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?