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I have an expression:

$\sum_{j=0}^{p} (-1)^j \frac{(|\ell|+p)!}{(p-j)!(|\ell|+j)!j!} \Big(\frac{\sqrt{2}}{w}\Big)^{2j+\ell}\Big(\frac{2}{a}\Big)^{2j+\ell+1} \int_0^{\infty} x^{2j+\ell+1} e^{-x^2/a} J_{2j+\ell} (x y) dx$

where $p=0,1,2,...$, $\ell = 0, \pm1, \pm2, ...$. I have tested it in MMA and figured out that for any $j$ I can just drop $2j$ in the order of the Bessel function with negligible accuracy loss: $J_{2j+\ell} (x y) \rightarrow J_{\ell} (x y)$. Check it out:

w = 1; a = 4/w^2;
Trr[l_, p_] :=(-1)^p Sum[(-1)^i (Abs[l] + p)!/((p - i)! (Abs[l] + i)! i!) (Sqrt[2]/w)^(2 i + l) (2/a)^(2 i + l + 1) Integrate[x^(2 i + l + 1) Exp[-x^2/a] BesselJ[l, x r], {x, 0, \[Infinity]}], {i, 0, p}]
TrrAct[l_,p_] :=(-1)^p Sum[(-1)^i (Abs[l] + p)!/((p - i)! (Abs[l] + i)! i!) (Sqrt[2]/w)^(2 i + l) (2/a)^(2 i + l + 1) Integrate[x^(2 i + l + 1) Exp[-x^2/a] BesselJ[2 i + l, x r], {x, 0, \[Infinity]}], {i, 0, p}]

Plot[{Evaluate[Abs[TrrAct[0, 2]]^2 - Abs[Trr[0, 2]]^2]}, {r, 0, 5},PlotRange -> All]

which gives very small numbers:

enter image description here

Essentially the difference is just zero, but I need an analytical proof of this. Any ideas would be highly appreciated!

Interesting fact is that the integral itself differs a lot for $J_{\ell} (x y)$ and $J_{2j+\ell} (x y)$, especially for small y. However, when considering the entire expression they are just the same.

In my attempts I went over the recursion relations, tried to flip the summation and integration order for particular values of $p$, expand the Bessel... These didn't seem to work for me.

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  • $\begingroup$ Nemo, yeah, but its rather 6.631.1., not 4. Anyways, I don't need to take an integral, I need to prove that $2j$ in the order of the Bessel function does not change the result for the whole sum. $\endgroup$
    – MsTais
    Commented Jan 9, 2017 at 16:39
  • $\begingroup$ @Nemo Good point, I have corrected the code. It is still small $\sim 10^{-15}$ for any $w$. $\endgroup$
    – MsTais
    Commented Jan 9, 2017 at 18:11

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