Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

As a follow up to this question I am also interested in a symbolic closed form for this integral

$$\int_0^\infty d r \,r^2\, j_{n_1}( k_1 r)\, j_{n_2}( k_2 r)\, j_{n_3}( k_3 r)\,, $$ where $j_n(r)$ is the $n^{\rm th}$ order spherical Bessel function, $k_1$,$k_2$ and $k_3$ are real positive numbers and $n_1,n_2$ and $n_3$ are positive integers.

The spherical Bessel function $j_n$ can be defined by $$ j_n(x) = (-x)^n \left(\frac{1}{x}\frac{d}{dx}\right)^n\,\frac{\sin x}{x}.$$


As an answer to this question, @joriki provided a nice solution for $n_1=n_2=n_3=0$.

If I am to believe Mathematica again, for instance $$\int_0^\infty d r \,r^2\, j_2( r)\, j_2( 2 r)\, j_2( 3 r)=-\frac{\pi}{48}$$ and $$\int_0^\infty d r \,r^2\, j_2( r)\, j_4( 2 r)\, j_4( 3 r)=-\frac{\pi}{48}$$ so the integral seems possible. On the other hand, if some $n_i$ are odd the integral seems ill defined.

I would guess that for odd indices the answer is $\pi/(8k_1 k_2 k_3)$ times some function of the signs of $n_1$, $n_2$ and $n_3$.


My guess seems to be wrong. Symbolic integration for the first $8\times8\times 8 $ values of $(n_1,n_2,n_3)$ yields (with $k_1=k_2=k_3=1$)


share|improve this question
You should try to use integration by parts on the Bessel function given by the mentioned formula. –  Phira Nov 27 '12 at 17:22
@Phira thanks for the advice; it seems difficult to do in practice since 3 such functions are involved?? –  chris Nov 27 '12 at 17:26

1 Answer 1

R. Mehrem gives this as equation 5.14 in the paper The Plane Wave Expansion, Infinite Integrals and Identities involving Spherical Bessel Functions (arXiv:0909.0494v4). It looks like this:

Mehrem paper screenshot

The symbols in curly braces are Wigner 6j-symbols, and the angle-bracketed letters are Clebsch-Gordan coefficients. The P_l is a Legendre polynomial, and Delta is k1^2+k2^2-k3^2/(2k1k2), coming from applying law of cosines to a triangle in k-space (it is just cos theta_12)). The beta function is zero unless the k1, k2, k3 combination is such that it forms a closed triangle (i.e. the side lengths satisfy the triangle inequality).

share|improve this answer
Please summarize the relevant information from the paper. –  Null Dec 17 at 7:01
Thank you for your answer. –  chris Dec 17 at 9:33

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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