I wonder if it is possible to derive an analytical form for the Fourrier transform of the indicative 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 $\mathcal{B}_+(R) = \{ \underline{X} \in \mathbb{R}^3 : \|\underline{X}\|_2 \leq R, X_3 \geq 0\}$ a demi-ball. The Fourrier transform of the unit-ball is itself well-known (as well as the unit-sphere): $$g_0(\underline{\xi}) = \int_{\mathcal{B}(R)} e^{i \underline{\xi} \cdot \underline{X} } \mathrm{d}\underline{X} = \frac{4\pi}{\xi^3}\left(\sin(R\xi)-R\xi \cos(R\xi)\right) $$
Unfortunately, the symmetries applicable for these two calculations do not hold for a semi-ball. Note that $g_1(\xi) + \overline{g_1(\xi)} = g_1(\xi) + g_1(-\xi) = g_0(\xi)$ so the only unknown lie 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. @Shannon_Starr did suggest a number of leads, but I could not conclude.

A one-dimension integral have been obtained, through the following manipulation, but it is probably a dead end: \begin{align}g_1 &= \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?