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I need to find the Fourier transform of $e^{-2 \pi |x|}$. Normally, I can do something like this, but the absolute value is kind of confusing me.

doraemonpaul
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Richard Carpenter
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1 Answers1

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Hint $$f(x)=\begin{align}e^{-2\pi x}\;\;\;\;\;\;x>0\\ e^{2\pi x}\;\;\;\;\;\;\;x<0 \end{align}$$ Thus the fourier transform is $$\hat{f}{(\psi)}=\int_{-\infty}^0e^{-2\pi i\psi x}e^{2\pi x}d\psi+\int_0^{\infty}e^{-2\pi i\psi x}e^{-2\pi x}d\psi$$

  • So if I have $f(x) = e^{-2 \pi x}$ for $x > 0$ and $0$ when $x < 0$. Also, $g(x) = 0$ when $x > 0$ and $e^{2 \pi x}$ when $x < 0$, then would the Fourier transform of $e^{-2 \pi |x|}$ be the sum of the two Fourier transforms I just said? That would mean flipping the inequalities of what you have above. – Richard Carpenter Apr 26 '13 at 00:18
  • yes. but it seems, mistakenly, I swapped signs above. you are right. –  Apr 26 '13 at 00:21