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Fourier Transform--Exponential Function

The Fourier Transform of $e^{-k_0\vert x\vert}$ is given by

$\displaystyle {\mathcal F}[e^{-k_0\vert x\vert}]$ $\textstyle =$ $\displaystyle \int_{-\infty}^\infty e^{-k_0\vert x\vert}e^{-2\pi ikx}\,dx$  
  $\textstyle =$ $\displaystyle \int_{-\infty}^0 e^{-2\pi ikx}e^{2\pi xk_0}\,dx+\int_0^\infty e^{-2\pi ikx}e^{-2\pi k_0x}\,dx$  
  $\textstyle =$ $\displaystyle \int_{-\infty}^0 [\cos(2\pi kx)-i\sin(2\pi kx)]e^{2\pi k_0x}\,dx+ \int_0^\infty [\cos(2\pi kx)-i\sin(2\pi kx)]e^{-2\pi k_0x}\,dx.$ (1)

Now let $u\equiv -x$ so $du=-dx$, then
$\displaystyle {\mathcal F}[e^{-k_0\vert x\vert}]$ $\textstyle =$ $\displaystyle \int_0^\infty [\cos(2\pi ku)+i\sin(2\pi ku)]e^{-2\pi k_0u}\,du]$  
  $\textstyle \phantom{=}$ $\displaystyle +\int_0^\infty [\cos(2\pi ku)-i\sin(2\pi ku)]e^{-2\pi k_0u}\,du]$  
  $\textstyle =$ $\displaystyle 2\int_0^\infty \cos(2\pi ku)e^{-2\pi k_0u}\,du,$ (2)

which, from the Damped Exponential Cosine Integral, gives
{\mathcal F}[e^{-2\pi k_0\vert x\vert}] = {1\over\pi} {k_0\over k^2+{k_0}^2},
\end{displaymath} (3)

which is a Lorentzian Function.

See also Damped Exponential Cosine Integral, Exponential Function, Lorentzian Function

© 1996-9 Eric W. Weisstein