Calculating sum of BesselJ[n, x]

Mathematica returns

Sum[BesselJ[n, x]^2/(n - k), {n, -Infinity, Infinity}]

unevaluated. However, BesselJ[n, x]^2 and BesselJ[-n, x]^2 are equal, so the Sum can be rewritten as

Simplify[-BesselJ[0, x]^2/k + 2 k Sum[BesselJ[n, x]^2/(n^2 - k^2), {n, 1, Infinity}]]
(* -π BesselJ[-k, x] BesselJ[k, x] Csc[k π] *)

which is the desired result. Its plot, here for k == .5, is

Plot[Evaluate[% /. k -> .5], {x, 0, 10}, PlotRange -> All, AxesLabel -> {x, sum}]

The sum converges quite rapidly, at least for k == .5. Define

f[nmax_, x_, k_] := Sum[BesselJ[n, x]^2/(n - k), {n, -nmax, nmax}]
Plot[Evaluate[Table[f[nm, x, .5], {nm, 0, 5, 1}]], {x, 0, 10}, 
    PlotRange -> All, AxesLabel -> {x, sum}]

The curve for nm == 3 is essentially identical to that for the symbolic Sum. Convergence probably is not so rapid for larger k.

Using Laplace transform: $$\mathcal{L}_t[J_0(a \sinh (b t))](s)=\frac{I_{\frac{s}{2 b}}\left(\frac{a}{2}\right) K_{-\frac{s}{2 b}}\left(\frac{a}{2}\right)}{b}$$ where I'm found in this Book on page 276 example 1.

Using identity:

$$a \sin (b t)=-i a \sinh (i b t)$$

all combining together:

 (-I)*(BesselI[s/(2 b), a/2] BesselK[-(s/(2 b)), a/2])/b /. a -> -I*2*x /. 
 b -> I/2 /. s -> I*k

-2 BesselI[k, -I x] BesselK[-k, -I x]

Sum[BesselJ[n, x]^2/(n - k), {n, -Infinity, Infinity}] == Re[-2 BesselI[k, -I x] BesselK[-k, -I x]]

$$\sum _{n=-\infty }^{\infty } \frac{J_n(x){}^2}{n-k}=\Re(-2 I_k(-i x) K_{-k}(-i x))$$

Comment:

Formula works for:$x\in \mathbb{R}$ and $k=c+\frac{1}{2}$ where $c\in \mathbb{Z}$