Let $s(t)$ be a Gaussian process with zero mean, unit variance and flat power spectrum given by
$ S_{ss}(f)=\frac{1}{2(f_2-f_1)} $ for $ \vert f \vert \in [f_1,f_2]$,
where $ f_1 $ and $f_2$ are two given frequencies.
I am trying to understand why $ \sigma^2(t) = \mathbb E \left[\int_0^t \int_0^t du dv s(u)s(v)\right]$ tends to be constant for large times.
Changing the order of integration I get $ \sigma^2(t) = \int_0^t \int_0^t du dv K_{ss}(u-v)dudv$, where $K_{ss}(u-v)$ denotes the correlation function.
Then, after a change of integration variables, and applying the Wiener Khintchine theorem we can express the correlation function as the Fourier transform of the power spectrum:
$ \sigma^2(t) = 2\int_0^t d\tau\int_{-\infty}^\infty df e^{-i2\pi f \tau} S_{ss}(f) (t-\tau)$.
Finally, integrating over $\tau$ and recalling that the power spectrum is symmetric, we obtain
$ \sigma^2(t) = \int_{-\infty}^\infty df S_{ss}(f) \frac{1-cos(2\pi ft)}{2\pi^2f^2}$.
Now, the text from which I am studying says that in the limit for $t\to\infty$ we should have
$ \lim_{t\to\infty} \sigma^2(t) = \frac{1}{2\pi^2 f_2f_1}$.
I sincerely don't get why, since I can easily calculate the first part of the integral, $S_{ss}(f)\frac{1}{2\pi^2 f^2}$ (and it would equal the limit above), but as far as I know the limit of the cosine function does not exist.
I guess I should consider some concepts from signal theory, which I miss. Thus I kindly ask you any advice. Thank you in advance for your help!