# $\mathrm{Bessel}^3$ Integral

I'm trying to calculate the following integral: $\int_0^\infty \mathrm{BesselJ}[l_0,k_0r] \cdot \mathrm{BesselJ}[l_1,k_1r] \cdot \mathrm{BesselJ}[l_0-l_1,kr] \cdot r\,dr$

($\mathrm{BesselJ}[n,x]$ is the Bessel function of the first kind of order $n$) I assume that $l_0,l_1$ are integers, and that $k_0,k_1,k>0$.

The result has nice implications in quantum mechanics - to explain selection rules.

I was able to prove that the integral is zero when $k,k_0,k_1$ cannot be the lengths of a triangle. So there is a non-zero result when $|k_0-k_1| < k < k_0+k_1$ However, I don't know how to calculate the integral when it is not zero. Can anyone help?

I tried using Mathematica to get a numerical answer (which is fine for me), but I do not think I can count on it. Mathematica is giving a non-zero result for the regime where the integral should be zero.

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You should look up Watson's book: pages 411-412 have a very similar vanishing theorem, plus resulting 3-fold product integrals evaluated in terms of the area $\Delta$ of your triangle. (In particular when $l_0=l_1=0$ the integral is $1/2\pi\Delta$.) – Francois Ziegler Jan 13 '14 at 1:33
And when $l_0=l_1=n$, the cited paper of MacDonald (p. 144) gives the result in a nicer form than both Watson (p. 412) and Auluck (eq. 17):$$\int_0^\infty J_n(ar)J_n(br)J_0(cr)r\,dr=\begin{cases}0 &\text{if } c^2<(a-b)^2\\\dfrac{\cos(nC)}{\pi ab\sin C} &\text{if }c^2 = a^2+b^2-2ab\cos C\\0&\text{if }c^2>(a+b)^2\end{cases}$$ where I have used that $P^{\frac12}_{n-\frac12}(\cos C) = \sqrt{\frac 2{\pi\sin C}}\cos(nC)$. – Francois Ziegler Jan 13 '14 at 6:29

In fact, precisely your integral has been computed in closed form, in: Annie Gervois and Henri Navelet, Some integrals involving three Bessel functions when their arguments satisfy the triangle inequalities, J. Math. Phys. 25 (1984), no. 11, 3350–3356. Their result is $$\int_0^\infty J_m(ar)J_n(br)J_{m+n}(cr)r\,dr = \begin{cases} 0&\text{if }c^2 < (a-b)^2\\ \dfrac{\cos(mB-nA)}{\pi ab\sin C}&\text{if }c^2 = a^2+b^2-2ab\cos C\\ 0&\text{if }c^2 > (a+b)^2 \end{cases}$$ where in the second case, $A, B, C$ are the angles opposite to sides $a, b, c$ of the resulting triangle. (They give formulas valid for real $m, n$, which reduce to the above when $m, n\in\mathbf Z$ as you assume.)

To put this in your notation, take $(m,n,a,b,c)=(l_0,-l_1,k_0,k_1,k)$ and use that $A+B+C=\pi$ and $J_n=(-1)^nJ_{-n}$. We get $$\int_0^\infty J_{l_0}(k_0r)J_{l_1}(k_1r)J_{l_0-l_1}(kr)r\,dr = \begin{cases} 0&\text{if }k < |k_0-k_1|\\ \dfrac{\cos([l_0-l_1]K_1-l_1K)}{\pi k_0k_1\sin K}&\text{if }|k_0-k_1|<k<|k_0+k_1| \\ 0&\text{if }k > |k_0+k_1| \end{cases}$$ where $K_0,K_1,K$ are the angles opposite to $k_0,k_1,k$. The denominator is $2\pi$ times the triangle's area.

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This is even better than I expected! So beautiful that there is an exact analytic solution. Thank you – Mikhael Jan 17 '14 at 6:47
@Mikhael I agree, that formula is pretty cool. Glad you sent me looking for it. By the way, I'm curious about your application. Could you post links to it (or background) if/when available? – Francois Ziegler Jan 17 '14 at 6:58

This triple-Bessel integral was studied in much detail by S.K.H. Auluck, On the Integral of the Product of Three Bessel Functions over an Infinite Domain. Closed form expressions are given if two the three orders coincide. For the more general case, a reliable way to evaluate this integral using Mathematica is described and tested. The numerical difficulties arise because the integral diverges when $k\rightarrow|k_0\pm k_1|$.

When considered as a function of $k_0,k_1,k$, the triple-Bessel integral is sharply peaked at $k\rightarrow|k_0\pm k_1|$, and can be approximated (in the sense of a distribution) by a sum of delta functions at these points:

Applying that to your special case (where one order is the sum of the other two orders), I arrive at the following delta-function approximation:

$$\int_0^\infty \mathrm{BesselJ}[l_0,k_0r] \cdot \mathrm{BesselJ}[l_1,k_1r] \cdot \mathrm{BesselJ}[l_0-l_1,kr] \cdot r\,dr\approx$$

$$\frac{1}{2\sqrt{k_1 k}}\left[\delta(k_0-k_1-k)+(-1)^{l_0-l_1}\delta(k_0-k_1+k)H(k_1-k)+(-1)^{l_1}\delta(k_0-k+k_1)H(k-k_1)\right]$$

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Thanks for the help. I found both the special case by Francois Ziegler and the approximation of the full problem by Carlo Beenakker very helpful. – Mikhael Jan 15 '14 at 4:33