# Bessel integral solution or simplification

I am trying to verify the following formula involving Bessel functions of the first kind and am having no luck. The formula is

$$\int{\omega} J_n(\rho \omega)\mathrm d\omega = \frac {1} {\rho} \left\{ -\omega J_{n-1} (\rho \omega) + n \int{J_{n-1}(\rho \omega)\mathrm d\omega } \right\}$$

I apologize if this is painfully obvious with integration by parts but I couldn't see it. Moreover, I get the impression from this other post about a nearly identical integral that the above may not be right.

Any help is greatly appreciated. Also, if there is a simpler way to express/solve this integral, I would also be very grateful for that.

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You can set $\rho=1$ without loss of generality. More importantly: did you try to prove all these relations at once, relying on the generating functions of the Bessel functions of the first kind? – Did May 6 '11 at 23:04
I'm sorry, but I don't follow. I am not aware of any generating function for J_n. – anonstudent007 May 6 '11 at 23:13
See the Laurent series here: en.wikipedia.org/wiki/Bessel_function#Properties – Did May 6 '11 at 23:58
Again I am sorry, perhaps I am missing something obvious, but that is not a generating function for J_n. Rather, it's a generating function for $e^{(x/2)(t-1/t)}$ that uses J_n(x) as it's coefficients. – anonstudent007 May 7 '11 at 0:13
The idea was to sum over $n$ the relations you tried to prove multiplied by $t^n$ and to see what could be said about the resulting generating function. However, see below. – Did May 7 '11 at 0:28

One can set $\rho=1$ without loss of generality. According to this page (see the paragraph $p+1$ dependency), $$\omega J_{n}(\omega)=(n-1)J_{n-1}(\omega)-\omega (J_{n-1})'(\omega)=-(\omega J_{n-1}(\omega))'+nJ_{n-1}(\omega).$$ Hence a primitive of $\omega J_{n}(\omega)$ is $-\omega J_{n-1}(\omega)$ plus $n$ times a primitive of $J_{n-1}(\omega)$. This is your formula.
@anon: the formula Didier quotes is also the second formula in formula 10.6.2 in the DLMF; replace $\nu$ with $n-1$ and $\mathscr{C}$ with $J$. – J. M. May 7 '11 at 0:44