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Numerical experiments suggest that the following integral identity holds for Bessel functions of the first kind, $$J_2(t) = 12 \int_0^{1/2}\mathrm{d}x\,\cot \pi x \int_0^x \mathrm{d}y\, \cot \pi y \, J_0(ty)\big[J_0(tx)\,J_0(t(1-x-y))-J_0(t(1-x))\,J_0(t(x-y))\big],$$ but so far I have been unable to prove this. Does anyone have a suggestion how to go about this?

Note that by symmetry one could equivalently drop the second term in the brackets and extend the range of $x$ to $(0,1)$, but then one should take the principal value of the integration.

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  • $\begingroup$ remarkable identity; to order $t^2$ both sides of the equation indeed give $t^2/8$. $\endgroup$ May 28, 2021 at 9:51
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    $\begingroup$ With the integral representation of the Bessel function $$J_{2}(t)=\frac{2^{5} t^{2}}{ 3 \pi} \int_{0}^{\pi/2} \frac{\cos^{3/2}(\theta)}{\sin^{5}( \theta)}\sin\left(t-2 \theta+\frac{\theta}{2}\right) e^{-2\ t \cot \theta} d\theta$$ (as found in, e.g., Gradshteyn & Ryzhik) one can get rid of the outer integral and compare the inner integral with (essentially) the integrand of this formula. $\endgroup$ Jun 9, 2021 at 10:29
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    $\begingroup$ After transforming the integration variable to $x=\theta/\pi$ you have a $\int_{0}^{1/2} dx$ , so everything else (except the $\cot \pi x =\cos(\pi x)/\sin(\pi x)$, but including the prefactors) has to be equal to the inner integral. (Sorry for the shortness. Comments are just too short.) $\endgroup$ Jun 9, 2021 at 13:18
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    $\begingroup$ Using the change of variable $\theta=\pi x$ and making $12 \cot (\pi x)$ appear we get an expression for the $y-$ integral. (oops Johannes already answered). $\endgroup$
    – Archie
    Jun 9, 2021 at 13:27
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    $\begingroup$ Sorry, but I still do not understand. That the definite integrals agree, does not mean that the integrands do. $\endgroup$ Jun 9, 2021 at 15:14

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With hindsight the identity is not that magical: the Bessel functions play only a secondary role, in the sense that there is a more general identity for arbitrary differentiable functions $f : [0,1] \to \mathbb{R}$, namely \begin{align} &12 \int_0^{1/2}\mathrm{d}x\,\cot \pi x \int_0^x \mathrm{d}y\, \cot \pi y \, f(y)\big[f(x)\,f(1-x-y)-f(1-x)\,f(x-y)\big] \\ &= -f(1)\,f(0)^2+ 2 \int_0^1\mathrm{d}x\int_0^{1-x} \mathrm{d}y\, f(x)f(y)f(1-x-y).\tag{1} \end{align} Substituting $f(x) = J_0(t\,x)$ and using the Laplace transform $\mathcal{L}\{J_0(t\, x)\}(\xi) = (t^2 + \xi^2)^{-1/2}$, we find that the double convolution integral satisfies $$\int_0^1\mathrm{d}x\int_0^{1-x} \mathrm{d}y\, J_0(t\,x)J_0(t\,y)J_0(t(1-x-y)) = -\frac{1}{t}J_0'(t).$$ Together with $J_2(t) = -J_0(t)-2 J_0'(t)/t$ and $J_0(0)=1$, this reproduces the claimed identity $$J_2(t) = 12 \int_0^{1/2}\mathrm{d}x\,\cot \pi x \int_0^x \mathrm{d}y\, \cot \pi y \, J_0(ty)\big[J_0(tx)\,J_0(t(1-x-y))-J_0(t(1-x))\,J_0(t(x-y))\big].$$

Let's see why (1) holds. For this it is convenient to introduce the integral $$I(\epsilon):=2\int_\epsilon^{1-2\epsilon}\mathrm{d}x\int_\epsilon^{1-x-\epsilon} \mathrm{d}y\, f(x)f(y)f(1-x-y)$$ with $0<\epsilon <1/4$. The meat of the argument is Hermite's cotangent identity, $$\cot(\pi x)\, \cot(\pi y) + \cot(\pi y)\,\cot(\pi(1-x-y)) + \cot(\pi(1-x-y))\cot(\pi x)= 1,$$ which after insertion in the integrand and using the symmetry under permutation of $x$, $y$ and $1-x-y$, implies $$I(\epsilon) = 6\int_\epsilon^{1-2\epsilon}\mathrm{d}x\int_\epsilon^{1-x-\epsilon} \mathrm{d}y\,\cot(\pi x)\cot(\pi y)\, f(x)f(y)f(1-x-y).$$ Since the integrand is symmetric in $x$ and $y$ we can further reduce the domain to $$I(\epsilon) = 12\int_\epsilon^{1-2\epsilon}\mathrm{d}x\int_\epsilon^{\min(1-x-\epsilon,x)}\!\!\!\! \mathrm{d}y\,\cot(\pi x)\cot(\pi y)\, f(x)f(y)f(1-x-y).$$ Folding the region $1/2< x < 1$ onto $0 < x < 1/2$ via the substitution $x\to 1-x$ gives \begin{align} I(\epsilon) =& 12\int_{2\epsilon}^{1/2}\mathrm{d}x\int_\epsilon^{x-\epsilon}\! \mathrm{d}y\,\cot(\pi x)\cot(\pi y)\, f(y)\big[f(x)f(1-x-y)-f(1-x)f(x-y)\big]\\ &+ 12 \int_\epsilon^{1/2}\mathrm{d}x\int_{\max(x-\epsilon,\epsilon)}^{\min(1-x-\epsilon,x)}\! \mathrm{d}y\,\cot(\pi x)\cot(\pi y)\, f(x)f(y)f(1-x-y). \end{align} Both $I(\epsilon)$ and the first term on the right-hand side have limits given by the corresponding terms in (1) as $\epsilon \to 0$. It remains to examine the last integral on the right-hand side. By the change of variables $x = \epsilon\, \hat{x}$ and $y = \epsilon\,\hat{y}$ we see that it has a limit as $\epsilon\to 0$ given by \begin{align} &\frac{12 f(1)f(0)^2}{\pi^2}\int_1^\infty \mathrm{d}x\int_{\max(x-1,1)}^x \mathrm{d}y \frac{1}{x\,y} \\ &=\frac{12 f(1)f(0)^2}{\pi^2}\int_1^\infty \mathrm{d}y\int_{y}^{y+1} \mathrm{d}x \frac{1}{x\,y} \\ &= \frac{12 f(1)f(0)^2}{\pi^2} \int_{1}^\infty \frac{1}{y^2}\log\left(\frac{y+1}{y}\right) = f(1)f(0)^2. \end{align} This finishes the proof of (1).

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  • $\begingroup$ Well done! I still find it noticeable in that $J_2$ pops up alone, without any t-dependent factor. $\endgroup$
    – Archie
    Jun 9, 2021 at 21:27
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    $\begingroup$ perhaps not magical, but certainly impressive! $\endgroup$ Jun 9, 2021 at 21:47

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