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In the context of a cosmology text, I have found the following function, called a spatial top-hat filter:

$$W_{TH,R}(r)=\dfrac{3\theta(R-r)}{4\pi R^3}$$

It is claimed that this leads to:

$$W_{TH,R}(r)=3\dfrac{\sin(kR)-(kR)\cos(kR)}{(kR)^3}$$

which I assume is the Fourier transform of the expression given above. I want to be able to reproduce this result, so I attempted the following:

$$W_{TH,R}(k)=\int d^3r\ e^{-i\vec{k}\cdot\vec{r}}\cdot W_{TH,R}(r)=\int d^3r\ e^{-i\vec{k}\cdot\vec{r}}\cdot\dfrac{3\theta(R-r)}{4\pi R^3}=$$

$$=\int_0^{2\pi}d\phi\int_0^\pi d\theta\int_0^\infty dr\ r^2\sin\theta\cdot e^{-ikr\cos\theta}\cdot\dfrac{3\theta(R-r)}{4\pi R^3}=$$

$$=2\pi\dfrac{3}{4\pi R^3}\int_0^\pi d\theta\int_0^\infty dr\ r^2\theta(R-r)\sin\theta\cdot e^{-ikr\cos\theta}=$$

$$\dfrac{3}{2R^3}\int_0^\infty dr \int_0^\pi d\theta\ r^2\theta(R-r)\sin\theta\cdot e^{-ikr\cos\theta}$$

But I cannot imagine how to proceed any further. I would like to be able to change variables as $u=\cos\theta$, but the extra $\theta$ in front of the exponential makes it seem like this would not solve anything.

Any help would be greatly appreciated!

Wild Feather
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    It is not the same $\theta$, one is the Heavyside function – user619894 Jul 12 '23 at 14:25
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    I may suppose that $\theta(R-r)=1$ for $r\in[0;R]$, and zero otherwise. If this supposition is correct, this theta has nothing to do with the polar angle in the spherical system of coordinates. – Svyatoslav Jul 12 '23 at 14:27
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    Assuming $r \ge 0$, you might start with the substitution $\Theta(R-r) = 1 - \Theta(r-R)$. – Andy Walls Jul 13 '23 at 22:45
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    And you might want to use this formula for the Fourier Transform when there is n-dimensional symmetry: https://math.stackexchange.com/a/3029986/441161 – Andy Walls Jul 13 '23 at 22:51

2 Answers2

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Presumably $\Theta(R-r)$ signifies the indicator function of the 3D ball of radius $R$ ( using Heaviside function notation). If so, the Fourier transform is described here:Fourier transform of the indicator of the unit ball

Then the fractional-order Bessel function that so arises can be represented using this technique: Bessel function $J_{3\over 2}(x)$

MathFont
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The Heaviside step function $\theta(R-r)$ is defined here. It is not to be confused with the integration variable $\theta$ (which you should rename in this particular calculation, e.g. $\alpha$). The Heaviside step function basically allows you to change the integration bounds of your $r$ integral if you think about it:

\begin{equation} \int_{0}^{\infty}dr\, f(r)\, \theta(R-r) = \int_{0}^{R}dr\, f(r). \end{equation}

So your integral actually turns out to actually simplify a lot:

\begin{align} \tilde{W}_{TH,R}(k) &= \,...\\ & = \frac{3}{2R^3} \int_{0}^{R}dr\, \int_{0}^{\pi}d\alpha\, r^2 \sin(\alpha)e^{-ikr\cos(\alpha)} \\ & = \frac{3}{kR^3} \int_{0}^{R}dr\, r\sin(kr) \\ & = 3 \frac{\sin(kR) - kR\cos(kR)}{(kR)^3}, \end{align}

using the properties of the Heaviside step function in the first step, your proposed substitution in the second step and integration by parts in the third step.

silas
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