Reducing Bessel Function Integral

[Jacob Nie]

Homework Statement

Any integral of the form $\int x^m J_n(x) dx$ can be evaluated in terms of Bessel functions and the indefinite integral $\int J_0(x)dx,$ which cannot be simplified further. Use the identities $J_p'(x) = J_{p-1}(x) - \dfrac{p}{x}J_p(x)$ and $J_p'(x) = \dfrac{p}{x} J_p(x) - J_{p+1}(x)$ to evaluate $\int x^2J_0(x)dx.$

Relevant Equations

Should be able to do it with just the two properties listed in the problem.

I tried integration by parts with both $u = x^2, dv = J_0 dx$ and $u = J_0, du = -J_1 dx, dv = x^2 dx.$ But neither gets me in a very good place at all. With the first, I begin to get integrals within integrals, and with the second my powers of $x$ in the integral would keep growing instead of getting smaller.

The book's answer is $x^2J_1(x) + xJ_1(x) - \int J_0(x)dx + C.$ (BTW the book is Elementary DE by Edwards and Penney)

How do I get to that? I appreciate any help!

 

[Jacob Nie]

Sorry - I have solved it.

I gave the wrong identities - these are more directly useful: $\dfrac{d}{dx}\left(x^pJ_p(x)\right) = x^pJ_{p-1}(x)$ and $\dfrac{d}{dx}\left(x^{-p}J_p(x)\right) = -x^{-p}J_{p+1}(x).$ (Although, the identities I gave are attained simply by carrying out the differentiations here!)

With $p=1,$ we have
$$\dfrac{d}{dx}(xJ_1) = xJ_0\implies \int xJ_0 dx = xJ_1.$$
Then the integration by parts is better carried out with $u=x, du=dx, dv=xJ_0dx, v=xJ_1$:
$$\int x^2J_0 dx = x^2J_1 - \int xJ_1dx.$$
Then with $p=0,$ we have
$$\dfrac{d}{dx}J_0 = -J_1\implies \int J_1dx = -J_0.$$
Integration by parts again: $u=x, du=dx, dv=J_1dx, v=-J_0$:
$$\int x^2J_0dx = x^2J_1 + xJ_0 - \int J_0 dx,$$
as desired.
(Sorry that was a typo in the original post.)