Inverse FT of $Z(\omega) = a [- \frac{1}{i\omega}+\pi \delta(\omega)]$ (Contour integration) Given is the Fourier transform of some function $z(t)$:
$Z(\omega) = a [- \frac{1}{i\omega}+\pi \delta(\omega)]$
I now want to invert the tranform using contour integrals.
How can I do that?
I noticed that $Z$ has a singularity at $\omega =0$, but I don't yet know how to use this yet.
Can anyone help?
 A: The inverse Fourier transform is assumed to take the form
$$z(t) = \frac{1}{2 \pi} \int_{-\infty}^{\infty} d\omega \, Z(\omega) \, e^{-i \omega t}$$
Let's consider the two terms of $Z$ separately.  The delta term just gives a constant:
$$z_1(t) = \frac{1}{2 \pi} a \pi \int_{-\infty}^{\infty} d\omega \, \delta(\omega) \, e^{-i \omega t} = \frac{a}{2}$$
The $1/\omega$ piece we can handle in the complex plane, although keep in mind that we will be finding its Cauchy principal value.  Consider the contour integral:
$$-\frac{a}{i 2 \pi}\oint_C dz \frac{e^{-i t z}}{z}$$
where $t \gt 0$.  In this case, $C$ is a semicircular contour of radius $R$ in the lower half-plane, with a small semicircular deformation of radius $\epsilon$ about the origin into the lower half-plane so as to avoid the pole at the origin.  We will eventually take the limits as $R \to \infty$ and $\epsilon \to 0$.  We write this contour integral out as follows:
$$-\frac{a}{i 2 \pi} \int_{-\epsilon}^{-R} dx  \frac{e^{-i t x}}{x} -\frac{a}{i 2 \pi} i R \int_{-\pi}^0 d\theta \, e^{i \theta} \frac{e^{t R \sin{\theta}} e^{-i t R \cos{\theta}}}{R e^{i \theta}} \\ -\frac{a}{i 2 \pi} \int_{R}^{\epsilon} dx  \frac{e^{-i t x}}{x} -\frac{a}{i 2 \pi} i \epsilon \int_0^{-\pi} d\phi \, e^{i \phi} \frac{e^{-i t \epsilon e^{i \phi}}}{\epsilon e^{i \phi}}$$
As $R \to \infty$, we may show that the second integral vanishes, because its magnitude is bounded by
$$\frac{a}{2 \pi} 2 \int_0^{\pi/2} d\theta \, e^{-t R \sin{\theta}} \le \frac{a}{\pi} \int_0^{\pi/2} d\theta \, e^{-2 t R \theta/\pi} \le \frac{a}{2 R t}$$
The first and third integrals combine to form the Cauchy principal value of the sought-after integral.  By Cauchy's theorem, the contour integral is zero (recall that we avoided the pole at the origin), so that we now have
$$-PV \int_{-\infty}^{\infty} d\omega \frac{e^{-i \omega t}}{\omega} - i \pi = 0$$
Now, remember that this was for $t \gt 0$.  For $t \lt 0$, we must use a contour in the upper half-plane.  Performing similar steps as above, we find that for $t \lt 0$:
$$PV \int_{-\infty}^{\infty} d\omega \frac{e^{-i \omega t}}{\omega} - i \pi = 0$$
That is, 
$$z_2(t) = \left (-i \pi \text{sgn}(t)\right ) \left (- \frac{a}{i 2 \pi} \right ) = \frac{a}{2} \text{sgn}(t)$$
where sgn is the signum (sign) function.  The transform we seek is then
$$z(t) = z_1(t) + z_2(t) = \frac{a}{2} [1+\text{sgn}(t)] = a \theta(t)$$
where $\theta$ is the Heaviside step function (and is not to be confused with the angular integration variables used above).
