Let $J_k$ denote the $k$-th order Bessel function of the first kind. I know that $$\sum_{k\in\mathbb{Z}} J_{\mu-k}(x) J_{\nu-k}(y) = J_{\mu-\nu}(x-y) \quad \forall x,y\in\mathbb{R},\mu,\nu\in\mathbb{Z},$$ so in particular, $\sum_{k\in\mathbb{Z}} J_k(x)^2 = 1 \ \forall x\in\mathbb{R}$. Now I was wondering whether there is a closed form expression for $$\sum_{k\in\mathbb{Z}} J_k(x)^4,$$ if $x\in\mathbb{R}$. Or, more generally, for $$\sum_{k\in\mathbb{Z}} J_{\mu-k}(x)^2 J_{\nu-k}(y)^2,$$ if $x,y\in\mathbb{R}, \, \mu,\nu\in\mathbb{Z}$.

Obviously, the result must be non-negative and $\leq 1$ in both cases, but I have neither been able to deduce a result myself nor find anything online.


1 Answer 1


After having tried different plausible generalisations of formula (20) and a non-labeled equation towards the bottom of page 6 in this article (kindly pointed out by Gary in the comments), I conclude that it seems to be true that $$\sum_{k\in\mathbb{Z}} J_{\mu-k}(x)^2 J_{\nu-k}(y)^2 = \frac 1 \pi \int_0^\pi J_{\mu-\nu}\left(\sqrt{x^2 + y^2 - 2xy \cos \vartheta}\right)^2 \mathrm{d} \vartheta$$ if $x, y > 0$ (for $x, y \leq 0$ just use $J_k(x)^2 = J_k(-x)^2$) and $\mu,\nu\in\mathbb{Z}$. Note that I have not derived this equation, but checked that it agrees with explicit numerical evaluations for various arguments.


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