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Fourier Transform Inverse Function

The Fourier Transform of the function $1/x$ is given by

$\displaystyle {\mathcal F}\left({-{1\over \pi x}}\right)$ $\textstyle =$ $\displaystyle -{1\over\pi}\int_{-\infty}^\infty {e^{-2\pi ikx}\over x}\,dx$ (1)
  $\textstyle =$ $\displaystyle PV\int_{-\infty}^\infty {\cos(2\pi kx)-i\sin(2\pi kx)\over x} \,dx$ (2)
  $\textstyle =$ $\displaystyle \left\{\begin{array}{ll} -{2i\over\pi} \int_0^\infty {\sin(2\pi k...
...} \int_0^\infty {\sin(2\pi kx)\over x}\,dx & \mbox{for $k>0$}\end{array}\right.$ (3)
  $\textstyle =$ $\displaystyle \left\{\begin{array}{ll}-i & \mbox{for $k<0$}\\  i & \mbox{for $k>0$,}\end{array}\right.$ (4)

which can also be written as the single equation
{\mathcal F}\left({-{1\over \pi x}}\right)= i[1-2H(-k)],
\end{displaymath} (5)

where $H(x)$ is the Heaviside Step Function. The integrals follow from the identity
$\displaystyle \int_0^\infty {\sin(2\pi kx)\over x}\,dx$ $\textstyle =$ $\displaystyle \int_0^\infty {\sin(2\pi kx)\over 2\pi kx}\, d(2\pi kx)$  
  $\textstyle =$ $\displaystyle \int_0^\infty \mathop{\rm sinc}\nolimits z\,dz = {\textstyle{1\over 2}}\pi.$ (6)

© 1996-9 Eric W. Weisstein