[Math] Integral $\int_{-\infty}^\infty J^3_0(x) e^{i\omega x}\mathrm dx $

calculusdefinite integralsintegrationreal-analysisspecial functions

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…

Best Answer

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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