DTFT of the unit step function If i apply the DTFT on unit step function, then i get follow:
$$DTFT\{u[n]\}=\sum_{n=-\infty}^{\infty}u[n]e^{-j\omega n}=\sum_{n=0}^{\infty}e^{-j\omega n} = \frac{1}{1-e^{-j\omega}}$$.
Now i have the problem, if $|e^{-j\omega}|$ = 1, the sum diverges.To handle this case, i know that $e^{-j\omega}$ is $2\pi$ periodic, i get 
$$\frac{1}{1-e^{-j\omega}}+\underbrace{e^{-j0}}_1
\sum_{k=-\infty}^{\infty}\delta(\omega+2\pi k)$$.
In books i found that the DTFT of the unit step is 
$$\frac{1}{1-e^{-j\omega}}+\pi
\sum_{k=-\infty}^{\infty}\delta(\omega+2\pi k)$$.
Can me anyone explain why get the $\pi$ in the DTFT of the unit step?

Hi guys I think this way is better than others,What's Ur idea?
$u[n]=f[n] + g[n] $
Where:
$f[n]= {1\over2}$  for  $-\infty<n<\infty $
and
$g[n]=\left\{ 
\begin{array}{c}
{1\over2} \text{ for } n\ge 0 \\ 
{-1\over2} \text{ for } n<0
\end{array}
\right.$
do:
$ \delta [n] = g[n] - g[n-1]$
U Know  DTFT of $\delta[n]$ is $1$
and DTFT of $g[n] - g[n-1]\to G(e^{j\omega})-e^{-j\omega}G(e^{j\omega})$
so:
$1=G(e^{j\omega})-e^{-j\omega}G(e^{j\omega})$
therefore
$G(e^{j\omega})={1\over 1-e^{-j\omega}}$
and we know that the DTFT of $f[n]\to F(e^{j\omega})=\pi\sum_{k=-\infty}^\infty\delta(\omega -2\pi k)$
finally :
$u[n] = f[n]+g[n] \to U(e^{j\omega})=F(e^{j\omega})+G(e^{j\omega})$ 
$U(e^{j\omega})={1\over 1-e^{-j\omega}}+\pi\sum_{k=-\infty}^\infty\delta(\omega -2\pi k)$
