Discrete-time Fourier Transform of the unit step sequence $u[n]$
Cedron Dawg posted an interesting initial point in this answer. It begins with these steps:

$$ \begin{align}
U(\omega) &amp;= \sum\limits_{n=0}^{+\infty}  e^{-j \omega n} \\
&amp;= \lim_{ N \to \infty } \sum\limits_{n=0}^{N-1}  e^{-j \omega n}\\
&amp;= \lim_{ N \to \infty } \left[  \frac{  1 -  e^{-j \omega N}  }{ 1 -  e^{-j \omega }   } \right] \\
&amp;= \frac{ 1 }{ 1 -  e^{-j \omega } } - \lim_{ N \to \infty } \left[  \frac{  e^{-j \omega N}  }{ 1 -  e^{-j \omega } } \right]
\end{align} $$

It turns out the term inside the limit can be expanded as follows:

$$\begin{aligned}
\frac{  e^{-j \omega N}  }{ 1 -  e^{-j \omega } }
={}
&amp;\frac{1}{\sin^2(\omega)+(1-\cos(\omega))^2} \cdot \\
&amp;\left[-\cos(\omega)\cos(N\omega)+\cos(N\omega)-\sin(\omega)\sin(N\omega)+ \\
j(-\sin(\omega)\cos(N\omega)+\cos(\omega)\sin(\omega)-\sin(N\omega))\right]
\end{aligned}
$$

The common factor outside the brackets can be expressed as:

$$\frac{1}{\sin^2(\omega)+(1-\cos(\omega))^2} = \frac{1}{4\sin^2(\omega/2)}$$

The real part inside the brackets also equals:

$$-\cos(\omega)\cos(N\omega)+\cos(N\omega)-\sin(\omega)\sin(N\omega)= 2\sin(\omega/2)\sin[\omega(-N+1/2)]$$

On the other hand, the imaginary part can be rewritten as:

$$-\sin(\omega)\cos(N\omega)+\cos(\omega)\sin(\omega)-\sin(N\omega)=-2\sin(\omega/2)\cos[\omega(-N+1/2)]$$

Rewritting the original term we get that:

$$\begin{align}
\frac{  e^{-j \omega N}  }{ 1 -  e^{-j \omega } }
&amp;=\frac{2\sin\left(\frac \omega 2\right)}{4\sin^2\left(\frac \omega 2\right) }
\left( \sin[\omega(-N+1/2)] - j\cos[\omega(-N+1/2)]\right) \\
&amp;=-\frac{\sin[\omega(M+1/2)]}{2\sin\left(\frac \omega 2\right)} - j\frac{\cos[\omega(M+1/2)]}{2\sin\left(\frac \omega 2\right)}
\end{align}
$$

where I used $M=N-1$ and the limit stays unaffected as $M\to \infty$ as well.

According to the 7th definition in this site:

$$\lim_{M\to \infty} -\frac{1}{2\sin(\omega/2)}\sin[\omega(M+1/2)] = -\pi \delta(\omega)$$

So far we have that:

$$\lim_{ M \to \infty }   \frac{  e^{-j \omega (M+1)}  }{ 1 -  e^{-j \omega } } =-\pi\delta(\omega)
-j\lim_{M\to \infty} \frac{\cos[\omega(M+1/2)]}{2\sin(\omega/2)}
$$

If we could prove that the second term on the right of the equality is $0$ in some sense, then we are done. I asked it at math.SE and, indeed, that sequence of functions tends to the zero distribution. So, we have that:

$$ \begin{align}
U(\omega) &amp;= \frac{ 1 }{ 1 -  e^{-j \omega } } - \lim_{ N \to \infty } \left[  \frac{  e^{-j \omega N}  }{ 1 -  e^{-j \omega } } \right] \\
&amp;=\frac{ 1 }{ 1 -  e^{-j \omega } }+\pi\delta(\omega)
+j\lim_{M\to \infty} \frac{\cos[\omega(M+1/2)]}{2\sin(\omega/2)} \\
&amp; =\frac{ 1 }{ 1 -  e^{-j \omega } }+\pi\delta(\omega)
\end{align} $$