Fourier Transform--Exponential Function

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

$${\mathcal F}[e^{-k_0\vert x\vert}] = \int_{-\infty}^\infty e^{-k_0\vert x\vert}e^{-2\pi ikx}\,dx$$

$$= \int_{-\infty}^0 e^{-2\pi ikx}e^{2\pi xk_0}\,dx+\int_0^\infty e^{-2\pi ikx}e^{-2\pi k_0x}\,dx$$

$$= \int_{-\infty}^0 [\cos(2\pi kx)-i\sin(2\pi kx)]e^{2\pi k_0x}\,dx+ \int_0^\infty [\cos(2\pi kx)-i\sin(2\pi kx)]e^{-2\pi k_0x}\,dx.$$ (1)

Now let $u\equiv -x$ so $du=-dx$, then

$${\mathcal F}[e^{-k_0\vert x\vert}] = \int_0^\infty [\cos(2\pi ku)+i\sin(2\pi ku)]e^{-2\pi k_0u}\,du]$$

$$\phantom{=} +\int_0^\infty [\cos(2\pi ku)-i\sin(2\pi ku)]e^{-2\pi k_0u}\,du]$$

$$= 2\int_0^\infty \cos(2\pi ku)e^{-2\pi k_0u}\,du,$$ (2)

which, from the Damped Exponential Cosine Integral, gives

$${\mathcal F}[e^{-2\pi k_0\vert x\vert}] = {1\over\pi} {k_0\over k^2+{k_0}^2},$$ (3)

which is a Lorentzian Function.

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