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The question asks to find the Fourier transform of $(x^2 + b^2)^{-1}$ given that $F[e^{-b\vert x\vert}](k) = \frac{1}{\sqrt{2\pi}}\frac{2b}{k^2+b^2} $ $$\sqrt{\frac{\pi}{2b^2}}e^{-b\vert x\vert} = I\left[\frac{1}{k^2+b^2}\right](x).$$ Making the transformation $x\leftrightarrow k$ $$I\left[\frac{1}{x^2+b^2}\right](k)= \sqrt{\frac{\pi}{2b^2}}e^{-b\arrowvert k\arrowvert}$$ $$F[f(x)](k) = I[f(x)](-k) \;\;\;(*).$$ Making the transformation $k\rightarrow -k$ $$F\left[\frac{1}{x^2+b^2}\right](k) = \sqrt{\frac{\pi}{2b^2}}e^{-b\vert k\vert}$$

What I don't understand is how (*) was obtained from the previous step. Can I also ascertain the notation that $F[f(x)](k)$ means taking $\frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty}e^{-ikx}f(x)dx$ and that the answer obtained will be in terms of $k$, which explains the $(k)$?

Davide Giraudo
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user91820
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  • $I[g]$ is the inverse Fourier transform? Then $(\ast)$ is general, applying the Fourier transform twice reflects the function in $0$. – Daniel Fischer Sep 02 '13 at 15:16
  • Yes, I[g] is the inverse Fourier transform, although I don't understand how (*) is a manifestation of having applied the Fourier transform twice. – user91820 Sep 02 '13 at 15:47
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    The integral representation $Ig = \frac{1}{\sqrt{2\pi}}\int g(t)e^{i\xi t},dt$ means $Ig = Fg$. Now if we write $g = F[f]$, we get $f(\xi) = IF[f] = FF[f]$, almost everywhere, i.e. $F^2[f]$ is the reflection of $f$ in $0$. Conversely, if you know that $F^2[f]$ is the reflection, the integral of the inverse Fourier transform follows. – Daniel Fischer Sep 02 '13 at 15:54
  • Ok I follow your explanation up to the point where $F^2[f]$ is the reflection of f in 0. But I don't understand what is meant by "if $F^2[f]$ is the reflection, the integral of the inverse Fourier transform follows" and how all these links back to the question. Apologies if you find my questions very dumb. – user91820 Sep 05 '13 at 14:23

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