My friend has a sum in his research paper that looks like this

$$ \sum_{n=-\infty}^{\infty}\frac{J_n^2(x)}{n-\kappa}. $$

He was able to calculate this sum analytically, by substituting the denominator with an integral. I wonder, is there a way to calculate this in Mathematica by adjusting some options in the Sum[BesselJ[n, x]^2/(n - К), {n, -Infinity, Infinity}].

Thank you in advance!

UPD People were interested in how to take this sum analytically. So if you substitute

$$ \sum_{n=-\infty}^{\infty}\frac{J_n^2(x)}{n-\kappa} = \sum_{n=-\infty}^{\infty}J_n^2(x)(-i)\int_0^\infty e^{-i(\kappa-n)t}\mathrm{d}t =, $$

where we can now exchange the integration and summation and use $n\to-n$ symmetry

$$ =(-i)\int_0^\infty\mathrm{d}t\left\{\sum_{n=1}^{\infty}J_n^2(x)\left(e^{-i(\kappa-n)t}+e^{-i(\kappa+n)t}\right)+J_0^2(x)e^{-i\kappa t}\right\}. $$

Then we take out the $e^{-i\kappa t}$ and get $2\cos{nt}$, so the whole sum becomes

$$ \sum_{n=-\infty}^{\infty}\frac{J_n^2(x)}{n-\kappa} = (-i)\int_0^\infty\mathrm{d}t\left[e^{-i\kappa t}\left\{2\sum_{n=1}^{\infty}J_n^2(x)\cos{nt}+J_0^2(x)\right\}\right], $$

and now can be easily taken

$$ \sum_{n=-\infty}^{\infty}\frac{J_n^2(x)}{n-\kappa} =(-i)\int_0^\infty e^{-i\kappa t}J_0\left(2x\sin{\left({t\over 2}\right)}\right)\mathrm{d}t. $$

Now we have a geometrical progression $a_{n+1}=qa_{n}$, where

$$ a_0=\int_0^{2\pi} e^{-i\kappa t}J_0\left(2x\sin{\left({t\over 2}\right)}\right)\mathrm{d}t,~~\text{and}~~q =e^{-2\pi i \kappa}. $$


$$ \sum_{n=-\infty}^{\infty}\frac{J_n^2(x)}{n-\kappa} =(-i)\sum_{n=0}^{\infty}a_n=i\frac{a_0}{q-1}. $$ $a_0$ can be found using the integrals shown below on pic, but still Mathematica is not helpful to take it.


  • 1
    $\begingroup$ What do you mean by "substituting the denominator with an integral"? I mean, you can do Integrate @@ Sum[ ... ], but that probably won't be equal to the original expression and in any event Mathematica doesn't evaluate that either. $\endgroup$ May 31, 2016 at 19:25
  • $\begingroup$ I added the brief explanation in the upd. $\endgroup$
    – hayk
    Jun 1, 2016 at 11:27
  • $\begingroup$ @HaykHakobyan your link of $a_0$ may be broken. Please check it. $\endgroup$
    – tanghe2014
    Oct 10, 2017 at 10:57
  • $\begingroup$ @tanghe2014 fixed, thnx. $\endgroup$
    – hayk
    Oct 10, 2017 at 21:28

2 Answers 2


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}]

enter image description here

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}]

enter image description here

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]]

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


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


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