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I have been doing some physics and after a long calculation (tell me if you would like more context) I have to calculate the following integral (which could be interpreted as a distribution)

$$\int_{-\infty}^\infty d\omega e^{i\omega t} \frac{1}{-M\omega^2+\eta i \omega |\omega|^{2\alpha}},$$

where $M, \eta > 0$. I tried doing a substitution $ \omega \to -i\omega$, which gives $$i\int_{-\infty}^\infty d\omega e^{-\omega t} \frac{1}{-M\omega^2-\eta \omega |\omega|^{2\alpha}} = i\int_{0}^\infty d\omega e^{-\omega t}\frac{1}{-M\omega^2-\eta \omega \omega^{2\alpha}} + i\int_{-\infty}^0d\omega e^{-\omega t}\frac{1}{-M\omega^2+\eta \omega \omega^{2\alpha}} .$$ Where I would think that the first integral would vanish by using complex analysis (since there are no poles) but the second integral diverges I think, which maybe could be seen as some kind of distribution? I am not sure if these last steps even simplified the problem.

Thanks for the help!

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Partial answer:

  • You should always first clean/simplify the question a maximum, for example asking for the inverse Fourier transform amounts to ask for the Fourier transform with $t$ replaced by $-t$ up to a constant factor. Also, you can define $a = 2\alpha$ and remove the constants $M$ and $\eta$ since by the change of variable $\omega = -\lambda\,x$ your integral becomes $$ I(t) = \int_{-\infty}^\infty e^{-i\,\lambda\,x\,t} \frac{1}{-M\lambda^2 |x|^2 - i\,\eta\,\lambda^{a+1}\,|x|^a\,x} \,\lambda\,\mathrm d x $$ and so by taking $\lambda^{1-a} = \eta/M$, it gives $M\lambda^2 = \eta\lambda^{a+1} = M^\frac{a+1}{1-a}\eta^\frac{2}{1-a}$ and so $$ I(t) = \frac{-1}{M\lambda}\int_{-\infty}^\infty e^{-i\,\lambda\,x\,t} \frac{1}{|x|^2 + i\,|x|^a\,x} \,\mathrm d x $$ and so what you really want to know is the Fourier transform of the function $h_a(x) = \frac{1}{|x|^2 + i\,|x|^a\,x}$. Then your integral will just be $-\frac{1}{M\lambda}\,\widehat{h_a}(\lambda\,t)$. Now decomposing the function in a real and an imaginary part you can also write $h_a = f_a - i\,g_a$ where $$ f_a(x) = \frac{|x|^2}{|x|^4+|x|^{2a+2}}, \qquad g_a(x) = \frac{|x|^a\,x}{|x|^4+|x|^{2a+2}}. $$

  • In the case when $a=1$, then $$ f_1(x) = \frac{1}{|x|^2}, \qquad g_a(x) = \frac{x}{|x|^3}. $$ and you can get a result using the fact that the function are homogeneous as in the answers here The Fourier transform of $1/p^3$, Fourier transform of $ |x|^{s} $ and $\log|x| $. In this case, to go faster, let me use WolframAlpha to get indeed that $\widehat{h_1}(t) = t\left(\ln(|t|)+\gamma-1\right) -\frac{\pi\,|t|}{2}$.

  • I would bet that there is unfortunately no simple expression of these Fourier transforms in terms of standard functions in the general case of a real number $a$.

  • Now the true question is what do you want to know precisely about this function? Indeed, the behavior of the function gives a lot of information on the behavior of the Fourier transform. For example the decay of the function gives you informations about the regularity of its Fourier transform, and reciprocally the regularity gives you informations about the decay of the Fourier transform ...

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  • $\begingroup$ Thanks for the answer and it helped already, and I will keep the cleaning up in my mind, it is a very good habit which I still need to build. I am interested in the time evolution of this function convoluted with another function (that has a horible specific form haha) more specifically, for small $t$ and for large $t$ (if there is some asymptotic behavior) . $\endgroup$
    – Audrique
    Dec 13, 2022 at 22:12
  • $\begingroup$ Hi @LL3.14 ... The Fourier transform of $f_1=1/|x|^2$ is $-\pi |t|/2$. If fact, WA does not give the other terms (i.e., "$t(\log(|t|)+\gamma-1)$"). $\endgroup$
    – Mark Viola
    Dec 14, 2022 at 0:38
  • $\begingroup$ Yes, it was for the whole $h_1$, not $f_1$, but it strangely does not indicate a real and imaginary part ... $\endgroup$
    – LL 3.14
    Dec 14, 2022 at 9:37
  • $\begingroup$ Then I will add the behavior later if Ihave tine $\endgroup$
    – LL 3.14
    Dec 14, 2022 at 9:39

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