I have the following correlation function in time domain defined as
$$ \left<\xi(t)\xi(s)\right> = \frac{D}{\tau}e^{-|t-s|/\tau} $$
I wish to take the fourier transform of the correlation function above. I first perform a change in variable such that $t-s=T$. My correlation function simplifies to
$$ \left<\xi(T+s)\xi(s)\right> = \frac{D}{\tau}e^{-|T|/\tau} $$
Next, I take a fourier transform in $s$ such that $$ S[\omega] = \int_{-\infty}^{\infty}ds e^{i\omega s}\left<\xi(T+s)\xi(s)\right> = \frac{D}{\tau}\int_{-\infty}^{\infty}ds e^{i\omega s} e^{-|T|/\tau} $$ Since $s=t-T$, $ds=-dT$. Changing the integration with respect to $T$ gives $$ S[\omega] = \frac{D}{\tau}\int_{-\infty}^{\infty}dT e^{i\omega(t-T)}e^{-|T|/\tau} $$
Now, I split the integral into two parts and remove the absolute value in the exponent
$$ S[\omega] = \frac{D}{\tau}\left(\int_{-\infty}^{0}dTe^{i\omega(t+T)}e^{T/\tau} + \int_{0}^{\infty}dTe^{i\omega(t-T)}e^{-T/\tau}\right) $$
By letting $T\rightarrow -T$ for the first term, the final integral simplifies as
$$ S[\omega] = \frac{2D}{\tau}\int_{0}^{\infty}dT e^{i\omega(t-T)}e^{-T/\tau} = \frac{2De^{i\omega t}}{1 + i\omega\tau} $$
This is incorrect since $S[\omega]$ is an observable spectrum so it should be real. In fact, the correct answer is given as
$$ S[\omega] = \frac{2D}{1 + \omega^{2}\tau^{2}} $$
What was my mistake here?
Edit: I realized that one possible issue stems from the fact that I'm treating $t$ as a constant which subsists throughout the integration, post substitution in terms of $T$. However, I don't see how I can eliminate it.