How do you integrate a Bessel function? I don't want to memorize answers or use a computer, is this possible?

Question

I am attempting to integrate a Bessel function of the first kind multiplied by a linear term: $\int xJ_n(x)\mathrm dx$ The textbooks I have open in front of me are not useful (Boas, Arfken, various Schaum's) for this problem. I would like to do this by hand. Is it possible? I have had no luck with expanding out $J_n(x)$ and integrating term by term, as I cannot collect them into something nice at the end.

If possible and I just need to try harder (i.e. other methods or leaving it alone for a few days and coming back to it) that is useful information.

Thanks to anyone with a clue.

This post imported from StackExchange Mathematics at 2014-06-12 20:39 (UCT), posted by SE-user Jennifer

Comments

This is a good question, but I think it belongs on math.SE. I'm going to flag it for the moderators to move it there, but I'm interested in the answer myself. This would be good to know!

This post imported from StackExchange Mathematics at 2014-06-12 20:40 (UCT), posted by SE-user Colin K

Answer 1

At the very least, $\int u J_{2n}(u)\mathrm du$ for integer $n$ is expressible in terms of Bessel functions with some rational function factors.

To integrate $u J_0(u)$ for instance, start with the Maclaurin series:

$$u J_0(u)=u\sum_{k=0}^\infty \frac{(-u^2/4)^k}{(k!)^2}$$

and integrate termwise

$$\int u J_0(u)\mathrm du=\sum_{k=0}^\infty \frac1{(k!)^2}\int u(-u^2/4)^k\mathrm du$$

to get

$$\int u J_0(u)\mathrm du=\frac{u^2}{2}\sum_{k=0}^\infty \frac{(-u^2/4)^k}{k!(k+1)!}$$

thus resulting in the identity

$$\int u J_0(u)\mathrm du=u J_1(u)$$

For $\int u J_2(u)\mathrm du$, we exploit the recurrence relation

$$u J_2(u)=2 J_1(u)-u J_0(u)$$

and

$$\int J_1(u)\mathrm du=-J_0(u)$$

(which can be established through the series definition for Bessel functions) to obtain

$$\int u J_2(u)\mathrm du=-u J_1(u)-2J_0(u)$$

and in the general case of $\int u J_{2n}(u)\mathrm du$ for integer $n$, repeated use of the recursion relation

$$J_{n-1}(u)+J_{n+1}(u)=\frac{2n}{u}J_n(u)$$

as well as the additional integral identity

$$\int J_{2n+1}(u)\mathrm du=-J_0(u)-2\sum_{k=1}^n J_{2k}(u)$$

should give you expressions involving only Bessel functions.

On the other hand, $\int u J_{\nu}(u)\mathrm du$ for $\nu$ not an even integer cannot be entirely expressed in terms of Bessel functions; if $\nu$ is an odd integer, Struve functions are needed ($\int J_0(u)\mathrm du$ cannot be expressed solely in terms of Bessel functions, and this is where the Struve functions come in); for $\nu$ half an odd integer, Fresnel integrals are needed, and for general $\nu$, the hypergeometric function ${}_1 F_2\left({{}\atop b}{a \atop{}}{{}\atop c}\mid u\right)$ is required.

This post imported from StackExchange Mathematics at 2014-06-12 20:40 (UCT), posted by SE-user Ｊ. Ｍ.

Answer 2

There are analytic answers involving hypergeometric functions (which you could look up e.g. on http://functions.wolfram.com). I don't find hypergeometric functions very enlightening or intuitive, but at least they would give a closed-form answer.

At large arguments, Bessel functions are approximately proportional to cosines multiplied by $1/\sqrt{x}$, so maybe you could patch together an approximate answer for a definite integral using the power series expansion for small $x$ and the integrals of expressions involving a cosine at large $x$.

This post imported from StackExchange Mathematics at 2014-06-12 20:40 (UCT), posted by SE-user Matt Reece

Answer 3

I would do it numerically. Also for x not too large, you could use the power series expansion, but this will run into convergence issues before x gets too high. Numerical integration can be done to high accuracy without too much computation using Gaussian quadrature. Good luck looking for an analytic solution (although maybe one exists?). Possibly you can use Bessel's equation, and by substituting your integral you can derive a new differential equation, but I think you'd be very lucky if this allowed an analytic solution. But, a numerical solution should be straightforward (at least for fixed N).

This post imported from StackExchange Mathematics at 2014-06-12 20:40 (UCT), posted by SE-user Omega Centauri