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I begin with Wikipedia's identity

\begin{equation*} \int_{0}^{\infty} J_{\alpha}(z) J_{\beta}(z) \frac{dz}{z} = \frac{2}{\pi}\frac{sin(\frac{\pi}{2}(\alpha - \beta))}{\alpha^2 - \beta^2} \end{equation*}

found at https://en.wikipedia.org/wiki/Bessel_function#Properties 8 equations down. Now using $j_{\alpha}(z) = \sqrt{\frac{\pi}{2z}} J_{\alpha + \frac{1}{2}}(z)$, we get

\begin{align*} \int_{0}^{\infty} j_{\alpha}(z)j_{\beta}(z)dz &= \frac{\pi}{2} \int_{0}^{\infty} J_{\alpha+\frac{1}{2}}(z)J_{\beta+\frac{1}{2}}(z)\frac{dz}{z} \\ &= \frac{\pi}{2} \frac{2}{\pi} \frac{sin(\frac{\pi}{2}(\alpha - \beta))}{(\alpha+\frac{1}{2})^2 - (\beta+\frac{1}{2})^2} \\ &= \frac{1}{\alpha+\beta+1} \frac{sin(\frac{\pi}{2}(\alpha-\beta))}{\alpha - \beta} \end{align*}

In the limit as $\alpha \to \beta$, calculus gives a limit of $\frac{\pi}{2(\alpha+\beta+1)}$, in general nonzero which would be desirable for an "orthogonality relationship." BUT this doesn't seem to be an orthogonality relationship to me because $sin(\frac{\pi}{2}(\alpha-\beta)) \neq 0$ for $\alpha - \beta$ odd. Only for $\alpha - \beta$ even. Did I do this calculation wrong or is this the sense of "orthogonality" that the spherical Bessel functions have? In other words "nonzero with a certain sign if their difference is odd and zero whenever their difference is even." I could see this being useful then because in practice that would mean, say that when solving for coefficients, you multiply both sides of an equation by $j_0$ which the orthogonality condition would tell you only odd coefficients survive.

1 Answers1

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You are right in your conclusion, however there is a resolution to the apparent paradox.

I base my reasoning on the analogous result in Gradshteyn and Ryzhik, Eq. 6.538.2 in the book. (I use LateX notation for the formulae below, as I don't use Mathjax and I don't have time).

Write $\alpha = \nu + 2n + 1, \beta = \nu + 2m + 1$ where $m,n$ are non-negative integers and $\nu > -1$. We are interested only in $\nu = \pm 1/2$. Then

$\alpha + \beta + 1 = 2\nu + 2(n + m + 1)$, $\alpha - \beta = 2(n - m)$.

Consequently the numerator is always zero unless $n = m$. But in that case the denominator vanishes as well, and your own equation is to be taken in the sense of letting $\alpha \to \beta$ continuously.

Choose $\nu = -1/2$ to generate the relation for all $N = 2n$, choose $\nu=1/2$ for all $N = 2n + 1$. So the orthogonality of the spherical Bessels is confirmed for any integer $N \geq 0$.

  • I think this reasoning just makes clear that the $\alpha$ and $\beta$ of the spherical Bessel functions are orthogonal no matter if both indices are odd or even as long as both are of the same type (odd or even). But the original post is right. Orthogonality is only present within the functions with odd or even indices. – user5415068 Nov 29 '23 at 20:02