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When I was deriving the Fourier part of Ewald summation (see P5 in this pdf or P287 in this book), I am stuck on the charge density part $\rho(\mathrm k)$ in Fourier space. The primitive formula is as followed,

$$ \begin{aligned} \rho(\mathrm k) & =\int_V \mathrm d\mathbf r\rho(\mathbf r)e^{-\mathrm i\mathbf k\cdot\mathbf r}\\ &=\int_{\rm all\ space} \mathrm d\mathbf r e^{-\mathrm i\mathbf k\cdot\mathbf r} \sum\limits_{j=1}^{N} q_j{(\frac{\alpha}{\pi})}^{3/2} e^{-\alpha|\mathbf r-\mathbf r_j|}\\ & =\int_{\rm all\ space} \mathrm d \mathbf r \sum\limits_{j=1}^N q_j(\frac{\alpha}{\pi})^{3/2} e^{-(\frac{\mathrm i\mathbf k}{2\sqrt{\alpha}}+\sqrt{\alpha}(\mathbf r-\mathbf r_j)^2)}e^{-\mathrm i\mathbf k \mathbf r_j} e^{-\frac{k^2}{4\alpha}} \end{aligned} $$

There is a vector $\mathbf r=(x,y,z)$, and how to compute the integral above, which can be simplified as

$$\int_{\rm{all\ space}} e^{-\mathbf r^2} {\mathrm d}{\mathbf r}$$

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  • $\begingroup$ Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or closed. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers. $\endgroup$ Commented Jul 28, 2022 at 8:21
  • $\begingroup$ I guess I found the key to solving the integral. Is the key the Fubini's theorem? $\endgroup$
    – yongyouhe
    Commented Jul 28, 2022 at 12:48

1 Answer 1

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Using Fubini's theorem, we can solve the problem.

$$ \begin{aligned} \int_{\rm all\ space} e^{-(\sqrt \alpha \mathbf r)^2} \mathrm d \mathbf r &=\int_{\rm all\ space} e^{-(\sqrt \alpha \mathbf r)^2} \mathrm d(\frac{1}{\sqrt \alpha}\sqrt \alpha\mathbf r) \\&=\iiint e^{-\alpha(x^2+y^2+z^2)} \mathrm d(\frac{1}{\sqrt \alpha}\sqrt \alpha x,\frac{1}{\sqrt \alpha}\sqrt \alpha y,\frac{1}{\sqrt \alpha}\sqrt \alpha z) \\&=\int\frac{1}{\sqrt \alpha}\mathrm d(\sqrt \alpha z) \int\frac{1}{\sqrt \alpha}\mathrm d(\sqrt \alpha y) \int\frac{1}{\sqrt \alpha}\mathrm d(\sqrt \alpha x) e^{-\alpha x^2} e^{-\alpha y^2} e^{-\alpha z^2} \\&=(\frac{\pi}{\alpha})^{3/2} \end{aligned} $$

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