_1-J_3 = 2J'_2\quad\quad(1)$$ $$J_1 = -J'_{0}\quad\quad(2)$$ So taking the difference $(2)-(1)$ you get $$J_1-(J_1-J_3) = J_3 = -J'_0 - 2J'_2$$ And therefore using integration and the fundamental theorem of calculus $$\int J_3(z) dz = \int \Big(-J'_0(z) - 2J'_2(z)\Big) dz = -J_0(z) - 2J_2(z) + C$$
And your equation b) is IMO wrong (typo?): e.g. ask Wolfram Alpha for int(z^3*BesselJ(0,z),z) and int(z^3*BesselJ(1,z),z). It should read
$$\int z^3 J_0(z)dz = z^3J_1(z)-2z^2J_2(z)+C$$
and can be proved by differentiating and simplifying the RHS giving
$$\frac{d}{dz}\Big(z^3J_1(z)-2z^2J_2(z)\Big) = z^3 J_0(z)$$
