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Fourier Transform--Inverse Function

The Fourier transform of the generalized function 1/x is given by

F_x(-PV1/(pix))(k)=-1/piPVint_(-infty)^infty(e^(-2piikx))/xdx
(1)
=PVint_(-infty)^infty(cos(2pikx)-isin(2pikx))/xdx
(2)
={-(2i)/piint_0^infty(sin(2pikx))/xdx for k<0; (2i)/piint_0^infty(sin(2pikx))/xdx for k>0
(3)
={-i for k<0; i for k>0,
(4)

where PV denotes the Cauchy principal value. Equation (4) can also be written as the single equation

F_x(-PV1/(pix))(k)=i[1-2H(-k)],
(5)

where H(x) is the Heaviside step function. The integrals follow from the identity

int_0^infty(sin(2pikx))/xdx=int_0^infty(sin(2pikx))/(2pikx)d(2pikx)
(6)
=int_0^inftysinc(z)dz
(7)
=1/2pi.
(8)

See also

Fourier Transform

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Cite this as:

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

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