prove $Θ(t)=⁡\lim_{ϵ\to0+}\frac{1}{2πi} \int_{-∞}^{+∞}dβ\frac{e^{iβt}}{β-iϵ}$ for $Θ(t) =1 \text{ if t>0, and 0 else t<0} $ How to prove the following:
$$Θ(t)=⁡\lim_{ϵ\to0+}\frac{1}{2πi} \int_{-∞}^{+∞}dβ\frac{e^{iβt}}{β-iϵ}$$
for
$$Θ(t) =
\begin{cases}
1  & \text{if } t>0 \\
0 & \text{else } t<0
\end{cases}$$
I tried to solve the integral using exponential integral function using a substitution $u=iβt+ϵt$ which gave me;
$$Θ(t)=⁡\lim_{ϵ\to0+}\frac{e^{-ϵt}}{2πi}\left.Ei(iβt+ϵt)\right\rvert_{-∞}^{+∞} =⁡\lim_{ϵ\to0+}⁡e^{-ϵt}$$
But it gives $1$ at both cases. Can someone help me please?
 A: Your integral can be rewritten as:
\begin{align}\frac{1}{2\pi i}\int_\mathbb{R}\frac{\beta}{\beta^2+\varepsilon^2}e^{\beta i t}\,d\beta +\frac{\varepsilon}{2\pi}\int_\mathbb{R}\frac{e^{\beta i t}}{\beta^2+\varepsilon^2}\,d\beta\tag{0}\label{zero}
\end{align}
The second term in \eqref{zero} can be estimated as follows:
$$\frac{\varepsilon}{2\pi}\int_\mathbb{R}\frac{e^{\beta i t}}{\beta^2+\varepsilon^2}\,d\beta=\frac{1}{2\pi \varepsilon}\int\frac{e^{\beta i t}}{(\beta/\varepsilon)^2+1}\,d\beta=\frac{1}{2\pi }\int \frac{e^{\beta\varepsilon i t}}{\beta^2+1}\,d\beta=\frac{1}{2}e^{-|t\varepsilon|}
$$
The last identity follows from the relationship between the Cauchy distribution and the doubly exponential distribution.
Similarly, the first term in \eqref{zero}
\begin{align}
\frac{1}{2\pi i}\int_\mathbb{R}\frac{\beta}{\beta^2+\varepsilon^2}e^{\beta i t}\,d\beta&=\frac{1}{2\pi i}\int \frac{\beta}{\beta^2+1}e^{\beta\varepsilon i t}=\frac{1}{i\varepsilon}\partial_t\Big(\frac{1}{2i}e^{-|t\varepsilon|}\Big)\\
&=\operatorname{sign}(t)\frac{1}{2}e^{-|t\varepsilon|}
\end{align}
The conclusion follows immediately.
