Inverse fouriertransform problem

[Aedrha2]

Hi there, I am studying a course where we currently are examining fouriertransforms. I got stuck on an exercise with a inverse fouriertransform.

I am supposed to find time discrete inverse fourier transform to [MATH]x(\omega)=cos^2(\omega)[/MATH]
I make use of : [MATH]x(n)=\frac{1}{2\pi}\int_{-\pi}^{\pi}cos^2(\omega)e^{j\omega n} d\omega[/MATH]
Eulers formula gives: [MATH]cos^2(\omega)= (\frac{1}{2}e^{j\omega}-\frac{1}{2}e^{-j\omega})^2=\frac{1}{4}(e^{2j\omega}+2+e^{-2j\omega}) x(n)=\frac{1}{2\pi}\int_{-\pi}^{\pi}\frac{1}{4}(e^{2j\omega}+2+e^{-2j\omega})e^{j\omega n} d\omega\\ x(n)=\frac{1}{8\pi}\int_{-\pi}^{\pi}e^{j\omega (n+2)}+2e^{j\omega n}+e^{j\omega (n-2)} d\omega\\ \\ x(n)=\frac{1}{8\pi}[\frac{e^{j\omega (n+2)}}{n+2}+\frac{e^{j\omega n}}{n}+\frac{e^{j\omega (n-2)}}{n-2}]_{-\pi}^\pi\\ \\ x(n)=\frac{1}{8\pi}(\frac{e^{(n+2)}}{n+2}+\frac{e^{n}}{n}+\frac{e^{(n-2)}}{n-2})[e^{j\omega}]_{-\pi}^\pi\\ \\ x(n)=\frac{1}{8\pi}(\frac{e^{(n+2)}}{n+2}+\frac{e^{n}}{n}+\frac{e^{(n-2)}}{n-2})(e^{j\pi}-e^{-j\pi})\\ \\ x(n)=\frac{1}{8\pi}(\frac{e^{(n+2)}}{n+2}+\frac{e^{n}}{n}+\frac{e^{(n-2)}}{n-2})\cdot0\\ \\ x(n)=0[/MATH]
This is not correct, and I don't quite know where i made a mistake.
Any help or insights is appreciated.
Please and thank you.

 

[Cubist]

This seems to be the dodgy step...

[MATH]x(n)=\frac{1}{8\pi}\left[\frac{e^{j\omega (n+2)}}{n+2}+\frac{e^{j\omega n}}{n}+\frac{e^{j\omega (n-2)}}{n-2}\right]_{-\pi}^\pi[/MATH]
[MATH]x(n)=\frac{1}{8\pi}\left(\frac{e^{(n+2)}}{n+2}+\frac{e^{n}}{n}+\frac{e^{(n-2)}}{n-2} \right) \left[e^{j\omega}\right]_{-\pi}^\pi [/MATH]
that last line is actually equivalent to

[MATH]x(n)=\frac{1}{8\pi}\left[\frac{e^{j\omega\color{red} + \color{black}(n+2)}}{n+2}+\frac{e^{j\omega\color{red} + \color{black}n}}{n}+\frac{e^{j\omega \color{red}+ \color{black}(n-2)}}{n-2}\right]_{\pi}^\pi[/MATH]
note the "+" that I highlighted in red! It took me a while to see this

--

There also seems to be a small mistake between these lines...

[MATH] x(n)=\frac{1}{8\pi}\int_{-\pi}^{\pi}e^{j\omega (n+2)}+2e^{j\omega n}+e^{j\omega (n-2)} d\omega [/MATH]
[MATH]x(n)=\frac{1}{8\pi} \left[\frac{e^{j\omega (n+2)}}{n+2}+\frac{e^{j\omega n}}{n}+\frac{e^{j\omega (n-2)}}{n-2} \right]_{-\pi}^\pi[/MATH]
...see here(click) for the first term's integral. You forgot the -ve and also need to times by "j".