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I am trying to find the integrals $$\large\int\limits_{\mathbb{R}^3} e^{-\beta \left|\vec{r}\,\right|}e^{i \vec{q} \cdot\,\vec{r}} \mathop{d^3r}$$

$$\large\int\limits_{\mathbb{R}^3} \left|\vec{r}\,\right|e^{-\beta \left|\vec{r}\,\right|}e^{i \vec{q} \cdot\,\vec{r}} \mathop{d^3r}$$

Both integrated over all of $\mathbb{R}^3 $. These are essentially 3D Fourier transforms. Unfortunately the first factor looks nice in spherical coordinates, but not the second factor. I was able to integrate $e^{i \vec{q} \cdot\,\vec{r}} $ over the azimuthal angle using Mathematica, but it was not able to proceed integrating the polar angle. The actual integral I'm trying to solve is $$\large\int\limits_{\mathbb{R}_3} e^{-3 \left|\vec{r}\,\right|/(2 a)} \left(2-\frac{ \left|\vec{r}\,\right|}{a}\right)e^{i \vec{q}\, \cdot\, \vec{r}}\mathop{d^3r}$$

This appeared as part of a physics textbook problem, so I think this should somehow be doable.

WhatIAm
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  • Perhaps the thing to note is that the functions $e^{-\beta r}$ and $r e^{-\beta r}$ are symmetric. I may be able to simplify the problem by choosing the z axis to be in the direction of $\vec{q}$. – WhatIAm Mar 25 '14 at 02:04

1 Answers1

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Inserting the Jacobian $r^2\sin\theta$ where $\theta$ is the polar angle and assuming that $|\mathbf{q}|=q$ is parallel to the polar direction \begin{equation} \int \exp(-\beta r)e^{i\vec{q}\cdot \vec{r}}d^3r \end{equation} \begin{equation} =\int_{r=0}^\infty r^2 dr \int_0^{2\pi}d\phi \int_0^\pi \sin \theta d\theta \exp(-\beta r)e^{iqr\cos\theta} =2\pi \int_{r=0}^\infty r^2 dr \int_0^\pi \sin \theta d\theta \exp(-\beta r)e^{iqr\cos\theta} \end{equation} with $\cos\theta =z$, $dz=-\sin\theta d\theta$ \begin{equation} =-2\pi \int_{r=0}^\infty r^2 dr \int_{1}^{-1} dz \exp(-\beta r)e^{iqrz} \end{equation} Substitute $t=iqrz$, $dz=dt/(iqr)$ \begin{equation} =2\pi \int_{r=0}^\infty r^2 dr \int_{-1}^1 dz \exp(-\beta r)e^{iqrz} =\frac{2\pi}{iq} \int_{r=0}^\infty r \exp(-\beta r)dr \int_{-iqr}^{iqr} dt e^{t} \end{equation} \begin{equation} =\frac{2\pi}{iq} \int_{r=0}^\infty r \exp(-\beta r)[e^{iqr}-e^{-iqr}]dr \end{equation} \begin{equation} =\frac{4\pi}{q} \int_{r=0}^\infty r \exp(-\beta r)\sin(qr) dr \end{equation} \begin{equation} =\frac{4\pi}{q} \frac{2\beta q}{(\beta^2+q^2)^2} \end{equation} ...all very similar to 3D Fourier transform

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R. J. Mathar
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