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What is multidimensional Fourier transform of $e^{-\mathrm{abs}(x)}$?

I know the answer for 1-dimension case. However, I cannot do integration for higher dimensional case.

I used spherical coordinate to arrive at the following, but I can't go further.

$\begin{align}\int_{\mathbb{R}^n}e^{-a |x| -i\xi x}dx &= \int_0^\infty dx \int_0^\pi d\theta \int d\Omega_{n-2} x^{n-1}\sin^{n-2}(\theta) e^{-ax-i|\xi|x \cos(\theta)} \\ &= C \int_0^\infty \int_0^\pi x^{n-1}\sin^{n-2}(\theta) e^{-ax-i|\xi|x \cos(\theta)}\end{align}$

where by $d\Omega_{n-2}$ I mean measure on $S^{n-2}$

For $n = 3$ case, it is simple, but I don't know how to do this integral for higher dimension.

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  • $\begingroup$ Try using spherical symmetry. $\endgroup$
    – hordubal
    Oct 12, 2021 at 9:24
  • $\begingroup$ Let me just point out as a comment that, through the Hankel transform, what you are asking for is essentially the value of $\int_0^{+\infty} r^{(n/2)-1} \, \exp(-r) \, J_{(n/2)-1}(k r) \, dr$ where $J_\nu$ is the Bessel function of the first kind of order $\nu$. $\endgroup$
    – Gro-Tsen
    Oct 14, 2021 at 19:55

2 Answers 2

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$$\frac{1}{(2\pi)^n} \int e^{i\xi\cdot x}e^{-|\xi|}d\xi= \frac{\Gamma((n+1)/2)}{\pi^{(n+1)/2}}\frac{1}{(|x|^2+1)^{(n+1)/2}}.$$ Hint: up to a constant factor, the RHS is the value of the Poisson kernel for the upper half-space at the point $(0,0,...0,1)$. The LHS is what you obtain when you solve the Laplace equation for the upper half-space by Fourier method.

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  • $\begingroup$ Can I get more explanations about how it is derived?? $\endgroup$
    – user409452
    Oct 12, 2021 at 14:58
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    $\begingroup$ The usual subordination trick: use the fact that $e^{-|\xi|} = \pi^{-1/2} \int_0^\infty e^{-t |\xi|^2/4} t^{-3/2} e^{-1/t} dt$ and Fubini's theorem to transform the left-hand side. $\endgroup$ Oct 12, 2021 at 20:30
  • $\begingroup$ I have trouble in checking that formula. Can you give me explanation on that formula? $\endgroup$
    – user409452
    Oct 13, 2021 at 0:22
  • $\begingroup$ One way to prove the formula in the comment ($e^{-|\xi|}$ as an integral of Gaussians) is to show that the Fourier transform of the two sides agree. $\endgroup$ Oct 13, 2021 at 5:33
  • $\begingroup$ @user409452: Oh, that's one of the standard examples of "differentiate under the integral" problems. I gave more details in another answer. $\endgroup$ Oct 14, 2021 at 19:09
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Upon request of user409452, here is an extended version of my comment to Alexandre Eremenko's answer.


First of all, for $a > 0$ we evaluate $$ \phi(a) = \int_0^\infty e^{-a^2 t/4} t^{-3/2} e^{-1/t} dt . $$ By the dominated convergence theorem, $$ \phi'(a) = -\frac{a}{2} \int_0^\infty e^{-a^2 t/4} t^{-1/2} e^{-1/t} dt . $$ On the other hand, substituting $t = 4/(a^2 s)$ we find that $$ \phi(a) = \frac{a}{2} \int_0^\infty e^{-1/s} s^{-1/2} e^{-a^2 s / 4} ds . $$ It follows that $$ \phi'(a) = -\phi(a) , $$ and hence $$ \phi(a) = e^{-a} \phi(0) . $$ By a substitution $t = 1/s$, $$ \phi(0) = \int_0^\infty t^{-3/2} e^{-1/t} dt = \int_0^\infty s^{-1/2} e^{-s} ds = \Gamma(\tfrac12) = \sqrt{\pi} , $$ and so we finally have $$ \phi(a) = \sqrt{\pi} e^{-a} . $$


Now we write $$ f(\xi) = e^{-|\xi|} = \pi^{-1/2} \phi(|\xi|^2) = \frac{1}{\sqrt{\pi}} \int_0^\infty e^{-|\xi|^2 t / 4} t^{-3/2} e^{-1/t} dt . $$ Evaluating the inverse Fourier transform of both sides and using Fubini's theorem, we find that $$ \check{f}(x) = \frac{1}{\sqrt{\pi}} \int_0^\infty \frac{1}{(\pi t)^{n/2}} e^{-|x|^2 / t} t^{-3/2} e^{-1/t} dt = \frac{1}{\pi^{(n + 1)/2}} \int_0^\infty t^{-(3 + n) / 2} e^{-(1 + |x|^2) / t} dt . $$ Once again we substitute $t = 1 / s$: $$ \check{f}(x) = \frac{1}{\pi^{(n + 1)/2}} \int_0^\infty s^{(n - 1)/2} e^{-(1 + |x|^2) s} ds = \frac{1}{\pi^{(n + 1)/2} (1 + |x|^2)^{(n + 1)/2}} \, \Gamma(\tfrac{n + 1}{2}) . $$ Finally, $$ \hat{f}(x) = (2 \pi)^n \check{f}(x) = 2^n \Gamma(\tfrac{n + 1}{2}) \pi^{(n - 1)/2} \, \frac{1}{(1 + |x|^2)^{(n + 1)/2}} \, . $$

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