Fourier transform of Autocorrelation function, what am I missing here? (Power spectrum) Given this problem


I know that the autcorrelation and power spectrum are fourier pairs, so when taking taking fourier transform of Rxx, one should end up with Sxx 

However, when I take the transform of $R_{xx}(m)$, I end up with $S_{xx}(\omega)=\frac{2\alpha }{\alpha ^{2} + \omega ^{2}}$, which also matches up with fourier transform pairs in the book. Here's by hand: 

$S_{xx}(\omega )= \int_{-\infty}^{\infty} R_{xx}(m)e^{-i\omega m} dm=\widehat{R}_{xx}(\omega)$

$=\int_{-\infty}^{\infty} R_{xx}(m)e^{-i\omega m} dm=\int_{-\infty}^{\infty} e^{-\alpha |m|}e^{-i\omega m} dm=$
$=\int_{-\infty}^{0} e^{\alpha m}e^{-i\omega m} dm + \int_{0}^{\infty} e^{-\alpha m}e^{-i\omega m} dm=$

$ =\int_{-\infty}^{0} e^{m(\alpha -i\omega)} dm + \int_{0}^{\infty} e^{-m(\alpha +i\omega )} dm=$

$=\frac{1}{(\alpha -i\omega)} - \frac{0}{(\alpha -i\omega)}+\frac{0}{(\alpha +i\omega)}-\frac{1}{(\alpha +i\omega)}$

$=\frac{1}{(\alpha -i\omega)}+\frac{1}{(\alpha +i\omega)}=\frac{\alpha +i\omega+\alpha -i\omega}{(\alpha -i\omega)(\alpha +i\omega)}=\frac{2\alpha}{\alpha^{2} +\omega^{2}} $

Which doesn't look like equation given in the problem, unless I'm missing some identities or something here. 

Also, as for problem h), that should be pretty straight forward. Avg. power of $X=R_{xx}(0)$ right?
That question was posted quite some time ago! Have you tried the discrete time Fourier transform of $R_{XX}[m]$, instead of the Fourier transform? I believe this is what was meant, there (the text talks about a "sequence"). 

An "easy" way to solve this, according to me, is to compute the Z-transform, then identify $S_{XX}(\omega) = Z[R_{XX}](z=e^{j\omega})$. The Z-transform is given as:
\begin{align}
Z[R_{XX}](z) &amp;= \sum_{m=-\infty}^{\infty} R_{XX}[m] z^{-m} \\
   &amp;= \sum_m e^{-\alpha |m|} z^{-m} \\
   &amp;= \frac{1 - e^{-2\alpha}}{(1-e^{-\alpha}z^{-1})(1 - e^{-\alpha}z)} 
\end{align}

Hope that still helps!