$$ \begin{eqnarray}
x[n] = \frac{1}{2\pi} \int_{<2\pi>}X(\Omega)e^{i\Omega n}d\Omega\\\\
X(\Omega) = \sum_{k=-\infty}^{\infty}x[k]e^{-i\Omega k}
\end{eqnarray}$$
DFT FOR PERIODIC CASE
Consider a periodic signal $x[n]$, $x[n+N]=x[n]$.
$$ \begin{eqnarray}
X(\Omega) &=& \sum_{k=-\infty}^{\infty}x[k]e^{-i\Omega k}\\\\
&=& \sum_{n=-\infty}^{\infty}\sum_{k=nN}^{(n+1)N-1}x[k]e^{-i\Omega k}\\\\
&=& \sum_{n=-\infty}^{\infty} \sum_{k=0}^{N-1}x[k+nN]e^{-i\Omega (k+nN)}\\\\
&=& \sum_{n=-\infty}^{\infty} \sum_{k=0}^{N-1}x[k]e^{-i\Omega k}e^{-i\Omega Nn}\\\\
&=&\left( \sum_{k=0}^{N-1}x[k]e^{-i\Omega k}\right)\left(\sum_{n=-\infty}^{\infty}e^{-i\Omega Nn}\right)
\end{eqnarray}$$