Confusion around $f(\tau) = \frac{\sin(\pi x)}{(\pi x)}$ as a covariance function for a weakly stationary process.

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I've just started an intro course on stationary stochastic processes. The main concept of the course is weakly stationarity, i.e. processes with a constant mean value function:

$$E[X(t)]=c$$

and where the covariance function $f(\tau)$ only depends on the time distance. Im at a chapter that goes over spectral representations of covariance functions.

There is this one theroem that states the following: A continuous function $f(t)$ with $ \int_{-\infty}^\infty \lvert f(t)\rvert \, dt < \infty$ is a covariance function if its Fourier transform is symmetric and non-negative.

Now is where the confusion comes. If i consider the covariance function $f(\tau) = \frac{\sin(\pi x)}{(\pi x)}$ which i know is the covariance function for a low frequency white noise process. How can this be the case?

The above theorem clearly states that $f(\tau)$ has to be a $L^1$-function in order for it to be the covariance function for a weakly stationary stochastic process, and since

$$\int_{-\infty}^\infty \lvert \frac{\sin(\pi x)}{(\pi x)}\rvert \, dt$$

does not converge then it shouldn't be possible for it to be a covariance function, yet it is the covariance function for low-frequency white noise, so i must be wrong with my thinking, but i have no idea where.