Are the r.v.s with following characteristic functions infinitely divisible?
- $e^{it}e^{-|t|}e^{-t^2}$
- $\left(\frac{1}{2}(e^{it}+e^{-it})\right)^n$
The second one is easy because it is just a $\cos(t)^n$ but $\cos(\frac{\pi}{2}) = 0$ and characteristic functions of infinitely divisible distributions never vanish. So the answer is no.
What about the first one? This seems like convolution of independent $\delta_{1} * N(0, \sqrt2) * Cauchy$ so $$ \varphi_3(t)=e^{it}e^{-|t|}e^{-t^2} = \varphi_{X_1 + X_2 + X_3}(t) = \varphi_{X_1}\varphi_{X_2}\varphi_{X_3} $$
Both Cauchy and Gaussian are infinitely divisible. I am not sure if Dirac's delta is. If it isn't then can I conclude that $\varphi_3(t)$ is not infinitely divisible? Any tips appreciated.
But $\delta_k = \delta_1 + \dots + \delta_1$ (k times) so I think Dirac's delta is infinitely divisible.
Your arguments are correct except that what you have is $\delta_1$. To show that it is i.d. just note that it is $\delta_{1/n} *\delta_{1/n}*...*\delta_{1/n}$ ($n$ factors) for each $n$.