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
$$\int_{\rm{all\ space}} e^{-\mathbf r^2} {\mathrm d}{\mathbf r}$$
Using Fubini's theorem, we can solve the problem.
$$ \begin{aligned} \int_{\rm all\ space} e^{-(\sqrt \alpha \mathbf r)^2} \mathrm d \mathbf r &=\int_{\rm all\ space} e^{-(\sqrt \alpha \mathbf r)^2} \mathrm d(\frac{1}{\sqrt \alpha}\sqrt \alpha\mathbf r) \\&=\iiint e^{-\alpha(x^2+y^2+z^2)} \mathrm d(\frac{1}{\sqrt \alpha}\sqrt \alpha x,\frac{1}{\sqrt \alpha}\sqrt \alpha y,\frac{1}{\sqrt \alpha}\sqrt \alpha z) \\&=\int\frac{1}{\sqrt \alpha}\mathrm d(\sqrt \alpha z) \int\frac{1}{\sqrt \alpha}\mathrm d(\sqrt \alpha y) \int\frac{1}{\sqrt \alpha}\mathrm d(\sqrt \alpha x) e^{-\alpha x^2} e^{-\alpha y^2} e^{-\alpha z^2} \\&=(\frac{\pi}{\alpha})^{3/2} \end{aligned} $$