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I am trying to calculate the following integral:

$ \int \frac{d k_x d k_y d k_z}{(2 \pi)^3} \left[ \exp( - \frac{(k_x^2 + k_y^2 + k_z^2) \sigma^2}{2}) + \frac{1}{2} H(\sqrt{k_x^2 + k_y^2 + k_z^2} - k_0) \right] \exp{(- \text{i} (k_x, k_y, k_z) \cdot (x , y ,z))}$

where $H$ stands for the Heaviside function and $k_0$ and $\sigma$ are known constants.

While I am quite sure about the first half (it becomes another Gaussian), I don't know how to deal with the second part, the one with the Heaviside function. How can I deal with that?

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So what you do is to realize that you can replace $(k_x,k_y,k_z)\cdot(x,y,z)$ by $\kappa r \cos(\theta)$, where $\kappa = \sqrt{k_x^2 + k_y^2 + k_z^2}$ and $r = \sqrt{x^2+y^2+z^2}$ and $\theta$ is the angle between $(k_x,k_y,k_z)$ and $(x,y,z)$. Also, you have that the Fourier transform of the constant function is the Dirac delta function, so we can consider $1 - H(\sqrt{k_x^2 + k_y^2 + k_z^2} - k_0) = H(k_0 - \sqrt{k_x^2 + k_y^2 + k_z^2})$ instead. Then using spherical coordinates as shown in the first picture on this website https://en.wikipedia.org/wiki/Spherical_coordinate_system, making $(k_x,k_y,k_z)$ parallel to the $z$-axis in this same picture, and replace $r$ with $\kappa$: $$ \int \frac{d k_x d k_y d k_z}{(2 \pi)^3} H(k_0 - \sqrt{k_x^2 + k_y^2 + k_z^2}) \exp{(- \text{i} (k_x, k_y, k_z) \cdot (x , y ,z))} $$ is equal to $$ \int_{\theta=0}^{\pi}\int_{\kappa=0}^\infty \int_{\varphi=0}^{2\pi} \frac{\kappa^2 \sin\theta \, d\kappa \, d\theta \, d\varphi}{(2\pi)^3} H(k_0-\kappa) e^{-i\kappa r \cos\theta} $$ which equals $$ \int_{\kappa=0}^{k_0} \int_{\theta=0}^{\pi} \frac{2 \pi \kappa^2 \sin\theta \, d\kappa \, d\theta}{(2\pi)^3} e^{-i\kappa r \cos\theta} $$ Being lazy, we plug this into Wolfram Alpha: http://www.wolframalpha.com/input/?i=Integrate%5B2+Pi+k%5E2+Sin%5Bt%5D+Exp%5B-I+k+r+Cos%5Bt%5D%5D%2F%282+Pi%29%5E3%2C+%7Bk%2C+0%2C+k0%7D%2C+%7Bt%2C+0%2C+Pi%7D%5D to get $$ \frac{ \sin (k_0 r ) - k_0 r \cos (k_0 r )}{2 \pi ^2 r ^3} .$$

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  • $\begingroup$ I follow your reasoning, the only thing I don't understand how you can get as a result a function of $\kappa$ and not $r$...maybe just a typo? $\endgroup$
    – johnhenry
    Commented Aug 16, 2015 at 21:37
  • $\begingroup$ Yes, it was a typo. Fixed it. $\endgroup$ Commented Aug 16, 2015 at 21:45

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