For discrete- -time sinc function, what is the inverse discrete-time Fourier transform of the function as shown in the figure?

Concept: Inverse disude time fourier transform is given by the formula

x(n) = \(\frac{1}{{2π }}\int\limits_{ - π }^π { \times \left( {{e^{jw}}} \right)} \,\,{e^{jwn}}\,dw\)

Calculation: Given DTFT

i.e. x(e^jn) = \(\left\{ {\begin{array}{*{20}{c}} 1&{\left| \Omega \right| < w} \\ 0&{w\, < \,\,\left| \Omega \right| \leqslant π } \end{array}} \right.\)

Applying the formulae for inverse DTFT we get

x[n] = \(\frac{1}{{2π }}\int\limits_{ - π }^π { \times \left( {{e^{j\Omega }}} \right)} \,\,{e^{j\Omega n}}\,d\Omega \)

        = \(\frac{1}{{2π }}\int\limits_{ - w}^w {{e^{j\Omega n}}\,d\Omega } \,\,\)

       = \(\frac{{{e^{j\Omega n}}}}{{2π jn}}\int_{ - w}^w {} = \,\frac{{{e^{jwn}} - {e^{ - jwn}}}}{{\left( {2j} \right)π n}}\,\, = \,\,\frac{{\sin \,(wn)}}{{π n}}\)

Converting into sinc function:   

x[n] = \(\frac{sin(Wn)}{Wn} .\frac{Wn}{πn}\) = \(\frac{{\sin π \frac{{Wn}}{π }}}{{π \,.\frac{{Wn}}{π }}}\,\,.\,\,\frac{{π \frac{{Wn}}{π }}}{{π n}}\)

x [n] = \(\frac{w}{π} sinc(\frac{w}{π}n)\)