I've got a problem with proving this Bessel's equality: $$x^2 = 2\sum_{n=1}^{\infty} (2n)^2 J_{2n}(x)$$
The Bessel generating function is $\exp(\frac{x}{2}(t-t^{-1})) = \sum_{n=-\infty}^{\infty}J_{n}(x)t^n$.
I think the solution should is replace $t$ with some other forms and proving the equality by comparing the coefficient, but I can't find a way out.
Use the following: $$ e^{ix\cos t}=\sum_{n=-\infty}^\infty J_n(x)e^{int} $$ Differentiate both sides twice with reapect to $t$: $$ -(x^2\sin^2t+ix\cos t)e^{ix\cos t}=-\sum_{n=-\infty}^\infty n^2 J_n(x)e^{int} $$ taking $t=\dfrac\pi 2$ and combine $n$ with $-n$ terms and using $J_{-n}(x)=(-1)^nJ_n(x)$ we get: $$ x^2=\sum_{n=1}^\infty n^2 J_n(x)[e^{in\pi/2}+(-1)^ne^{-in\pi/2}] $$ The expression in the square bracket vanishes for odd $n$ and is $2$ otherwise, so we get: $$ x^2=2\sum_{n=1}^\infty (2n)^2 J_{2n}(x) $$