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.