5
$\begingroup$

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

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

So

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

pics

$\endgroup$
  • 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. $\endgroup$ – Michael Seifert May 31 '16 at 19:25
  • $\begingroup$ I added the brief explanation in the upd. $\endgroup$ – Hayk Hakobyan Jun 1 '16 at 11:27
  • $\begingroup$ @HaykHakobyan your link of $a_0$ may be broken. Please check it. $\endgroup$ – tanghe2014 Oct 10 '17 at 10:57
  • $\begingroup$ @tanghe2014 fixed, thnx. $\endgroup$ – Hayk Hakobyan Oct 10 '17 at 21:28
8
$\begingroup$

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.

$\endgroup$
3
$\begingroup$

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 $c\in \mathbb{Z}$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.