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I'm using the convention that the fourier transform of a function $f(\vec x$) is$$\frac{1}{(2\pi)^{3/2}}\int{d^3x\: e^{-i\vec k\cdot \vec x}f(\vec x)}$$ I'm now fourier transforming the function $$f(\vec x)=\frac{1}{\sqrt{\pi a^3}}e^{-r/a}$$ with $r:= \sqrt{x^2+y^2+z^2} $. The result I get is $$\frac{\sqrt{(2a)^3}}{\pi}\frac{1}{(1+(ka)^2)^2}, \: k:=\sqrt{k_1^2+k_2^2+k_3^2} $$ which I think should be correct. Once I do the backwards fourier transform (which looks exactly as the fourier transform just with an integration over $\vec k$ and $e^{i\vec k\cdot \vec x}$ instead of the negative exponent), I'm running into trouble. When I first start the inverse transformation I switch into spherical coordinates and after a few steps I arrive at: $$\frac{\sqrt{(2a)^3}}{\pi} \sqrt{\frac{2}{\pi}}\frac{1}{r} \frac{2}{a^4}\int _0^\infty{ dk \frac{\sin(kr) k}{(k-\frac{i}{a})^2(k+\frac{i}{a})^2}}$$ I now want to solve this using the residue theorem. Since the function is even I can simply put an $\frac{1}{2}$ in front and integrate from $-\infty$ to $+\infty$ instead. Then the 2-fold degenerate complex zeros of the denominator are of course $z_1 = \frac{i}{a}$ and $z_2=-\frac{i}{a}$, with the following residues $$Res(z_{1/2})=\mp \frac{1}{4}i a r \cosh{r/a}$$ If I understand the way to do these kind of integrals correctly, then I have to close the integral in the complex plane making it a contour integral and depending on wheter I close the loop in the positive or negative complex plane I will have to use the residue of $z_1$ or $z_2$ respectively, since that's the one that gets enclosed by the contour. But if I do it like that I simply get some kind of $\cosh$ for my inverse fourier transform and not the $e^{-r/a}$ that I started with.

I assume I'm just missing something about how to correctly solve these kind of complex integrals, since if I could somehow combine the two residues I could get the exponential I want out of them.

Any help or tipps are greatly appreciated, thank you!

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    $\begingroup$ The problem is that your integral along the extended contour does not go to 0. In complex numbers $\sin kz$ is not bounded between $-1$ and $1$ $\endgroup$
    – Andrei
    Aug 21 '21 at 23:03
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    $\begingroup$ As a side comment, I just wrote here (math.stackexchange.com/questions/4220006/…) a method without complex analysis for the same integral ;) $\endgroup$
    – LL 3.14
    Aug 22 '21 at 21:31
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As mentioned in my comment, the integral over the part of the contour where $|z|\to\infty$ is not $0$. If you want to integrate over the upper contour, use $e^{ikr}$ instead of $\sin kr$. Then, in the upper half of the plane, $k$ becomes $|k|e^{i\phi}$. Then $$|e^{ikr}|=|e^{i|k|r(\cos\phi+i\sin\phi)}|=|e^{i|k|r(\cos\phi}e^{-|k|r\sin\phi}|$$ When $|k|\to \infty$ the second exponential goes to $0$. The absolute value of the first exponential is always $1$. Then notice $$\int_{-\infty}^\infty\frac{ke^{ikr}}{(1+(ka)^2)^2}dk=\int_{-\infty}^0\frac{ke^{ikr}}{(1+(ka)^2)^2}dk+\int_{0}^\infty\frac{ke^{ikr}}{(1+(ka)^2)^2}dk$$ In the first integral replace $k$ with $-k$. Can you continue from here?

EDIT $$\int_{-\infty}^\infty\frac{ke^{ikr}}{(1+(ka)^2)^2}dk=\int_0^\infty\frac{k(e^{ikr}-e^{-ikr})}{(1+(ka)^2)^2}dk=2i\int_0^\infty\frac{k}{(1+(ka)^2)^2}\frac{(e^{ikr}-e^{-ikr})}{2i}dk$$ The last fraction is $\sin(kr)$. Then using residue theorem, and the fact that on the upper contour the integral is zero, we get: $$2i\int_0^\infty\frac{k\sin(kr)}{(1+(ka)^2)^2}=2\pi i \frac 14 ar e^{-r/a}$$ Notice that you have $2i$ on both sides. So when multiply with your prefactors you get $$\frac{\sqrt{(2a)^3}}{\pi} \sqrt{\frac{2}{\pi}}\frac{1}{r} \frac{2}{a^4}\frac 14\pi are^{-r/a}=\frac2{\sqrt{\pi a^3}}e^{-r/a}$$ This is your original function, up to a factor of $2$, that either you or I have missed in the calculation.

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  • $\begingroup$ Thank you very much for your answer! So in order to extend my integral to the contour, what kind of property has to be fullfilled? I read online that the function in the integral times $k$ to the power of the zero in the denominator has to vanish at $|k|\to \infty$, is that correct? $\endgroup$
    – DeltaChief
    Aug 22 '21 at 17:21
  • $\begingroup$ And I get to the point where I have $\int_{-\infty}^\infty\frac{ke^{ikr}}{(1+(ka)^2)^2}dk$, but I don't get where your hint on separating the integral into two parts leads. When I just use the residue theorem for the Integral as it is and close in the upper half I almost get something correct, but I also are left with a factor of $i$, which doesn't match up with my starting function. My residue that I get for that Integral is $\frac{1}{4}a\exp^{-r/a}r$ $\endgroup$
    – DeltaChief
    Aug 22 '21 at 17:24
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    $\begingroup$ I've updated the answer. I am off by a factor of 2, that it might be from either my calculations or yours. $\endgroup$
    – Andrei
    Aug 22 '21 at 19:17
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    $\begingroup$ As for the property of the integral, the integrand has to be on the order less than $k^{-1}$. If the integrand is $C/k$, then the integral on the contour at infinity is a constant. $\endgroup$
    – Andrei
    Aug 22 '21 at 19:21
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    $\begingroup$ @DeltaChief You are welcome. I hope you can find the extra factor of 2. $\endgroup$
    – Andrei
    Aug 22 '21 at 20:03

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