I've been studying Fourier analysis of time series and in order to find an unbiased and consistent estimator for the spectrum, Barlett proposed a solution in 1953, the so called "smooth covariances estimator".
Problem is, for this estimator, there are some complex integrals with which I am not totally familiar yet. I am having trouble understanding a setp from the book. It says that:
$$ W(\lambda) = \int_{\infty}^{\infty} e^{-i\lambda \tau} \left(1 - \frac{|\tau|}{T}\right)d\tau $$
is the Fourier transform of this function $w(\tau) = 1 - |\tau|/T$, for $|\tau| \leq T $. Now the book states that:
$$ W(\lambda) = T\left[ \frac{sin(\pi T \lambda)}{\pi T \lambda}\right]^2 $$
and I cant figure out how I can get there. Can you guys give me any advice? Thanks!
EDIT
Trying to evaluate the integral, I have:
$$ \int_{-T}^{T}e^{-i\lambda \tau}(1-\frac{|\tau|}{T})d\tau = \int_{-T}^{T}e^{-i\lambda \tau}d\tau - \frac{1}{T} \int_{-T}^{T} |\tau|e^{-i\lambda \tau}d\tau $$
Now what I get is that:
$$ \int_{-T}^{T} e^{-i\lambda \tau}d\tau = \frac{e^{i\lambda T} - e^{-i\lambda T}}{i\lambda} = 2\frac{sin(\lambda T)}{\lambda T}$$ which seems quite reasonable.
Breaking piecewise $\frac{1}{T} \int_{-T}^{T} |\tau|e^{-i\lambda \tau}d\tau $:
$$\frac{1}{T} \int_{-T}^{T} |\tau|e^{-i\lambda \tau}d\tau = \frac{1}{T} \int_{0}^{T} \tau e^{-i\lambda \tau}d\tau - \frac{1}{T} \int_{-T}^{0} \tau e^{-i\lambda \tau}d\tau $$
Since:
$$\frac{1}{T} \int \tau e^{-i\lambda \tau}d\tau = \frac{e^{-i\lambda \tau}(1+i\lambda \tau)}{\lambda^2 T} $$
The resulting integral is:
$$ \frac{1}{T} \int_{-T}^{T} |\tau|e^{-i\lambda \tau}d\tau = \frac{2(\lambda T sin(\lambda T) + cos(\lambda T)-1)}{T \lambda^2 } $$
Computing both integrals together, still my result doesn't match what I expected. Where am I mistaken?