Fourier transform of 3D Sinc function - part 2

Question

I'd like to know the Fourier transform of

$$\frac{\sin(P|\mathbf{x}|)}{|\mathbf{x}|}$$

in 3 dimensions.

This was discussed before in a previous question, but the two answers are contradicting. One answer is $$ \frac {1}{4\pi P}\delta_{S_P}(\xi) . $$ The second one is $$ \sqrt{\frac{\pi}{2}}\frac{i}{|\mathbf{k}|} \delta(P-|\mathbf{k}|) , $$ which seems more correct to me.

Can anyone help me out on this? Which one is correct? Is the answer still something else? And how would one arrive at the correct answer?

Answer 1

$\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[0.5em,#ffe,border:2px groove navy]{% \iiint_{\mathbb{R}^{3}}{\sin\pars{pr} \over r} \,\expo{\ic\vec{k}\cdot\vec{r}}\,\dd^{3}\vec{r}} = \int_{0}^{\infty}{\sin\pars{pr} \over r}\ \overbrace{% \pars{\int_{\Omega_{\,\vec{r}}}\expo{\ic\vec{k}\cdot\vec{r}} \,{\dd\Omega_{\,\vec{r}} \over 4\pi}}}^{\ds{\sin\pars{kr} \over kr}}\ 4\pi r^{2}\,\dd r \\[5mm] = &\ 2\pi p\int_{-\infty}^{\infty}{\sin\pars{pr} \over pr} \,{\sin\pars{kr} \over kr}\,r^{2}\,\dd r = {2\pi \over p^{2}}\int_{-\infty}^{\infty}{\sin\pars{\xi} \over \xi} \,{\sin\pars{\tilde{k}\xi} \over \tilde{k}\xi}\,\xi^{2}\,\dd\xi\,,\qquad \bbox[0.5em,border:1px groove navy]{\tilde{k} \equiv {k \over p}} \end{align}

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\begin{align} &\bbox[0.5em,#ffe,border:2px groove navy]{% \iiint_{\mathbb{R}^{3}}{\sin\pars{pr} \over r} \,\expo{\ic\vec{k}\cdot\vec{r}}\,\dd^{3}\vec{r}} = {2\pi \over p^{2}}\int_{-\infty}^{\infty} \pars{{1 \over 2}\int_{-1}^{1}\expo{\ic q\xi}\,\dd q} \pars{{1 \over 2}\int_{-1}^{1}\expo{-\ic s\tilde{k}\xi}\,\dd s}\,\xi^{2}\,\dd\xi \\[5mm] = &\ {\pi \over 2p^{2}}\int_{-1}^{1}\int_{-1}^{1}\int_{-\infty}^{\infty} \expo{\ic\pars{q - s\tilde{k}}\xi}\,\,\xi^{2}\,\dd\xi\,\dd q\,\dd s = -\,{\pi \over 2p^{2}\tilde{k}^{\, 2}}\int_{-1}^{1}\int_{-1}^{1}\partiald[2]{}{s} \int_{-\infty}^{\infty}\expo{\ic\pars{q - s\tilde{k}}\xi} \,\dd\xi\,\dd q\,\dd s \\[5mm] = &\ -\,{\pi^{2} \over k^{2}}\int_{-1}^{1} \partiald[2]{}{s}\int_{-1}^{1}\delta\pars{q - s\tilde{k}}\,\dd q\,\dd s = -\,{\pi^{2} \over k^{2}}\int_{-1}^{1} \partiald[2]{\Theta\pars{1 - \tilde{k}\verts{s}}}{s}\,\dd s \\[5mm] = &\ \left. -\,{\pi^{2} \over k^{2}} \partiald{\Theta\pars{1 - \tilde{k}\verts{s}}}{s} \right\vert_{\ s\ =\ -1}^{\ s\ =\ 1} = \left. -\,{\pi^{2} \over k^{2}}\braces{% \delta\pars{1 - \tilde{k}\verts{s}} \bracks{-\tilde{k}\,\mathrm{sgn}\pars{s}}}\right\vert_{\ s\ =\ -1}^{\ s\ =\ 1} \\[5mm] = &\ -\,{\pi^{2} \over k^{2}}\bracks{% \delta\pars{1 - \tilde{k}}\pars{-\tilde{k}} - \delta\pars{1 - \tilde{k}}\tilde{k}} = \,{2\pi^{2} \over k^{2}}\,\tilde{k}\,\delta\pars{1 - {k \over p}} = {2\pi^{2} \over kp}\,{\delta\pars{k - p} \over 1/p} \end{align}

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$\underline{Finally}$: $$ \bbox[0.5em,#ffe,border:2px groove navy]{% \iiint_{\mathbb{R}^{3}}{\sin\pars{pr} \over r} \,\expo{\ic\vec{k}\cdot\vec{r}}\,\dd^{3}\vec{r}} = \bbox[0.5em,#ffe,border:2px groove navy]{{2\pi^{2} \over k}\,\delta\pars{k - p}} $$

Answer 2

$\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ I arrived at the same answer but in a different way: \begin{align} &\bbox[0.5em,#ffe,border:2px groove navy]{% \iiint_{\mathbb{R}^{3}}{\sin\pars{pr} \over r} \,\expo{\ic\vec{k}\cdot\vec{r}}\,\dd^{3}\vec{r}} = \int_{0}^{\infty}{\sin\pars{pr} \over r}\ \overbrace{% \pars{\int_{\Omega_{\,\vec{r}}}\expo{\ic\vec{k}\cdot\vec{r}} \,{\dd\Omega_{\,\vec{r}} \over 4\pi}}}^{\ds{\sin\pars{kr} \over kr}}\ 4\pi r^{2}\,\dd r \\[5mm] = &\ {2\pi \over k}\int_{-\infty}^{\infty}{\sin\pars{pr}} \,{\sin\pars{kr}}\,\dd r \, , \end{align} but then proceeded like \begin{align} &{2\pi \over k}\int_{-\infty}^{\infty}{\sin\pars{pr}} \,{\sin\pars{kr}}\,\dd r = {2\pi \over k}\int_{-\infty}^{\infty} {\expo{\ic p r} - \expo{- \ic p r} \over 2 \ic} {\expo{\ic k r} - \expo{- \ic k r} \over 2 \ic} \,\dd r \\[5mm] = &\ {-\pi \over 2 k}\pars{\int_{-\infty}^{\infty} \expo{\ic p r} \expo{\ic k r} \,\dd r - \int_{-\infty}^{\infty} \expo{\ic p r} \expo{-\ic k r} \,\dd r - \int_{-\infty}^{\infty} \expo{-\ic p r} \expo{\ic k r} \,\dd r + \int_{-\infty}^{\infty} \expo{-\ic p r} \expo{-\ic k r} \,\dd r} \\[5mm] = &\ {-\pi^2 \over k} \pars{\delta\pars{-p-k}-\delta\pars{-p+k}-\delta\pars{p-k}+\delta\pars{p+k}} \, . \end{align} WLOG, assuming $p>0$, and because $k\geq0$, this reduces to \begin{align} {2\pi^2 \over k} \delta\pars{k-p} \, . \end{align}