Inverse Fourier Transform
let $ \mathcal{X}(w) = w \times u(-w) $
then $ \mathcal{F}^{-1} = \frac{1}{2\pi} \int_{-\infty}^{+\infty} \mathcal{X}(w) e^{jwt}\, dw = \frac{1}{2\pi} \int_{-\infty}^{+\infty} wu(-w) e^{jwt} = \frac{1}{2\pi} \int_{-\infty}^{0} we^{jwt} $
using integration by parts:
let $ u = w, du = dw, dv = e^{jwt} dw, v = \frac{1}{jt}e^{jwt} $
$ \Rightarrow \mathcal{F}^{-1} = \frac{1}{2\pi} \times \left [ uv - \int v \, du \right ] = \left [\frac{w e^{jwt}}{jt} \right ]_{-\infty}^{0} - \int_{-\infty}^{0} \frac{1}{jt}e^{jwt} \, dw $
$ \left [\frac{w e^{jwt}}{jt} \right ]_{-\infty}^{0} = 0 $ and
$ - \int_{-\infty}^{0} \frac{1}{jt}e^{jwt} \, dw = - \frac{1}{jt} \left [\frac{e^{jwt}}{jt} \right ]_{-\infty}^{0} = \frac{1}{t} $
$ \Rightarrow \mathcal{F}^{-1} = \frac{1}{2\pi t} $
