Can Sum Of Exponentially Increasing sinusoids be square integrable? Apologies if this is a trivial question.... im still learning
Suppose we have an exponentially increasing sinusoidal signal
$$f(x)=e^{\alpha x}\cos(\beta x)+e^{\alpha x}\sin(\beta x)$$
with $0<\alpha<1$ and $\beta\in\mathbb{R}$
Is it ever possible to dampen the signal with countably (perhaps infinitely) many other potentially exponentially increasing sinusoids such that the sum is square integrable?
$$\int_\mathbb{R} [f(x)+\sum[\omega_i e^{\alpha_i x}\cos(\beta_i x)+\omega_i e^{\alpha_i x}\sin(\beta_i x)]]^2<\infty$$
where, for the elements in the summation, $\omega_i$ is the amplitude of the i'th element, $0\leq\alpha_i<1$ and $\beta_i\neq \beta$.
my instinct says it is not possible, but im not sure.
edit: to make the question a bit more general, i'll add the possibility of both sin and cos.
 A: Define
$$g(x)=\sum_{i=1}^n\omega_i e^{(\alpha_i+2i\pi\beta_i)x}$$
where $\omega_i\in\mathbb R$, $0\leq \alpha_i<1$, and $\beta_i\in\mathbb R$. Then the finite case of your question is equivalent to asking whether $g$ is square integrable.
If $1,2\pi\beta_1, 2\pi\beta_2, \dots, 2\pi\beta_n$ are linearly independent over the rationals, we can use the multivariate form of Kronecker's theorem. See for example Hardy and Wright, theorem 442.
Kronecker's theorem: if $1,\beta_1, \beta_2, \dots, \beta_n$ are linearly independent over the rationals and $\delta_1, \dots, \delta_n$ are arbitrary, $N$ and $\epsilon$ are positive, then there are integers $n>N$ and $p_1, \dots, p_k$ such that
$$|n\beta_i-p_i-\delta_i|<\epsilon\,\forall i\in\{1,2,\dots n\}.$$
So, assuming the linear independence, set $\delta_i=0$ if $\omega_i\geq 0$ and $\delta_i=\pi$ otherwise. Now choose $\epsilon$ small and apply the theorem to obtain $n$ and $(p_i)_i$ with the properties as above. Then for $x=-n$ we have $\omega_i\cos(2\pi\beta_i x)>|\omega_i|\cos(\epsilon)$, so
$$\Re(g(x))>\cos(\epsilon)\sum_{i=1}^n \omega_i e^{\alpha_i x}$$
which is unbounded. With a bit more work, you can show the integral diverges.
If $1,2\pi\beta_1, 2\pi\beta_2, \dots, 2\pi\beta_n$ are not linearly independent, there is a $k<n$ and $\theta_1, \theta_2, \dots\theta_k$ so that
$$\beta_i=\frac{p_{i1}}{q_{i1}}\theta_1+\dots+\frac{p_{ik}}{q_{ik}}\theta_k$$
and the thetas are linearly independent. Letting
$$\lambda_j=\frac{\theta_j}{ q_{1j}\dots q_{nj}}$$
we have
$$\beta_i=r_{i1}\lambda_1+\dots+r_{ik}\lambda_k$$
where the $r_{ij}$ are integers. Then you can apply Kronecker's theorem again, to find an $x$ such that all the terms $\omega_i\cos(2\pi\beta_i x)$ are positive, and use the same argument again.
A: I think this question is equivalent to Laplace transform and stability of an LTI system. For complex $s$ $$\mathcal{L}\{f(t)\}(s) = \int_0^\infty f(t)e^{-st} \, dt$$
the question then reduces to stability of an LTI system if any pole is in the RHS of a zero-pole plot. my understanding is that if any pole is on the RHS (with positive exponential growth) then the entire system is unstable.
