Added 2/20/2010. It took me a while to derive this. The textbook was not clear. So I am it more clear below.
Give a unit step, over some window, as in
\[ u\left [ n\right ] =\left \{ \begin{array} [c]{ccc}1 & & \left \vert n\right \vert \leq M\\ 0 & & o.w. \end{array} \right . \]
Find its DTFT. There is a good trick to remember
\[ X\left ( \omega \right ) ={\sum \limits _{n=-M}^{M}}e^{-j\omega n}\]
Let \(n^{\prime }=M+n\), then when \(n=-M,n^{\prime }=0\), and when \(n=M,n^{\prime }=2M\), hence the above becomes
\[ X\left ( \omega \right ) ={\sum \limits _{n^{\prime }=0}^{2M}}e^{-j\omega \left ( n^{\prime }-M\right ) }={\sum \limits _{n^{\prime }=0}^{2M}}e^{-j\omega n^{\prime }}e^{+j\omega M}=e^{j\omega M}{\sum \limits _{n^{\prime }=0}^{2M}}e^{-j\omega n^{\prime }}\]
but \(n^{\prime }\) is dummy variable, so the above can be written as
\[ X\left ( \omega \right ) =e^{j\omega M}{\sum \limits _{n=0}^{2M}}e^{-j\omega n}\]
Ok, now we use the standard geometric series, the above becomes
\[ X\left ( \omega \right ) =e^{j\omega M}\frac{1-\left ( e^{-j\omega }\right ) ^{2M+1}}{1-e^{-j\omega }}=e^{j\omega M}\frac{1-e^{-j\omega \left ( 2M+1\right ) }}{1-e^{-j\omega }}\]
Ok, so far, nothing too exciting has happened, just using standard definitions. Here is the trick, without which I had hard time. The trick is to pull put half the exponential from the numerator and denominator, we obtain
\[ X\left ( \omega \right ) =e^{j\omega M}\frac{e^{-\frac{j\omega \left ( 2M+1\right ) }{2}}\left [ e^{\frac{j\omega \left ( 2M+1\right ) }{2}}-e^{-\frac{j\omega \left ( 2M+1\right ) }{2}}\right ] }{e^{-\frac{j\omega }{2}}\left ( e^{\frac{j\omega }{2}}-e^{-\frac{j\omega }{2}}\right ) }\]
Now simplify
\begin{align*} X\left ( \omega \right ) & =e^{j\omega M-\frac{j\omega \left ( 2M+1\right ) }{2}}\frac{\left [ e^{\frac{j\omega \left ( 2M+1\right ) }{2}}-e^{-\frac{j\omega \left ( 2M+1\right ) }{2}}\right ] }{e^{-\frac{j\omega }{2}}\left ( e^{\frac{j\omega }{2}}-e^{-\frac{j\omega }{2}}\right ) }\\ & =e^{\frac{2j\omega M-2j\omega M-j\omega }{2}}\frac{\left [ e^{\frac{j\omega \left ( 2M+1\right ) }{2}}-e^{-\frac{j\omega \left ( 2M+1\right ) }{2}}\right ] }{e^{-\frac{j\omega }{2}}\left ( e^{\frac{j\omega }{2}}-e^{-\frac{j\omega }{2}}\right ) }\\ & =e^{-\frac{j\omega }{2}}\frac{\left [ e^{\frac{j\omega \left ( 2M+1\right ) }{2}}-e^{-\frac{j\omega \left ( 2M+1\right ) }{2}}\right ] }{e^{-\frac{j\omega }{2}}\left ( e^{\frac{j\omega }{2}}-e^{-\frac{j\omega }{2}}\right ) }\\ & =\frac{e^{\frac{j\omega \left ( 2M+1\right ) }{2}}-e^{-\frac{j\omega \left ( 2M+1\right ) }{2}}}{\left ( e^{\frac{j\omega }{2}}-e^{-\frac{j\omega }{2}}\right ) } \end{align*}
That was fun. Now it is all clear, we have a couple of sin functions sitting on top of each others, we obtain
\[ X\left ( \omega \right ) =\frac{\sin \left ( \frac{\omega \left ( 2M+1\right ) }{2}\right ) }{\sin \left ( \frac{\omega }{2}\right ) }\]