# Fourier transform of the indicator function of the semi-ball

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?

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.