Sum of square of Bessel functions I'm working on the 1-D continuous quantum walk and I encounter the following problem:
Let $J_n(x)$ be the Bessel function of the first kind of order $n$. (In this post, I consider only real $x$ and integers $n$. This makes $J_n(x)$ always real.)
By Neumann’s Addition Theorem (Eq. 10.23.3 in DLMF), it holds that for any real $x$
$$ \sum_{n=-\infty}^\infty J_n^2(x) = 1. $$
In other words, for a fixed $x$, $J_n^2(x)$ can be seen as a distribution over the integers.
I'm wondering whether one can prove something like $$ \sum_{n=-0.1x}^{0.1x} J_n^2(x) \leq 1-\Omega(1)  $$ as $x\to\infty$?
 A: Put another way, you're asking about $S_\infty(10)$, where $$S_\infty(x)=\lim_{n\to\infty}S_n(x),\qquad S_n(x)=\sum_{k=-n}^{n}J_k^2(nx).$$
In fact there is a closed form: $$\color{blue}{S_\infty(x)=\begin{cases}\hfill 1,\hfill&|x|\leqslant 1\\\dfrac2\pi\arcsin\dfrac1{|x|},&|x|>1\end{cases}}.$$
Here is a sketch of a proof, using the integral representation $$J_k^2(z)=\frac1\pi\int_0^\pi J_0(2z\sin t)\cos 2kt\,dt.\tag{*}\label{intrep}$$
In turn, this is obtained using known $f(x,t)=\sum_{k\in\mathbb{Z}}J_k(z)t^k=\exp\big[(z/2)(t-1/t)\big]$ as $$\sum_{k\in\mathbb{Z}}J_k^2(x)t^{2k}=\frac1{2\pi i}\oint f(x,tz)f(x,t/z)(dz/z)=I_0\big[x(t-1/t)\big],$$ so that $J_k^2(x)=(1/2\pi i)\oint I_0\big[x(z-1/z)\big]z^{-2k-1}\,dz$, and we put $z=e^{it}$ to get \eqref{intrep}. Thus,
\begin{align*}
S_n(x)&=\frac2\pi\int_0^{\pi/2}J_0(2nx\sin t)\sum_{k=-n}^n\cos 2kt\,dt
\\&=\frac2\pi\int_0^{\pi/2}J_0(2nx\sin t)\frac{\sin(2n+1)t}{\sin t}\,dt
\\&=\frac2\pi\int_0^{(n+\frac12)\pi}J_0\left(2nx\sin\frac{y}{2n+1}\right)\frac{\sin y}{(2n+1)\sin\frac{y}{2n+1}}\,dy
\end{align*}
and, after (not entirely trivial step of) taking the limit $n\to\infty$, $$S_\infty(x)=\frac2\pi\int_0^\infty J_0(xy)\frac{\sin y}y\,dy.$$
This can be evaluated using $J_0(z)=(2/\pi)\int_0^{\pi/2}\cos(z\cos t)\,dt$, interchanging the integrations (which is again not entirely trivial to justify) and $\int_0^\infty\frac{\sin xy}{y}\,dy=\frac\pi2\operatorname{sgn}x$, which yields $$S_\infty(x)=\frac1\pi\int_0^{\pi/2}\big[\operatorname{sgn}(1+x\cos t)+\operatorname{sgn}(1-x\cos t)\big]dt.$$
This is evaluated easily, giving the closed form stated above.
