Logarithmic spherical Bessel square integral (${\int }_0^{\infty }{j}_{\ell }(x{)}^{2}dx/x=1/[2\ell (\ell +1)]$)

For integer $\ell \ge 1$, put $I_{\ell }={\int }_0^{\infty }{j}_{\ell }(x{)}^{2}dx/x$ and $J_{\ell }={\int }_0^{\infty }{j}_{\ell }(x)j_{\ell -1}(x)dx$. Integrating $j_{\ell }^{\prime }=j_{\ell -1}-(\ell +1)j_{\ell }/x$ gives $J_{\ell }=(\ell +1)I_{\ell }$. For $\ell >1$, integrating $j_{\ell -1}^{\prime }=(\ell -1)j_{\ell -1}/x-j_{\ell }$ also gives $J_{\ell }=(\ell -1)I_{\ell -1}$. For $\ell =1$, the boundary values $j_0(0)=1$ and $j_0(\infty )=0$ give $J_1=1/2$, hence $I_1=1/4$. The recurrence proves the displayed identity. The integral without the square is different; for example ${\int }_0^{\infty }{j}_1(x)dx/x=\pi /4$.