The following sums pop up in diffraction theory and are related to Lommel's function of two variables. Let $u,v\in\mathbb{R}$. I claim that $$\sum_{n=0}^\infty i^n \left ( \frac{u}{v} \right )^n J_{n+1}(v)=\text{sign}(u)e^{iu/4}\sqrt{\frac{2\pi}{u}}\sum_{n=0}^\infty i^n (2n+1)J_{n+1/2}\left ( \frac{u}{4} \right )J_{2n+1}(v),$$ where $J_a(b)$ is the Bessel J.

Can anyone show this?

The left hand side is related to Lommel's original definition and the right hand side was derived by Zernike and Nijboer in 1947 (published 1949). However, Zernike and Nijboer are light on detail.

The expansion on the RHS may somehow be related to the Bauer/Rayleigh expansion: $$e^{ikr\cos\theta}=\sum_{n=0}^\infty i^n(2n+1)j_n(kr)P_n(\cos\theta),$$ where $j_n$ is the spherical Bessel j and $P_n$ is the $n$th Legendre polynomial.

Also, I added the $\text{sign}(u)$ myself in order to make things fit numerically. The expression published by Boersma (where I got the expressions) is missing the signum and doesn't seem to work out for negative u.


1 Answer 1


Let us introduce an integral kernel $$K(v,\theta)=\frac{v}{4}\,J_0\left(v\sin\frac{\theta}{2}\right),\qquad \theta\in[0,\pi].$$ The proof of your formula boils down to two identities involving $K(v,\theta)$.

Lemma. For $n\in\mathbb{Z}_{\ge0}$, one has \begin{align*}\displaystyle J_{2n+1}(v)&=\int_0^{\pi}K\left(v,\theta\right)P_n\left(\cos\theta\right)\sin\theta\,d\theta,\tag{A}\\ v^{-n}J_{n+1}(v)&=\int_0^{\pi}K\left(v,\theta\right)\frac{\cos^{2n}\frac{\theta}{2}}{2^n n!}\,\sin\theta\,d\theta.\tag{B} \end{align*}

Indeed, substituting (A) into the right side of your formula and using the Rayleigh plane wave expansion, we get (assuming that $u>0$) \begin{align*} \text{R.H.S.}&=e^{iu/4}\sqrt{\frac{2\pi}{u}}\sum_{n=0}^\infty i^n (2n+1)J_{n+1/2}\left ( \frac{u}{4} \right )J_{2n+1}(v)\stackrel{\color{green}{\text{(A) and Rayleigh}}}{=}\\ &=\int_0^{\pi}K\left(v,\theta\right)e^{iu\left(1+\cos\theta\right)/4}\sin\theta\,d\theta\;\;\;\quad\stackrel{\color{blue}{\text{expand in }u}}{=}\\ &=\sum_{n=0}^{\infty}i^nu^n\int_0^{\pi}K\left(v,\theta\right)\frac{\cos^{2n}\frac{\theta}{2}}{2^n n!}\,\sin\theta\,d\theta\;\stackrel{\color{red}{\text{use (B)}}}{=}\\ &=\sum_{n=0}^\infty i^n \left ( \frac{u}{v} \right )^n J_{n+1}(v)=\\ &=\text{L.H.S.} \end{align*}

Proof of the Lemma:

For example, to demonstrate (B), expand the right side in powers of $v$ using the series representation of $J_0(z)$. The corresponding coefficients will be given by integrals with respect to $\theta$ that can be expressed in terms of Euler's beta function after the change of variables $t=\sin^2\frac{\theta}{2}$. It then suffices to compare them with the coefficients of the expansion of $v^{-n}J_{n+1}(v)$.

The proof of (A) is analogous: expand the right side in powers of $v$, compute the corresponding integrals and identify them with the coefficients of the left side. $\blacksquare$

  • $\begingroup$ Thanks for a great answer. How about the sign(u) on the R.H.S? Does it belong? $\endgroup$
    – SDiv
    Jun 5, 2015 at 7:42
  • $\begingroup$ @SDiv Actually at both sides of the relation we have entire power series of $u$ so the sign(u) is actually absent. That it appears in your computations is, I think, related to some stupid mathematical software which does not treat multivalued functions correctly. For example, if you ask Mathematica to compute $\sqrt{ \frac{\pi}{2u} }J_{1/2}(u)$ for $u=\pm 0.00001$, it will return two different values $\pm 1$ . But this function is nothing but $\frac{\sin u}{u}$, so the correct value is $1$. So instead of dividing two factors of $i$ produced by the two square roots, Mathematica multiplies them. $\endgroup$ Jun 5, 2015 at 9:12
  • $\begingroup$ @L.G.: Instructive answer! (+1) $\endgroup$
    – epi163sqrt
    Jun 5, 2015 at 18:55

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