Define $f: R^3 \rightarrow R, f(x) = \frac{1}{\|x\|} \chi_{B_1(0)}(x)$ (with $f(0) = 0$).
I would like to calculate the fourier transform $g(\xi) = \int_{R^3} f(x) e^{-ix\cdot \xi}dx$.
I tried using polar coordinates, but this leads to enormous expressions containing sin, cos etc.
Can anybody help me with this problem?
To avoid confusion we will use the notation for the spherical coordinate system $\vec{r}=(r,\theta ,\varphi )$. In this notation the function becomes
$$f(\vec{r})=\frac{1}{r}\chi _{B_1(0)}(r) $$
It is clear that this is identical to
$$f(\vec{r})=\frac{1}{r}\Theta(1-r) $$
where $\Theta $ is the Heaviside step function (we use capital letter to distinguish from the polar coordinate $\theta $). The Fourier transform is
$$g(\xi )=\int _{\mathbb{R}^3}\frac{1}{r}\Theta(1-r)e^{-i\vec{r}\cdot \vec{\xi}}d^3r$$
We can take $\vec{r}\cdot \vec{\xi}=r\xi \cos \theta $, where $\theta $ since this only fixes the coordinate axes in the inverse $\xi$-space. The volume element in spherical coordinates is $d^3r=r^2dr\sin \theta drd\theta d\varphi $ and thus
$$g(\xi )=\int _{0}^{\infty }\int _0^{\pi }\int _0^{2\pi}\frac{1}{r}\Theta(1-r)e^{-i\vec{r}\cdot \vec{\xi}}r^2dr\sin \theta drd\theta d\varphi $$
The intergrand does not depend on $\varphi $ and we can integrate and simplify to
$$g(\xi)=2\pi \int _0^1\int _0^{\pi }re^{-i\vec{r}\cdot \vec{\xi}}\sin \theta drd\theta$$
Using $\vec{r}\cdot \vec{\xi}=r\xi \cos \theta $ we have
$$g(\xi)=2\pi \int _0^1r\left(\int _0^{\pi }e^{-ir\xi \cos \theta}\sin \theta d\theta \right)dr$$
The integral in the brackets can be evaluated easily
$$ \begin{align} \int _0^{\pi }e^{-ir\xi \cos \theta}\sin \theta d\theta&=-\int _0^{\pi }e^{-ir\xi \cos \theta} d(\cos \theta)=\int _{-1}^1e^{-ir\xi t}dt=\frac{e^{-ir\xi} -e^{ir\xi }}{ir\xi}\\ &=-\frac{2\sinh (ir\xi )}{ir\xi }=\frac{2\sin (r\xi )}{r\xi }, \end{align} $$
where in the last line we used the identity $\sinh(ix)=-i\sin(x)$. Therefore,
$$g(\xi)=\frac{4\pi}{\xi} \int _0^1\sin (r\xi )dr$$
Integrating gives
$$g(\xi) =\frac{4\pi}{\xi ^2}(1-\cos \xi)$$