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How do I find the inverse fourier transform of a function of the form $$\hat{f}(k)=\frac{e^{-ck}}{k},$$ with $c$ being some constant (can be complex)? The definition of the inverse fourier transform that I am using is $$f(x) = \frac{1}{2\pi} \int_{-\infty}^{\infty} e^{ikx}\hat{f}(k)\, dk$$ I have tried direct integration which has led nowhere, and I cannot come up with some function which gives this as its Fourier transform. Thanks :)

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    $\begingroup$ There are some integrability issues. Are you sure $\hat{f}$ is in the given form? My bet is on $$\widehat{f}(k) = \frac{e^{-c|k|}}{|k|},$$ instead. In such a case the inverse Fourier transform is related with $2\gamma+\log(1+s^2)$. $\endgroup$ Dec 14, 2016 at 22:40
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    $\begingroup$ I am fairly sure$\hat{f}$ is of that form, since it comes from taking the Fourier transform of the 2D Laplace equation with respect to one of the spatial co-ordinates, which gives $f''-k^2f=g$ where $g$ is just the Fourier transform of the forcing term. I then used Green's functions to solve which gives exponentials (some imaginary, some not). Can you see any mistakes in that? $\endgroup$
    – John Doe
    Dec 15, 2016 at 12:38
  • $\begingroup$ Also I am not familiar with the Fourier transform you mentioned: $$2\gamma + \log(1+s^2)$$ so could you please elaborate on this? Or send some link where I can see it? $\endgroup$
    – John Doe
    Dec 15, 2016 at 12:39
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    $\begingroup$ If $\int_{-\infty}^{+\infty}e^{isk}\hat{f}(k)\,dk$ is not a converging integral, there has to be a mistake somewhere. $\endgroup$ Dec 15, 2016 at 12:59
  • $\begingroup$ Hmm, OK thanks I shall run through the question again, hopefully the mistake is amended. :) $\endgroup$
    – John Doe
    Dec 15, 2016 at 13:35

2 Answers 2

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If $c$ is a pure imaginary number, then the transform can be computed in the sense of tempered distributions. Let $1/k$ be the distribution defined by $$\left( \frac 1 k, \phi \right) = \operatorname{v.\!p.} \int_{-\infty}^\infty \frac {\phi(k)} k dk,$$ then $$\mathcal {F} \!\left[ \frac 1 k \right] = \left( \frac 1 k, \frac 1 {2 \pi} e^{i x k} \right) = \frac i 2 \operatorname{sgn}x, \\ \mathcal {F} \!\left[ \frac {e^{-c k}} k \right] = \mathcal {F} \!\left[ \frac 1 k \right](x + i c) = \frac i 2 \operatorname{sgn} (x + i c).$$ If $c$ has a real component, the answer will depends on the chosen space of test functions. Gelfand and Shilov's book defines the Fourier transform for distributions acting on functions with finite support. Then $\mathcal {F}[e^k]$ is well-defined, but it is a distribution living in a different dual space (acting on entire functions with a certain condition on their rate of growth). For the Schwartz space of functions of rapid decay, $e^k$ is not a valid functional, because the integral of $e^k \phi(k)$ does not necessarily converge.

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  • $\begingroup$ Thanks, this is an interesting answer. I haven't really studied anything about distributions in much detail, so I don't fully understand what is going on here. I will hopefully be reading about this area soon though, after which this answer may be more accessible for me. What I can say is, I am fairly certain I wouldn't have been required to know about these things when given the original problem, so I probably made a prior computational error (unfortunately I can't quite remember or find the work). I will accept this answer since it would be the answer that future readers would be looking for $\endgroup$
    – John Doe
    Apr 7, 2018 at 3:39
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As in @JackD'Aurizio 's observations, there was probably a prior computational error: in solving $(\Delta-s^2) u=\delta$, taking Fourier transforms gives (up to constants) $(x^2+s^2)\widehat{u}=1$, and $\widehat{u}(x)=1/(x^2+s^2)$. The inverse Fourier transform can be evaluated by residues (unsurprisingly), and/but this technique gives slightly different results depending on whether $x$ is positive or negative...

For that matter, we should certainly expect that a Fourier transform of a tempered distribution is a tempered distribution, and something like $e^x$ is definitely not a tempered distribution, whatever virtues it may have. So any computation that produces this as an alleged FT of a tempered distribution is ... doubtful.

And, then, indeed, something like $e^{-|x|}$ (with constants) is surely the right thing...

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  • $\begingroup$ Unfortunately, I can't quite remember what the prior computations were. Looking at the question again now, I can see the issues with performing this integral too, so it is probably true that I was looking for a different Fourier transform. $\endgroup$
    – John Doe
    Apr 7, 2018 at 3:35
  • $\begingroup$ @paulgarret Another formal 'indeed.' Another hackey formalism. Lose that and I'll upvote. $\endgroup$ Nov 10, 2022 at 3:33

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