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In the context of reproducing some known expressions, I'm trying to simplify the integral $$ \int \dfrac{dx dy}{|x-y|} \delta(x-r_j)\delta(y-r_l) \;, $$ which I know it should give as result $$ \int \dfrac{dk}{k^2}e^{-i k r_j}e^{-i k r_l} \;. $$ [Here I reported the integrals I actually care about in a simplified form, so that the main point of the calculation is emphasised. More actual details are not so relevant.]

I think that the idea in going from the first to the second integral must involve the Fourier transform, but I was not able to find out how. I have the impression one has to FT the fraction, and then integrate out $x,y$, but by taking FT of $1/|x|$ I get stuck.

m137
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  • Using the Fourier transform, we have \begin{align} \int \frac{dx dy}{|x-y|} \delta(x-r_j)\delta(x-r_l) &= \int dx \delta(x-r_j) \int dy \frac{e^{-i k (y-r_l)}}{|y-x|}
    &= \int dx \delta(x-r_j) \int dk \frac{e^{-i k (r_l-r_j)}}{k^2}
    &= \int \frac{dk}{k^2} e^{-i k (r_l-r_j)} \int dx \delta(x-r_j)
    &= \int \frac{dk}{k^2} e^{-i k (r_l-r_j)}
    &= \int \frac{dk}{k^2} e^{-i k r_j} e^{-i k r_l}. \end{align    }     – kuspia Apr 02 '23 at 07:38

1 Answers1

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By definition of the Dirac delta, your integral is just $$ I = \int \frac{1}{|r_j-y|} \,\delta(y-r_l) \,\mathrm d y = \frac{1}{|r_j-r_l|} $$ Assuming that you works in dimension $3$, the Fourier transform of $1/|x|$ is $1/(2\pi^2|x|^2)$, so the above function is just the Fourier transform of $1/(2\pi^2|x|^2)$ evaluated at $r_j-r_l$, that is $$ I = \int e^{-i(r_j-r_l)k} \frac{1}{2\pi^2|k|^2} \,\mathrm d k. $$ In particular, you have a sign error in one of the exponentials.

LL 3.14
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  • Thanks! I'm just confused by one thing: How is the FT of $1/|x|$ giving $1/(2\pi^2|x|^2)$, could you please elaborate ? – m137 Apr 03 '23 at 14:56
  • Yrs, this is a well-known result that you can get just by scaling, see e.g. https://math.stackexchange.com/questions/3742640/fx-1-lvert-x-rvert2-x-in-mathbbr3-for-the-fourier-transform – LL 3.14 Apr 04 '23 at 07:11