# Sum of Bessel functions to the fourth power, $\sum_{k\in\mathbb{Z}} J_k(x)^4$

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.

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.