A similar question was asked before for an interval in $\mathbb{R}$. I wonder how to do it for a characteristic function of $\{x\in\mathbb{R}^3:|x|
Fourier transform of characteristic function in a sphere
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calculus
integration
fourier-analysis
characteristic-functions
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0@DavideGiraudo thank you for your advise. Does it make sense switch to spherical coordinates? Evantually, I should calculate \int_{|x|
and |x| represents here a sphere centered $(0,0,0)$.Could you please give me a clue, how to calculate it? – 2012-10-14
1 Answers
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Choosing the $z$ axis along $k$ and denoting $|k|$ by $q$, we have
$ \begin{align} \int_{\mathbb R^3}\chi_{|x|\lt r}(x)\exp(-\mathrm ikx)\,\mathrm dx &=\int_0^r R^2\,\mathrm dR\int_0^\pi\sin\theta\,\mathrm d\theta\int_0^{2\pi}\mathrm d\phi\exp(-\mathrm iqR\cos\theta) \\ &=2\pi\int_0^r R^2\,\mathrm dR\int_0^\pi\sin\theta\,\mathrm d\theta\exp(-\mathrm iqR\cos\theta) \\ &=2\pi\int_0^r R^2\,\mathrm dR\left[\frac1{\mathrm iqR}\exp(-\mathrm iqR\cos\theta)\right]_{\theta=0}^{\theta=\pi} \\ &=\frac{2\pi}{\mathrm iq}\int_0^r \mathrm dRR\left(\exp(\mathrm iqR)-\exp(-\mathrm iqR)\right) \\ &=\frac{4\pi}{q^3}\left(r\cos qr-\sin qr\right)\;. \end{align} $
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0I think this is wrong? $r\cos qr−\sin qr$ do not have the same units. I think the answer is $\frac{4 \pi}{k^3} (\sin (k R)-k R \cos (k R))$ – 2017-06-27