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GeezerVison is showing mathworld.wolfram.com — Fourier Transform--Exponential Function -- from Wolfram MathWorld

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Fourier Transform--Exponential Function

The Fourier transform of e^(-2pik_0|x|) for k_0>0 is given by

F_x[e^(-2pik_0|x|)](k)=int_(-infty)^inftye^(-2pik_0|x|)e^(-2piikx)dx=int_(-infty)^0e^(-2piikx)e^(2pixk_0)dx+int_0^inftye^(-2piikx)e^(-2pik_0x)dx
(1)
=int_(-infty)^0[cos(2pikx)-isin(2pikx)]e^(2pik_0x)dx+int_0^infty[cos(2pikx)-isin(2pikx)]e^(-2pik_0x)dx.
(2)

Now let u=-x so du=-dx, then

F_x[e^(-2pik_0|x|)](k) =int_0^infty[cos(2piku)+isin(2piku)]e^(-2pik_0u)du+int_0^infty[cos(2piku)-isin(2piku)]e^(-2pik_0u)du =2int_0^inftycos(2piku)e^(-2pik_0u)du,
(3)

which, from the damped exponential cosine integral, gives

F_x[e^(-2pik_0|x|)](k)=1/pi(k_0)/(k^2+k_0^2),
(4)

which is a Lorentzian function.

The transform is real because the original function is real and an even function. In contrast, one-sided decay has the generally complex transform

F_x[H(x)e^(-ax)](k)=1/(a+2piik)
(5)

for a>0, where H(x) is the Heaviside step function.

See also

Damped Exponential Cosine Integral, Exponential Function, Fourier Transform, Heaviside Step Function, Lorentzian Function

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Cite this as:

Weisstein, Eric W. "Fourier Transform--Exponential Function." From MathWorld--A Wolfram Resource. https://mathworld.wolfram.com/FourierTransformExponentialFunction.html

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