I am interested in the following coefficients that are related to the Fourier expansion of the constant function 1 in a Bessel basis.

Define $J_n:\mathbb{R}\to \mathbb{R}$ as the $n$'th order Bessel function of the first kind. Define $\alpha_i\in \mathbb{R}^+$ as the $i$'th positive root of $J_0$.

Let $$z_i = \int_{r = 0}^1 J_0(\alpha_i r) r\, \mathrm{d} r.$$ Mathematica says that

$$z_i = \frac{J_1(\alpha_i)}{\alpha_i}.$$ Is this as close to "closed form" as I can get?

I have looked in the Wolfram functions page for identities of this type but nothing seemed to fit the bill. Any ideas?

  • 1
    $\begingroup$ How much nicer of an expression could you want? :-) $\endgroup$ – parsiad Sep 24 '15 at 23:12

Recall the identities

$$\begin{align} xJ_0(x)&=xJ_1'(x)+J_1(x) \tag 1\\\\ J_0'(x)&=-J_1(x) \tag 2 \end{align}$$

Then, we have from $(1)$ and $(2)$

$$xJ_0(x)=\left(xJ_1(x)\right)' \tag 3$$

Using $(3)$ reveals

$$\begin{align} \int_0^1J_0(ar)\,r\,dr&=\frac{1}{a}\int_0^1\frac{d\,\left(rJ_1(ar)\right)}{dr}\,dr\\\\ &=\frac{1}{a}J_1(a) \end{align}$$

Letting $a=\alpha_i$ recovers the result sought in the OP.

As requested in comment, the evaluation of the integral


is easily facilitated by the identity

$$\frac{d}{dx} \left(\frac12 x^2 \left(J_0^2(x)+J_1^2(x)\right)\right)=xJ_0^2(x) $$

and the fact the $J_0(\alpha_i)=0$

  • $\begingroup$ Hi, thanks for the response, but this isn't exactly what I was looking for. I am interested in something like $J_1(\alpha_i)/\alpha_i = X$, where $X$ is somewhat closer to a closed form that $J_1(\alpha_i)/\alpha_i$. Does that make sense? I have been fooling around and I think that I have $$J_1(\alpha_i)^2 = 2 \int_{r=0}^1 J_0(\alpha_i r)^2 r \mathrm{d}r$$ Do you know where that could have come from? $\endgroup$ – fred Sep 24 '15 at 23:56
  • $\begingroup$ I don't see a need to delete it. It is still useful and true. I don't know if someone else will down-vote it thought, I won't. $\endgroup$ – fred Sep 25 '15 at 0:01
  • $\begingroup$ @fred Sure. I've added. Please let me know how else I can help. $\endgroup$ – Mark Viola Sep 25 '15 at 0:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.