Fourier Transform Inverse Function

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

$${\mathcal F}\left({-{1\over \pi x}}\right) = -{1\over\pi}\int_{-\infty}^\infty {e^{-2\pi ikx}\over x}\,dx$$ (1)

$$= PV\int_{-\infty}^\infty {\cos(2\pi kx)-i\sin(2\pi kx)\over x} \,dx$$ (2)

$$= \left\{\begin{array}{ll} -{2i\over\pi} \int_0^\infty {\sin(2\pi k... ...} \int_0^\infty {\sin(2\pi kx)\over x}\,dx & \mbox{for $k>0$}\end{array}\right.$$ (3)

$$= \left\{\begin{array}{ll}-i & \mbox{for $k<0$}\\ i & \mbox{for $k>0$,}\end{array}\right.$$ (4)

which can also be written as the single equation

$${\mathcal F}\left({-{1\over \pi x}}\right)= i[1-2H(-k)],$$ (5)

where $H(x)$ is the Heaviside Step Function. The integrals follow from the identity

$$\int_0^\infty {\sin(2\pi kx)\over x}\,dx = \int_0^\infty {\sin(2\pi kx)\over 2\pi kx}\, d(2\pi kx)$$

$$= \int_0^\infty \mathop{\rm sinc}\nolimits z\,dz = {\textstyle{1\over 2}}\pi.$$ (6)