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.
Best Answer
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 $N \geq 0$.