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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?

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

$$J_1^2(\alpha_i)=2\int_0^1J_0^2(\alpha_ir)d\,dr$$

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 $J_0(\alpha_i)=0$

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  • $\begingroup$ 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

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