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.