= Solution
The even <time-series spectral density> makes the sine part vanish. At lag zero its integral is the area of the triangle, so $\gamma_0=\pi^2$. For any nonzero integer $k$, integration by parts gives
$$
\begin{aligned}
\gamma_k&=2\int_0^\pi(\pi-\lambda)\cos(k\lambda)d\lambda\\
&=\frac2k\int_0^\pi\sin(k\lambda)d\lambda
=\frac{2(1-\cos k\pi)}{k^2}.
\end{aligned}
$$
Therefore the <triangular spectral density of a stationary time series> has <autocovariance function>
$$
\boxed{\gamma_k=\begin{cases}\pi^2,&k=0,\\4/k^2,&k\text{ odd},\\0,&k\ne0\text{ even}.\end{cases}}
$$
No additional factor $1/(2\pi)$ belongs in this inverse integral under the convention stated above.
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