Let $n\geq 0$ be an integer and let $J_n=J_n(r)$ denote the usual Bessel function (of the first kind) of order $n$ i.e. one of the solutions to Bessel's differential equation


I am interested in the quantities

$$\beta_n:=\int_0^\infty J_n(r)^2 J_0(r)^2 dr$$ which arise naturally in the study of certain classes of oscillatory integrals. In particular, I would like to explore the convexity of the sequence $\{\beta_n\}$. My question is the following: is it true that

$$\beta_n<\frac{\beta_{n-1}+\beta_{n+1}}{2}$$ for every $n\geq 1$? This inequality seems to be numerically verifiable (Mathematica) for small values of $n$. To prove it, I was trying to use the so-called ''Neumann's formula'' (see e.g. Watson's treatise (1966), p.32)

$$J_n(r)^2=\frac{1}{\pi}\int_0^\pi J_{2n}(2r\sin\theta)d\theta,$$ together with the well-known recurrence

$$J_{n-1}(r)-J_{n+1}(r)=2 \frac{d J_n}{dr}(r)$$ and some integration by parts. Unfortunately I wasn't able to make this approach work.

Ideas or/and references are welcome, thank you very much in advance.


I fully suspect the answer to your question is yes. The reason is as follows. \begin{eqnarray} \beta_{n-1} + \beta_{n+1} &=& \int [J_{n+1}^2 + J_{n-1}^2] J_0^2 \nonumber \\ &=& \int [(J_{n+1} + J_{n-1})^2 - 2 J_{n+1} J_{n-1}] J_0^2 \nonumber \\ &=& \int [\frac{4 n^2}{r^2} J_n^2 - 2 J_{n+1} J_{n-1}] J_0^2 \end{eqnarray} where we we use the recurrence relation $2 \frac{n}{2} J_n = J_{n-1} + J_{n+1}$ to get the last inequality. Notice that since $$\int 2J_{n+1} J_{n-1}J_0^2 \leq \beta_{n+1} + \beta_{n-1}$$ we have that $$ \beta_n + \int (\frac{n^2}{r^2} - 1) J_n^2 J_0^2 \leq \frac{\beta_{n-1} + \beta_{n+1}}{2} $$ You now just need that one equation to be non-negative to get the answer you want. This quantity should be non-negative by the simple fact that $\frac{n^2}{r^2} - 1$ is positive near $r=0$ and this is where the Bessel functions have most of their mass. Further, $\frac{n^2}{r^2} - 1$ is negative where the Bessel function is small. I'm not sure how to get this rigorously though.

I hope this helps.

  • $\begingroup$ Thank you! Your heuristics are insightful but I think you're missing one point: the function $J_n(r)$ starts oscillating only around $r\sim n$, until then it looks like a convex function with absolute value very close to 0. And in fact I just checked (numerically) that the integrals $\int (n^2/r^2-1)J_n^2 J_0^2$ are negative for small values of $n$. I also suspect that my original question should have an affirmative answer, but something rather subtle seems to be going on... Any further thoughts? Thanks again. $\endgroup$ – user17240 Apr 12 '14 at 15:23
  • 2
    $\begingroup$ That's interesting. I didn't think the estimate would have been that damaging. The only place where we estimated was to replace $2ab$ with $a^2 + b^2$. If we don't estimate, then we get $2ab = a^2 + b^2 - (a - b)^2$. Setting $a = J_{n+1}$ and $b=J_{n-1}$ and using the recurrence relation $2 J_n' = J_{n-1} - J_{n+1}$ we get $$\beta_n + \int\{(\frac{n^2}{r^2} - 1)J_n^2 + J_n'^2\}J_0^2 = \frac{\beta_{n-1} + \beta_{n+1}}{2}$$ It seems interesting that one gets that $(\frac{n^2}{r^2} - 1)J_n^2 + J_n'^2$ turns out to be non-negative in an averaged sense. I'll keep thinking about this. $\endgroup$ – k3thomps Apr 12 '14 at 20:19
  • $\begingroup$ I don't think a 2 is missing. I re-did the calculation and what I had above looks correct. $\endgroup$ – k3thomps May 21 '14 at 18:03
  • $\begingroup$ @Wurlitzer and then you have to divide by 4. $\endgroup$ – k3thomps May 22 '14 at 12:43

This is not an answer, just a comment following k3thomps computations in remarks, the question is equivalent to, it seems: why is $$ \int_0^\infty \left(\left(\frac{n^2}{x^2}-1\right) J_n^2 +(J_n^\prime)^2\right)J_0^2\text{d}x> 0 \,? $$ Below is an alternative formulation of the same question (I can't make it work, but here it is anyway). We have, by definition, $$J_n^{\prime\prime}+\frac{1}{x}J_n ^\prime-\left(\frac{n^2}{x^2}-1\right) J_n=0$$ Integrating against $J_nJ_0^2$, $n\geq1$, we obtain \begin{eqnarray*} &&\int_0^\infty \left(\left(\frac{n^2}{x^2}-1\right) J_n^2 + (J_n^\prime)^2\right) J_0^2 \text{d}x\\ &=& \frac{1}{2}\int_0^\infty -(J_n^2)^\prime(J_0^2)^\prime +\frac{1}{x}(J_n^2)^\prime J_0^2\text{d}x\\ &=& \frac{1}{2}\int_0^\infty J_n^2\left((J_0^2)^{\prime\prime} -\left(\frac{1}{x}J_0^2\right)^\prime \right)\text{d}x. \end{eqnarray*}

This is puzzling, because $$(J_0^2)^{\prime\prime} -\left(\frac{1}{x}J_0^2\right)^\prime$$ oscillates on $(0,\infty)$.

Nevertheless, using the idea suggested in the original post, let $Q_{n}\left(r\right)=J_{n}^{2}\left(r\right).$ We have \begin{align*} \int_{0}^{\infty}\left(\frac{1}{r}J_{n}^{\prime}J_{n}+\left(J_{n}J_{n}^{\prime}\left(r\right)\right)^{\prime}\right)J_{0}^{2}\text{d}r & =\frac{1}{2}\int_{0}^{\infty}\left(\frac{1}{r}Q_{n}^{\prime}+Q_{n}^{\prime\prime}\right)Q_{0}\text{d}r\\ & =\frac{1}{2}\int_{0}^{\infty}\left(-\frac{1}{r}Q_{0}+Q_{0}^{\prime}\right)^{\prime}Q_{n}\text{d}r. \end{align*} Now, \begin{align*} Q_{0}\left(r\right) & =\frac{1}{\pi}\int_{0}^{\pi}J_{0}\left(2r\sin\psi\right)\text{d}{\psi},\\ Q_{n}\left(r\right) & =\frac{1}{\pi}\int_{0}^{\pi}J_{0}\left(2r\sin\varphi\right)\cos\left(2n\varphi\right)\text{d}{\varphi}, \end{align*} and \begin{align*} Q_{n}^{\prime} & =\frac{1}{\pi}\int_{0}^{\pi}J_{0}^{\prime}\left(2r\sin\varphi\right)2\sin\varphi\cos\left(2n\varphi\right)\text{d}{\varphi}\\ Q_{n}^{\prime\prime} & =\frac{1}{\pi}\int_{0}^{\pi}J_{0}^{\prime\prime}\left(2r\sin\varphi\right)\left(2\sin\varphi\right)^{2}\cos\left(2n\varphi\right)\text{d}{\varphi} \end{align*} Recall that $$ rJ_{0}^{\prime\prime}(r)+J_{0}^{\prime}(r)+rJ_{0}=0, $$ thus \begin{align*} Q_{n}^{\prime\prime}= & \frac{1}{r}\frac{1}{\pi}\int_{0}^{\pi}J_{0}^{\prime\prime}\left(2r\sin\varphi\right)\left(2r\sin\varphi\right)\left(2\sin\varphi\right)\cos\left(2n\varphi\right)\text{d}{\varphi}\\ = & -\frac{1}{r}\frac{1}{\pi}\int_{0}^{\pi}J_{0}^{\prime}\left(2r\sin\varphi\right)\left(2\sin\varphi\right)\cos\left(2n\varphi\right)\text{d}{\varphi}\\ & -\frac{1}{\pi}\int_{0}^{\pi}J_{0}\left(2r\sin\varphi\right)\left(2\sin\varphi\right)\cos\left(2n\varphi\right)\text{d}{\varphi},\\ \left(\frac{1}{r}Q_{n}^{\prime}+Q_{n}^{\prime\prime}\right) & =-\frac{1}{\pi}\int_{0}^{\pi}J_{0}\left(2r\sin\varphi\right)\left(2\sin\varphi\right)\cos\left(2n\varphi\right)\text{d}{\varphi},\\ \left(\frac{1}{r}Q_{n}^{\prime}+Q_{n}^{\prime\prime}\right)Q_{0} & =-\frac{1}{\pi}\int_{0}^{\pi}J_{0}\left(2r\sin\varphi\right)2\sin\varphi\cos\left(2n\varphi\right)\text{d}{\varphi} \times \\ &\frac{1}{\pi}\int_{0}^{\pi}J_{0}\left(2r\sin\theta\right)d\theta,\\ \frac{1}{2}\int_{0}^{\infty}\left(\frac{1}{r}Q_{n}^{\prime}+Q_{n}^{\prime\prime}\right)Q_{0}\text{d}r & =-\frac{1}{\pi^{2}}\int_{0}^{\pi}\text{d}\theta\int_{0}^{\pi}2\sin\varphi\cos\left(2n\varphi\right)\text{d}{\varphi} \times \\ &\int_{0}^{\infty}J_{0}\left(2t\sin\theta\right)J_{0}\left(2t\sin\varphi\right)\text{d}t \end{align*} Now, it appears that for $\alpha,\beta>0$ $\int_{0}^\infty J_{0}(\alpha r)J_{0}(\beta r) \text{d} r $ is a known quantity, a so-called complete elliptic integral of the first type named $K$, given by $$ \int_{0}^\infty J_{0}(\alpha r)J_{0}(\beta r) \text{d} r = \frac{2}{\pi \max(\alpha,\beta)} K \left(\frac{\min(\alpha,\beta)}{\max(\alpha,\beta)}\right) $$

Now, more DLMF or algebraic manipulations to sort out the subsequent integrals on triangles..to be continued (one day).


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