I wonder if it is possible to derive an analytical form for the Fourier transform of the indicator function of a semi-ball: $$g_1(\underline{\xi}) = \int_{\mathcal{B}_+(R)} e^{i \underline{\xi} \cdot \underline{X} } \mathrm{d}\underline{X} $$ with $\underline{\xi}$ a vector of $\mathbb{R}^3$ and $\mathcal{B}_+(R) = \{ \underline{X} \in \mathbb{R}^3 : \lVert\underline{X}\rVert_2 \leq R, X_3 \geq 0\}$ a demi-ball. The Fourier transform of the unit-ball is itself well-known (as well as the unit-sphere): $$g_0(\underline{\xi}) = \int_{\mathcal{B}(R)} e^{i \underline{\xi} \cdot \underline{X} } \mathrm{d}\underline{X} = \frac{4\pi}{\xi^3}\left(\sin(R\xi)-R\xi \cos(R\xi)\right) $$
with $\xi = \lVert\underline{\xi}\rVert_2 $. Unfortunately, the symmetries applicable for these two calculations do not hold for a semi-ball. Note that $g_1(\underline{\xi}) + \overline{g_1(\underline{\xi})} = g_1(\underline{\xi}) + g_1(-\underline{\xi}) = g_0(\underline{\xi})$ so the only unknown lies in the imaginary part of $g_1$.

One month ago, I asked this question on math.stackexchange but it seems that this issue was not covered in existing Fourier transform literature and is therefore a research question more suitable to this site; sorry for the cross-posting. @ShannonStarr did suggest a number of leads, but I could not conclude.

A one-dimensional integral has been obtained, through the following manipulation, but it is probably a dead end: \begin{align}g_1(\underline{\xi}) &= \int_0^1 \left( \int_0^{r} J_0\big(\xi_\rho w\big) \exp\left(i \xi_3 \sqrt{r^2-w^2}\right) \frac{w/r}{\sqrt{r^2-w^2}}\, dw\right)\, dr \\ &= \int_0^{1} \left( \int_w^{1} \exp\left(i \xi_3 \sqrt{r^2-w^2}\right) \frac{w/r}{\sqrt{r^2-w^2}}\, dr\right) dw \\ &= \int_0^{1} \frac{w}{\sqrt{1-w^2}}\, \left(\int_0^1 \frac{e^{i a t}}{b^2+t^2}\, dt \right)dw \end{align} with $\xi_{\rho} = \sqrt{\xi_1^2+\xi_2^2}$, $a = \xi_3 \sqrt{1-w^2}$, $b = w/\sqrt{1-w^2}$ and the following Mathematica integral $$\int_0^1 \frac{e^{i a t}}{b^2+t^2}\, dt = \frac{i}{2b}[e^{-ab}(-\Gamma(0,-ab)+\Gamma(0,-a(b + i))-\ln(1-ib)+\ln(-ib)-\ln(-ab)+\ln(-a(b+i)) + e^{ab}(\Gamma(0,ab)+\Gamma(0,a(b -i))-\ln(1+ib)+\ln(ib)-\ln(ab)+\ln(a(b-i))].$$

Other methods have been discussed, with even less success. Any suggestions/answers?


1 Answer 1


Without loss of generality I can rotate the $x_1$ and $x_2$ axes so that the $x_2$ component of the vector $\xi$ is zero, $\xi=(\xi_1,0,\xi_3)$. For the integral over $x_1,x_2$ I use polar coordinates, $$g_1(\xi_1,\xi_3)=\int_0^1 dx_3\, e^{i\xi_3 x_3}\int_0^{\sqrt{1-x_3^2}}rdr\int_0^{2\pi}d\phi\, e^{i\xi_1r\cos{\phi}}$$ $$\qquad=\frac{2\pi }{\xi_1}\int_0^1 dx_3\, e^{i\xi_3 x_3}\sqrt{1-x_3^2} \,J_1\left(\xi_1 \sqrt{1-x_3^2}\right).$$ I don't think this integral over a Bessel function can be reduced any further in terms of known functions — but at least you have a compact expression.

  • 1
    $\begingroup$ Nice expression. I have the feeling that a series enables more efficiency for numerical computation so I would express that as follows: \begin{align}g_1(\xi_\rho,\xi_3) - g_0(\xi) &= \frac{2\pi}{\xi_\rho} \Im\left(\int_0^1 e^{i\xi_3\sqrt{1-t^2}}\frac{t^2}{\sqrt{1-t^2}}J_1(\xi_\rho t)\mathrm{d}t\right) \\ &= i\frac{2\pi}{\xi_\rho} \sum_{n=0}^{+\infty} (-1)^n\frac{(\xi_3)^{2n+1}}{(2n+1)!}\int_0^1 t^2(1-t^2)^nJ_1(\xi_\rho t)\mathrm{d}t \\ & = i\frac{2\pi\xi_3}{(\xi_\rho)^2} \sum_{n=0}^{+\infty} \left(-\frac{2(\xi_3)^2}{\xi_\rho}\right)^n\frac{n!}{(2n+1)!}J_{n+2}(\xi_\rho) \\ \end{align} $\endgroup$
    – CNS
    Commented Feb 20, 2022 at 13:55

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