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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.

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closed as off-topic by user223391, colormegone, Namaste, Shailesh, Strants May 7 '16 at 0:35

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    $\begingroup$ Where are you having trouble? $\endgroup$ – Ron Gordon Apr 30 '13 at 9:33
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    $\begingroup$ 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... $\endgroup$ – Graham Hesketh Apr 30 '13 at 9:53
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    $\begingroup$ The easiest way to get Fourier transform of this is to use contour integral. $\endgroup$ – achille hui Apr 30 '13 at 9:58
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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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  • $\begingroup$ @user60725 is this result valid if $a$ is complex? $\endgroup$ – Tom Jan 10 '16 at 12:04
  • $\begingroup$ What about having at the beginning $a = 0$? $\endgroup$ – Von Neumann Apr 7 '17 at 15:56
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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

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

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  • $\begingroup$ Already figured! $\endgroup$ – Deiota Apr 30 '13 at 17:38
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    $\begingroup$ @Amccds: I do not understand. $\endgroup$ – Ron Gordon Apr 30 '13 at 17:46
  • $\begingroup$ @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 :( $\endgroup$ – user87543 Oct 30 '13 at 9:35
  • $\begingroup$ @PraphullaKoushik: I have no idea what you are talking about. $\endgroup$ – Ron Gordon Oct 30 '13 at 10:04
  • $\begingroup$ 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? $\endgroup$ – user87543 Oct 30 '13 at 10:17

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