Fourier transform Heaviside step function

The definition of Heaviside step function is :

\[\theta(x)= \begin{cases}0 & x<0 \\ \frac{1}{2} & x=0 \\ 1 & x>0\end{cases}\]

It can be written as (If we substitute $x$ with $t-t’$ as we’ll be using them in time ordering operator):

\[\theta\left(t-t^{\prime}\right)=-\frac{1}{2 \pi i} \lim_{\eta \rightarrow 0^+}\int_{-\infty}^{\infty} d \omega \frac{e^{-i \omega (t-t^{\prime})}}{\omega+i \eta} \tag{1}\]

To show this relation is correct, we need to use the The Sokhotski-Plemelj formula:

\[\lim _{\epsilon \rightarrow 0} \int_{-\infty}^{\infty} \frac{f(x) d x}{x \pm i \epsilon}=\mathcal{P} \int_{-\infty}^{\infty} \frac{f(x) d x}{x} \mp i \pi f(0)\]

and Eq. 1 turns into:

\[\begin{aligned} \theta\left(t-t^{\prime}\right)&=-\frac{1}{2 \pi i} \lim_{\eta \rightarrow 0^+}\int_{-\infty}^{\infty} d x \frac{e^{-i x\left(t-t^{\prime}\right)}}{x+i \eta}\\ &= -\frac{1}{2 \pi i} \left [ \mathcal{P} \int_{-\infty}^{\infty} \frac{ e^{-i x (t-t^{\prime})} d x}{x} - i \pi \right] \end{aligned} \tag{2}\]

expanding the tems inside the braket:

\[\begin{aligned} & \mathcal{P} \int_{-\infty}^{\infty} \frac{ e^{-i x (t-t^{\prime})} d x}{x} - i \pi \\ =& \lim_{\delta \rightarrow 0} \left \{ \int_{-\infty}^{-\delta} \frac{ e^{-i x (t-t^{\prime})} d x}{x} + \int_{\delta}^{\infty} \frac{ e^{-i x (t-t^{\prime})} d x}{x}\right \} -i \pi \\ =& \lim_{\delta \rightarrow 0} \left \{ \int_{-\infty}^{-\delta} \frac{ \cos[x (t-t^{\prime})] - i \sin[x (t-t^{\prime})]}{x}dx + \int_{\delta}^{\infty} \frac{ \cos[x (t-t^{\prime})] - i \sin[x (t-t^{\prime})]}{x}dx \right \} -i\pi\\ \end{aligned}\]

Becasue $\cos$ is odd, two $\cos$ within the first and second integral vanishes, and in the limit of $\delta \rightarrow 0$ we can combine the $\sin$ terms, giving us:

\[\begin{aligned} & \mathcal{P} \int_{-\infty}^{\infty} \frac{ e^{-i x (t-t^{\prime})} d x}{x} - i \pi \\ =&- i \int_{-\infty}^{\infty} \frac{ \sin[x (t-t^{\prime})]}{x}dx -i \pi \\ \end{aligned} \tag{3}\]

Note that the integral can be re-written as:

\[\int_{-\infty}^{\infty} \frac{ \sin[x (t-t^{\prime})]}{x}dx = \mathrm{Si} (x[t-t^{\prime})] \big \vert_{x=-\infty}^{x=\infty} = \int_{0}^{x(t-t^{\prime})} \frac{ \sin(\omega)}{\omega} d\omega \big \vert_{x=-\infty}^{x=\infty}\]

where $\mathrm{Si}$ is the sine integral.

Assuming we have $t-t’ > 0$ we have:

\[\int_{-\infty}^{\infty} \frac{ \sin[x (t-t^{\prime})]}{x}dx =\int_{0}^{x(t-t^{\prime})} \frac{ \sin(\omega)}{\omega} d\omega \big \vert_{x=-\infty}^{x=\infty} = \pi\]

So that Eq. 3 is:

\[\mathcal{P} \int_{-\infty}^{\infty} \frac{ e^{-i x (t-t^{\prime})} d x}{x} - i \pi = 0 \tag{4}\]

and if $t-t’ < 0$ we have:

\[\int_{-\infty}^{\infty} \frac{ \sin[x (t-t^{\prime})]}{x}dx =\int_{0}^{x(t-t^{\prime})} \frac{ \sin(\omega)}{\omega} d\omega \big \vert_{x=-\infty}^{x=\infty} =- \pi\]

So that Eq. 3 is:

\[\mathcal{P} \int_{-\infty}^{\infty} \frac{ e^{-i x (t-t^{\prime})} d x}{x} - i \pi = -2i\pi \tag{5}\]

and if $t-t’ = 0$ we have:

\[\int_{-\infty}^{\infty} \frac{ \sin[x (t-t^{\prime})]}{x}dx =\int_{0}^{x(t-t^{\prime})} \frac{ \sin(\omega)}{\omega} d\omega \big \vert_{x=-\infty}^{x=\infty} = 0\]

So that Eq. 3 is:

\[\mathcal{P} \int_{-\infty}^{\infty} \frac{ e^{-i x (t-t^{\prime})} d x}{x} - i \pi = -i\pi \tag{6}\]

Multiply Eq. 4, 5 and 6 with $\frac{-1}{i 2 \pi}$ gives Eq. 2:

\[\theta(t-t')= \begin{cases}0 & t-t'<0 \\ \frac{1}{2} & t-t'=0 \\ 1 & t-t'>0\end{cases}\]