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I am trying to compute $$\int d^Dq \; e^{iq\cdot\left(r-r'\right)}\cos\left(c_L\left|t-t'\right|\left|q\right|\right),$$ that is, the $D$-dimensional inverse Fourier transform of $\cos\left(a\left|q \right|\right),\;a>0$. I am interested in $D=1,2,3$ and am unskilled with distributions. I have computed the $D=1$ case to be $$\pi\delta\left(\left|r-r'\right|-c_L\left|t-t'\right|\right)+\pi\delta\left(\left|r-r'\right|+c_L\left|t-t'\right|\right).$$

Edit: When transformed into spherical coordinates, the integrals become:

$D=1:$ $$2\int_0^{\infty}dq\;\cos\left(\left|r-r'\right|q\right)\cos\left(c_L\left|t-t'\right|q\right)$$ $D=2:$ $$2\pi\int_0^{\infty}dq\;q\,J_0\left(\left|r-r'\right|q\right)\cos\left(c_L\left|t-t'\right|q\right)$$ $D=3:$ $$4\pi\int_0^{\infty}dq\;q\,\sin\left(\left|r-r'\right|q\right)\cos\left(c_L\left|t-t'\right|q\right)$$

S. Thornton
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    Let $\mathcal F[\phi_D] = \int_{\mathbb R^D} \phi_D e^{i \xi \cdot q} dq$ for a test function $\phi_D$. The FT of $\cos r$ will be essentially the same as here: $$\mathcal F[\cos r] = -2 \pi (1 - \rho^2)^{-3/2} [\rho < 1], \quad D = 2, \ \mathcal F[\cos r] = 8 \pi^2 \delta'(1 - \rho^2), \quad D = 3.$$ The action of these on $\phi_D$ is defined by taking $\phi(x) = \int_{\mathcal S_{D - 1}(x)} \phi_D dS$ and applying $\delta'(1 - x^2)$ or $$((1- x^2)^{-3/2} [x < 1], \phi) = \int_0^1 (1 - x^2)^{-3/2} (\phi(x) - \phi(1)) , dx$$ to $\phi$. – Maxim Nov 29 '20 at 12:35

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Spherical coordinates are good candidates to evaluate these integrals. The $d$-dimensional measure is given by $d^d k = k^{d-1}d^{d-1}\Omega\,dk$, where $d^{d-1}\Omega$ is the surface measure on the sphere $S^{d-1}$ and $k=|k|$. For example, in 3 dimensions, the integral would be $$\int_0^\infty k^2 dk \int_0^{2\pi}d\phi \int_0^\pi \sin \theta \,d\theta \,e^{ik|r-r'|\cos\theta} \cos(c_L|t-t'|k),$$ where I have chosen the $z$-axis as parallel to the $\vec{r}-\vec{r}'$ vector so that $\vec{k} \cdot (\vec{r}-\vec{r}')= k|r-r'|\cos\theta$.

Edit: For $d=3$, your integral is $$-2\pi\frac{d}{d|r-r'|}\int_{-\infty}^\infty \cos\left(|r-r'|q\right)\cos\left(c_L|t-t'|q\right)\,dq.$$ This is formally the same integral as for the case $d=1$, with an additional derivative. In the final result, you will have $\delta'$, the distributional derivative of $\delta$, defined by $\langle-\delta', f\rangle = \langle\delta, f'\rangle$, where $\langle g, h\rangle$ denotes the inner product $\int_{-\infty}^\infty g(x)\,h(x)\,dx$, and $f\in C^\infty_c$ is a smooth function with compact support.

Diffusion
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  • Hi Zachary, thanks. I have tried transforming to spherical coordinates, and none of the resulting integrals converge (formally). They need to be done in a distributional sense. See my edit. – S. Thornton Nov 28 '20 at 19:53
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    Hi S. Thornton, I have edited my post to help address your uncertainties. Although you are right that these integrals should be understood in the distributional sense, they can be computed as regular Riemann integrals. – Diffusion Nov 28 '20 at 20:12
  • Great, thank you. I will think about this more. – S. Thornton Nov 28 '20 at 20:16
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    It's unclear though how you define $\rho^2 (1 - \rho^2)^{-1/2} \delta(1 - \rho^2)$ or $(1 - \rho^2)^{-3/2} [\rho < 1]$, that is, what the result of applying them to a test function $\phi: \mathbb R^2 \to \mathbb R$ is. – Maxim Nov 29 '20 at 21:52
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    If your distribution is $\phi(r)$, I define its action on a test function $f$ as $$\phi(f) = \int \phi,f d^dr.$$ The $\delta$ part of the inverse fourier transform for $d=2$ is rather singular near $v=c_L$, and would require that the test function satisfies $f=o\left(\sqrt{1-v^2/c_L^2}\right)$ as $v\to c_L$. I'm not sure if this is correct though, and I'd be happy to be corrected and learn something. – Diffusion Nov 29 '20 at 22:26
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    If $\cos r$ is understood as a tempered distribution, its Fourier transform is, by definition, a tempered distribution. Your result isn't, since its action on $f$ is defined only for some $f$ from the Schwartz space. – Maxim Nov 29 '20 at 23:12
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    I see yeah, I was concerned about that. Where can I find the derivation for the fourier transform of $\cos r$ so that I can see where it diverges from mine? Thank you for helping me with this. – Diffusion Nov 30 '20 at 00:55
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    I don't have a reference for these particular formulas. Gelfand, Shilov, Generalized functions, Vol. 1 has a section on the FT of distributions with finite support. Since $\mathcal F[\cos r]$ always has finite support, it might be easier to show that the inverse transform gives $\cos r$. – Maxim Nov 30 '20 at 02:43