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I need help to calculate the Fourier transform of this funcion

$$b(x) =\frac{1}{x^2 +a^2}\,,\qquad a > 0$$ Thanks.

Simon S
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Deiota
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    Where are you having trouble? – Ron Gordon Apr 30 '13 at 09:33
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    This is known as a Lorentzian function just to help you find material on this and its Fourier transform, as a hint: do you know what the Fourier transform of a decaying exponential ${{\rm e}^{-a \left| x \right| }}$ is? This result might be useful, and integrating that is easy if you split it over $+x$ and $-x$ regions... – Graham Hesketh Apr 30 '13 at 09:53
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    The easiest way to get Fourier transform of this is to use contour integral. – achille hui Apr 30 '13 at 09:58

2 Answers2

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Consider the function $f(x)=e^{-a|x|}$. Then \begin{align*} \hat{f}(\omega)&=\int_{-\infty}^{\infty}e^{-a|x|}e^{-i\omega x}\, dx= \int_{-\infty}^{0}e^{ax}e^{-i\omega x}\, dx+\int_{0}^{\infty}e^{-ax}e^{-i\omega x}\, dx = \\ &= \left[ \frac{e^{(a-i\omega)x}}{a-i\omega} \right]_{-\infty}^0-\left[ \frac{e^{-(a+i\omega)x}}{a+i\omega} \right]_{0}^{\infty}=\frac{1}{a-i\omega}+\frac{1}{a+i\omega}=\frac{2a}{a^2+\omega^2} \end{align*} Now, by the inversion forumla, we have \begin{equation*} e^{-a|x|}=\frac{1}{2\pi}\int_{-\infty}^{\infty}\frac{2a}{a^2+\omega^2}e^{i\omega x}\, d\omega \end{equation*} Changing the sign on $x$ and multiplying by $\frac{\pi}{a}$, we finally get \begin{equation*} \frac{\pi}{a} e^{-a|-x|}=\frac{\pi}{a} e^{-a|x|}=\int_{-\infty}^{\infty}\frac{e^{-i\omega x}}{a^2+\omega^2}\, d\omega \end{equation*} Thus, \begin{equation*} \hat{b}(\omega)=\frac{\pi}{a}e^{-a|\omega|} \end{equation*}

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One way to attack this is via the residue theorem. Consider

$$\oint_C dz \frac{e^{i k z}}{z^2+a^2}$$

where $C$ is a semicircular contour in the upper half plane of radius $R$. Note that, to use the residue theorem, we expect the integral over the circular arc to vanish as $R \to \infty$; however, this only happens when $k > 0$. (I leave it to the reader to show this.) The residue at the pole $z=i a$ is $e^{-k a}/(i 2 a)$, so, by the residue theorem,

$$\int_{-\infty}^{\infty} dx \frac{e^{i k x}}{x^2+a^2} = i 2 \pi \frac{e^{-a k}}{i 2 a} = \frac{\pi}{a} e^{-a k}$$

when $k > 0$. When $k < 0$, however, we must use the semicircular contour in the lower half plane, rather than the upper half plane. Thus, we now consider the pole at $z=-i a$, and the integral takes the value $(\pi/a) e^{a k}$ when $k < 0$. Putting this together, we have

$$\int_{-\infty}^{\infty} dx \frac{e^{i k x}}{x^2+a^2} = \frac{\pi}{a} e^{- a |k|}$$

Ron Gordon
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  • Already figured! – Deiota Apr 30 '13 at 17:38
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    @Amccds: I do not understand. – Ron Gordon Apr 30 '13 at 17:46
  • @RonGordon : I am afraid that residue theorem is used only for bounded domain where as in your answer you have used for unbounded domain... I am so confused :( –  Oct 30 '13 at 09:35
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    @PraphullaKoushik: I have no idea what you are talking about. – Ron Gordon Oct 30 '13 at 10:04
  • you said, by residue theorem $\int_{-\infty}^{\infty} dx \frac{e^{i k x}}{x^2+a^2} = i 2 \pi \frac{e^{-a k}}{i 2 a} = \frac{\pi}{a} e^{-a k}$, but residue theorem is used only for bounded domain where as our domain is unbounded right? –  Oct 30 '13 at 10:17
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    @PraphullaKoushik: I skipped a few steps. We take semicircles of radius $R$ and then consider the limit as $R \to \infty$. – Ron Gordon Oct 30 '13 at 10:20
  • Oh. In that case it is same as what i have thought.. I am new to this subject so i was afraid if i was missing something.. Thank you for clarification... Sorry for the inconvenience.. :) –  Oct 30 '13 at 10:24
  • @RonGordon: By the ML Lemma we can put a bound on the contour around the arc, i.e $$\left| \int_{arc} f(z) dz \right| \leq Max|F(z(t))|L,$$ where $Max|F(z(t))| = |e^{-Rk\sin t}|$ if $z = R(\cos t + i\sin t)$. Then, provided $k>0$, as $R \rightarrow \infty$, this vanishes. Is that correct? Also, if $k<0$ should you extend the semicircle in the lower half plane to $-\infty$ so that again, the term above vanishes? Thanks. – CAF Mar 26 '14 at 17:10
  • @CAF: that is right. – Ron Gordon Mar 26 '14 at 17:26
  • I have seen some fourier transforms where $\exp(-ikz)$ is taken instead. In that case, if $z=R(\cos t, i \sin t)$, $|\exp(-ikz)| = |\exp(kR\sin t)|$ in which case if $k<0$, then $R$ should be taken to $+\infty$ since $sint > 0$ above the plane. If $k>0$ then how to make this term vanish? If $R$ is taken to $-\infty$ since $\sin t $ is then negative, we end up with a net positive. If $R$ is taken to positive infinity, then too, we end up with a positive. Thanks. – CAF Mar 28 '14 at 09:32
  • Or alternatively, going back to the original example, when I said that if $k <0$ we should take $R \rightarrow -\infty$, then that means $\sin t < 0$ so we too end up with a net positive in the exponent and the result then diverges? – CAF Mar 28 '14 at 09:42
  • @RonGordon What can we conclude about the inverse fourier transform of b and the fourier transform of b - in relation to the inverse fourier transform of f and the fourier transform of f? – Monty May 06 '18 at 09:56