I have a function $$ \operatorname{g}\left(r\right) = \left\{\begin{array}{lcl} {\displaystyle 0} & \mbox{if} & {\displaystyle r < \rho} \\ {\displaystyle 1 + \mathrm{e}^{-kr}} && \mbox{otherwise} \end{array}\right. $$ And I have a question that asks: What is the value of $S(\vec{k}) = \int d^3 r [g(\vec{r})-1] e^{i \vec{k} \cdot \vec{r}}\ ?$.
The question said it was a tri-dimensional fourier transform, so I checked on internet what is that and a found that for a tri-dimensional transform
$f(\vec{x}) = \int d^3 k \space \space \mathscr{F}(k) e^{i \vec{k} \cdot \vec{r}} /(2 \pi)^3$
So I assumed that I only need to inverse transform $[g(r)-1]$ and multiply by $1/(2\pi)^3$
Is that right? Cause I got a non-infinity value for $S$, but I had to assume that $r>0$, and the value of $S$ is one-dimensional and only depends of the absolute value of the vector.
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ $\ds{\operatorname{g}\left(r\right) =} \left\{\begin{array}{lcl} {\displaystyle 0} & \mbox{if} & {\displaystyle r < \rho} \\ {\displaystyle 1 + \mathrm{e}^{-kr}} && \mbox{otherwise} \end{array}\right.$
\begin{align} \on{S}\pars{\vec{k}} & \equiv \bbox[5px,#ffd]{\iiint_{\mathbb{R}^{3}} \bracks{\on{g}\pars{r} - 1}\expo{\ic\vec{k}\,{\large\cdot}\,\vec{r}}\dd^{3}\vec{r}} \\[5mm] & = \int_{0}^{\infty}\bracks{\on{g}\pars{r} - 1}\ \underbrace{\pars{\int_{\Omega_{\,\vec{\!r}}} \expo{\ic\vec{k}\,{\large\cdot}\,\vec{r}}\,{\dd\Omega_{\,\vec{r}} \over 4\pi}}} _{\ds{\sin\pars{kr} \over kr}} 4\pi r^{2}\,\dd r \\[5mm] & = {4\pi \over k^{3}}\left[% \int_{0}^{k\rho}\pars{-1} \sin\pars{\theta}\,\theta\,\dd\theta\right. \\[2mm] & \phantom{{4\pi \over k^{3}}\left[\,\,\,\,\,\,\,\right.} \left. + \int_{k\rho}^{\infty}\expo{-\ic\theta} \sin\pars{\theta}\,\theta\,\dd\theta\right] \end{align}
The $\ds{\underline{\mbox{first}}}$ integral is equal to $\ds{\quad k\rho\cos\pars{k\rho} - \sin\pars{k\rho}\quad}$ while the $\ds{\underline{\mbox{second}}}$ one $\ds{\color{red}{\tt diverges}}$.