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A similar question was asked before for an interval in $\mathbb{R}$. I wonder how to do it for a characteristic function of $\{x\in\mathbb{R}^3:|x|<r\}$ i.e. I want to calculate $$ \frac{1}{(2\pi)^{\frac{3}{2}}}\int_{\mathbb{R^3}} \chi_{|x|<r}(x)\exp(-ikx) dx.$$ Could you please help me?

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Do you work with euclidian norm? – Davide Giraudo Oct 14 '12 at 20:15
@DavideGiraudo yes, it is euclidian norm. – pcepkin Oct 14 '12 at 20:24
Have you tried with spherical coordinates? – Davide Giraudo Oct 14 '12 at 20:49
@DavideGiraudo thank you for your advise. Does it make sense switch to spherical coordinates? Evantually, I should calculate $\int_{|x|<r} exp(-ikx)dx$ and $|x|<r$ represents here a sphere centered $(0,0,0)$.Could you please give me a clue, how to calculate it? – pcepkin Oct 14 '12 at 22:18
up vote 2 down vote accepted

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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thank you very much for your kind help. – pcepkin Oct 15 '12 at 8:08
@pcepkin: You're welcome. – joriki Oct 15 '12 at 14:55

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