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Hi I am trying to evaluate the integral $$ \mathcal{I}(\omega)=\int_{-\infty}^\infty J^3_0(x) e^{i\omega x}\mathrm dx $$ analytically. We can also write $$ \mathcal{I}(\omega)=\mathcal{FT}\big(J^3_0(x)\big) $$ which is the Fourier Transform of the cube of Bessel function. The Bessel function $J_0$ is given by $$ J_0(x)=\frac{1}{2\pi}\int_{-\pi}^\pi e^{-ix\sin t} \mathrm dt. $$ If it helps, we can represent the cube of the Bessel function by $$ J^3_0(x)=-3\int J^2_0(x) J_1(x) \mathrm dx, \ \ \ \ \ J_1(x)=\frac{1}{2\pi}\int_{-\pi}^\pi e^{i(t-x\sin t)} \mathrm dt. $$ In general $$ J_n(x)=\frac{1}{2\pi}\int_{-\pi}^\pi e^{i(nt-x\sin t)}\mathrm dt. $$ The Fourier Transforms of the Bessel function and its square is given by $$ \mathcal{FT}\big(J_0(x)\big)=\sqrt{\frac{2}{\pi}}\frac{\theta(\omega+1)-\theta(\omega-1)}{\sqrt{1-\omega^2}} $$ and $$ \mathcal{FT}\big(J^2_0(x)\big)=\frac{\sqrt{2}K\big(1-\frac{\omega^2}{4}\big)\big(\theta(-\omega-2)-1\big)\big(\theta(\omega-2)-1\big)}{\pi^{3/2}} $$ where K is the elliptic-K function and $\theta$ is the heaviside step function. However I need the cube...

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    $\begingroup$ You mean $-\infty$ in the lower limit? $\endgroup$ – Ron Gordon Mar 11 '14 at 20:23
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    $\begingroup$ @Jeff This is a very interesting question! I tried to find an answer to it some time ago, but failed. For a specific case $\omega=0$ (also not trivial) you may take a look at math.stackexchange.com/questions/404222/… $\endgroup$ – Vladimir Reshetnikov Apr 2 '14 at 20:13
  • $\begingroup$ @VladimirReshetnikov Well if you have failed, then I guess there is no closed form! This calculation came up in a physics problem so I figured there must be a closed form ;)! Thanks! $\endgroup$ – Jeff Faraci Apr 2 '14 at 22:16
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    $\begingroup$ @ O.L. Why did you delete your solution? $\endgroup$ – Jeff Faraci Apr 6 '14 at 23:07
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    $\begingroup$ Dear @Jeff we both know who does what. The downvote of my answers here and here, both at 15:54, followed by your comment above at 15:55, was particularly ridiculous. In contrast to the downvotes, I have no way of checking that subsequent random upvoting came from you as well - but if yes, this is not a healthy behaviour either. I believe that being fair in our judgements we make this site more useful to others. $\endgroup$ – Start wearing purple Apr 9 '14 at 19:40
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It turns out that the Fourier transform of $J_0^3$ can still be expressed in terms of complete elliptic integrals, but it's considerably more complicated than the formula for ${\cal FT}(J_0^2)$: for starters, it involves the periods of a curve $E$ defined over ${\bf C}$ but (except for a few special values of $\omega$) not over ${\bf R}$.

Assume $|\omega| < 3$, else $I(\omega) = 0$. Then the relevant curve is $$ E : Y^2 = X^3 - \bigl(\frac{3}{4} f^2 + \frac{27}{2} f - \frac{81}{4}\bigr) X^2 + 9 f^3 X $$ where $$ f = \frac12 \bigl( e + 1 + \sqrt{e^2-34e+1} \bigr) $$ and $$ e = \bigl( |\omega| + \sqrt{\omega^2-1} \, \bigr)^2. $$ Let $\lambda_1, \lambda_2$ be generators of the period lattice of $E$ with respect to the differential $dx/y$ (note that these are twice the periods that gp reports, because gp integrates $dx/2y$ for reasons coming from the arithmetic of elliptic curves). Then: if $|\omega| \leq 1$ then $$ I(\omega) = \left|\,f\,\right|^{5/2}\, \left|\,f-1\right| \frac{\Delta}{(2\pi)^2}, $$ where $\Delta = \bigl|{\rm Im} (\lambda_1 \overline{\lambda_2}) \bigr|$ is the area of the period lattice of $E$. If $1 \leq |\omega| \leq 3$ then $$ I(\omega) = \left|\,f\,\right|^{-4}\, \left|\,f-1\right|^5 (3/2)^{13/2} \frac{\Delta'}{(2\pi)^2}, $$ where $\Delta' = \bigl| {\rm Re}(\lambda_1 \overline{\lambda_2}) \bigr|$ for an appropriate choice of generators $\lambda_1,\lambda_2$ (these "appropriate" generators satisfy $|\lambda_1|^2 = \frac32 |\lambda_2|^2$, which determines them uniquely up to $\pm$ except for finitely many choices of $\omega$).

The proof, alas, is too long to reproduce here, but here's the basic idea. The Fourier transform of $J_0$ is $(1-\omega^2)^{-1/2}$ for $|\omega|<1$ and zero else. Hence the Fourier transforms of $J_0^2$ and $J_0^3$ are the convolution square and cube of $(1-\omega^2)^{-1/2}$. For $J_0^2$, this convolution square is supported on $|\omega| \leq 2$, and in this range equals $$ \int_{t=|\omega|-1}^1 \left( (1-t^2) (1-(|\omega|-t)^2) \right)^{-1/2} \, dt, $$ which is a period of an elliptic curve [namely the curve $u^2 = (1-t^2) (1-(|\omega|-t)^2)$], a.k.a. a complete eliptic integral. For $J_0^3$, we likewise get a two-dimensional integral, over a hexagon for $|\omega|<1$ and a triangle for $1 \leq |\omega| < 3$, that is a period of the K3 surface $$ u^2 = (1-s^2) (1-t^2) (1-(|\omega|-s-t)^2). $$ (The phase change at $|\omega|=1$ was already noted here in a now-deleted partial answer.) In general, periods of K3 surfaces are hard to compute, but this one turns out to have enough structure that we can convert the period into a period of the surface $E \times \overline E$ where $\overline E$ is the complex conjugate.

Now to be honest I have only the formulas for the "correspondence" between our K3 surface and $E \times \overline E$, which was hard enough to do, but didn't keep track of the elementary multiplying factor that I claim to be $\left|\,f\,\right|^{5/2}\, \left|\,f-1\right|$ or $\left|\,f\,\right|^{-4}\, \left|\,f-1\right|^5 (3/2)^{13/2}$. I obtained these factors by comparing numerical values for the few choices of $\omega$ for which I was able to compute $I(\omega)$ to high precision (basically rational numbers with an even numerator or denominator); for example $I(2/5)$ can be computed in gp in under a minute as

intnum(x=0,5*Pi,2*cos(2*x/5) * sumalt(n=0,besselj(0,x+5*n*Pi)^3))

There were enough such $c$, and the formulas are sufficiently simple, that they're virtually certain to be correct.

Here's gp code to get $e$, $f$, $E$, and generators $\lambda_1,\lambda_2$ of the period lattice:

e = (omega+sqrt(omega^2-1))^2
f = (sqrt(e^2-34*e+1)+(e+1)) / 2
E = ellinit( [0, -3/4*f^2-27/2*f+81/4, 0, 9*f^3, 0] )
L = 2*ellperiods(E)
lambda1 = L[1]
lambda2 = L[2]

NB the last line requires use of gp version 2.6.x; earlier versions did not directly implement periods of curves over $\bf C$.

For $\omega=0$ we have $e=1$, $f=3$, and $E$ is the curve $Y^2 = X^3 - 27 X^2 + 243 X = (X-9)^3 + 3^6$, so the periods can be expressed in terms of beta functions and we recover the case $\nu=0$ of Question 404222, How to prove $\int_0^\infty J_\nu(x)^3dx\stackrel?=\frac{\Gamma(1/6)\ \Gamma(1/6+\nu/2)}{2^{5/3}\ 3^{1/2}\ \pi^{3/2}\ \Gamma(5/6+\nu/2)}$? .

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  • $\begingroup$ This seems nice, but doesn't quite help me as I don't understand anything. How did you come up with your curve, what is GP? What two dimensional integral do you get for $J^3_0$? Perhaps you can put the integral I am trying to calculate, and what is equal to in terms of elliptic functions. Your proof is very deep, and above my head. I shall award you the bounty if you can clear these things up for me, and also explain it a bit simpler (if possible). Thanks $\endgroup$ – Jeff Faraci Apr 12 '14 at 23:54
  • $\begingroup$ I can try to answer some of these questions: The two-dimensional integral is $$ (2\pi)^{-2} \iint ds \, dt \bigg/ \left( (1-s^2) (1-t^2) (1-(\omega-s-t)^2) \right)^{1/2} $$ over the region where $|s|<1$, $|t|<1$ and $|\omega-s-t|<1$. gp is a computer algebra system, see en.wikipedia.org/wiki/PARI/GP . The gp code together with the formulas let you compute $I(\omega)$ for any $\omega$. Yes, the periods of $E$ can be expressed as multiples of the elliptic functions ${\bf K}(k)$, ${\bf K}'(k)$ for some complex $k$, but these formulas are even more complicated because [cont'd] $\endgroup$ – Noam D. Elkies Apr 13 '14 at 1:11
  • $\begingroup$ [cont'd] to put $E$ in the form $y^2 = (1-x^2) (1-k^2 x^2)$ you need to solve a quartic for $k$, and thus to extract two more square roots. Finally, the path from the K3 surface $$ u^2 = (1-s^2) (1-t^2) (1-(\omega-s-t)^2) $$ to $E \times \overline{E}$ is a sequence of three rational changes of variable whose existence is predicted by the general theory of K3 surfaces (which also says that here $E$ and $\overline{E}$ are $6$-isogenous) but I had to work out explicitly. I don't know how much more I can explain here; in any case I doubt that the formula for $I(\omega)$ can be made much simpler. $\endgroup$ – Noam D. Elkies Apr 13 '14 at 1:18
  • $\begingroup$ Ok, so are you stating this: $$ \mathcal{I}(\omega)=\int_{-\infty}^\infty J^3_0(x) e^{i\omega x}dx=(2\pi)^{-2} \iint ds \, dt \left( (1-s^2) (1-t^2) (1-(\omega-s-t)^2) \right)^{-1/2}? $$ Thanks. I am not sure what the large / means in your double integral that you posted in the comment. Is that a division sign? what is it? Thanks again. I think this problem is too advanced for me now that I am seeing the solution, although I will still award the bounty to you. Great. $\endgroup$ – Jeff Faraci Apr 13 '14 at 2:40
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    $\begingroup$ Yes, your formula is equivalent to what I wrote: the big / is a division sign as you guessed, and is needed because the Fourier transform of $J_0$ itself is $1/\sqrt{1-\omega^2}$, not $\sqrt{1-\omega^2}$; your use of the exponent $-1/2$ has the same effect. Thank you not just for the bounty but also for calling my attention to the Fourier-transform problem: I hadn't thought to look for K3 periods in this context. I should at least be able to post some more details of the numerical example of $\omega = 2/5$, and also to show how the calculation works for some $\omega\in(1,3)$ such as $4/3$. $\endgroup$ – Noam D. Elkies Apr 13 '14 at 3:00

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