 Solution

The signal is continuous at zero with $f(0)=1$. Its envelope is $e^{-\lambda |t|}$, but its angular frequency falls from $2$ on the incoming side to $1$ on the outgoing side. Its one-sided slopes are $f^{\prime }(0^{-})=\lambda$ and $f^{\prime }(0^{+})=-\lambda$, producing a cusp at the peak. Both tails decay to zero.

Use $\tilde{f}(\omega )={\int }_{\mathbb{R}}{e}^{-i\omega t}f(t)dt$. Expanding each cosine into two exponentials and integrating its half-line givesEquivalently,The positive $\lambda$ guarantees convergence of the Fourier transform integrals.