This math.SE answer calculated the following (inverse) Fourier transform in $n$-dimensions

$$ f(x) = \int_{\mathbb{R}^n} d^nk \frac{e^{ik\cdot x}}{|k|^2 + 1} \propto |x|^{-\frac{n}{2}+1} K_{\frac{n}{2}-1}(|x|) \tag{1} $$

where $K_\alpha(x)$ is the modified Bessel function of the second kind. But in some physics applications, one is only interested in the large-$|x|$ asymptotic behavior of $f(x)$. Using the asymptotic expansion

$$ K_\alpha(x) \propto x^{-1/2} e^{-x} [1 + O(x^{-1})] \quad \text{when } \mathrm{arg}(x) < 3\pi/2 $$

we obtain (which reproduces Eq. (14.23) in Assa Auerbach's Interacting Electrons and Quantum Magnetism)

$$ f(x) \sim |x|^{-(n-1)/2} e^{-x} \tag{2} $$

My question is: is there a simpler way than the cited math.SE answer that directly aims to find the asymptotic behavior of $f(x)$ for large $|x|$?

  • $\begingroup$ Most of the properties of Bessel potentials often follow more easily from the formula I wrote here math.stackexchange.com/questions/4662845/… $\endgroup$
    – LL 3.14
    May 15, 2023 at 9:30
  • $\begingroup$ @LL3.14 But is it inevitable to first derive the exactly transformed function like Eq. (1)? I hope that one can immediately study the $x \to \infty$ expansion without the exact formula. $\endgroup$ May 15, 2023 at 9:43
  • $\begingroup$ Yes, the formula I indicated you is not (1). But I do not know if it helps a lot. I will think about it. In general studying directly the behavior of the oscillatory integral that is given by the Fourier transform is difficult as one has to take care of the oscillations. Integrals of positive functions are easier to handle. Anyway, this is a good question, this asymptotic seems to be rather precise/nontrivial actually. $\endgroup$
    – LL 3.14
    May 15, 2023 at 12:16

1 Answer 1


$$f(x)=\int_{-\infty}^\infty...\int_{-\infty}^\infty\frac {e^{ik_1x_1+ik_2x_2+...+ik_nx_n}}{1+k_1^2+...+k_n^2}dk_1...dk_n$$ $$=\int_{-\infty}^\infty...\int_{-\infty}^\infty e^{ik_1x_1+ik_2x_2+...+ik_nx_n}dk_1...dk_n\int_0^\infty e^{-t(1+k_1^2+...+k_n^2)}dt$$ $$\int_0^\infty e^{-t}dt\int_{-\infty}^\infty e^{-t\big((k_1^2-\frac{ix_1}{2t})^2+\frac{x_1^2}{4t^2}\big)}dk_1...\int_{-\infty}^\infty e^{-t\big((k_n^2-\frac{ix_n}{2t})^2+\frac{x_n^2}{4t^2}\big)}dk_n$$ Denoting $x=\sqrt{x_1^2+...+x_n^2}$ and integrating with respect $k_1\,...\,k_n$ $$=\pi^\frac n2\int_0^\infty e^{-t-\frac{x^2}{4t}}t^{-\frac n2}dt\overset{t=\frac{xs}2}{=}\pi^\frac n2\left(\frac x2\right)^{1-\frac n2}\int_0^\infty e^{-\frac x2(s+\frac1s)}s^{-\frac n2}ds$$ Using Laplace' method we can find as many asymptotic terms as we want. The extremum point is $s=1$. For the leading term we get $$f(x)\sim\pi^\frac n2\left(\frac x2\right)^{1-\frac n2}e^{-x}\int_{-\infty}^\infty e^{-\frac x2(s-1)^2}ds=\pi^\frac {n+1}2\left(\frac x2\right)^{\frac{1-n}2}e^{-x}$$

  • 1
    $\begingroup$ Laplace's method, powerful, thanks. $\endgroup$
    – LL 3.14
    May 15, 2023 at 13:35

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .