If $e^{itx_n}$ converges for every $t\in\mathbb R$, then does $x_n$ converge? 
Question : Is the following true for any real number sequence $x_n$ ?
"If $e^{itx_n}$ converges for every $t\in\mathbb R$, then $x_n$ converges."
Note that $i^2=-1$.

Motivation : We know that the converse of the above proposition is true. However, I'm facing difficulty for the above proposition. Can anyone help?
 A: Let $G$ be a locally compact topological group. The dual $\hat G$ of $G$ is the group of continuous homomorphisms $G \to S^1$.
Given a real number $s$, we may define a continuous homomorphism $\varphi_s :\mathbf R \to S^1$ by $t \mapsto e^{ist}$. The map $s \mapsto \varphi_s$ is a map $\mathbf R \to \hat{\mathbf R}$. This map is an isomorphism of locally compact abelian groups.
Therefore, the sequence $\{x_n\}$ converges in $\mathbf R$ if and only if the sequence $\{\varphi_{x_n}\}$ converges in $\hat{\mathbf R}$ if and only if the sequence $e^{itx_n}$ converges for every $t \in \mathbf R$.
A: Yes, $x_n$ converges.
Denote 
$$f(t):=\lim_{n\to\infty}e^{itx_n},\quad t\in\Bbb R.$$
Clearly $f$ is Lebesgue measurable on $\Bbb R$, so by dominated convergence theorem,
$$\lim_{n\to\infty}\int_{t_1}^{t_2}e^{itx_n}dt=\int_{t_1}^{t_2}f(t)dt,\quad \forall~ t_1<t_2.$$
Note that $|f(t)|=1$, $\forall t\in\Bbb R$, so there exist $a<b$, such that
$$\int_a^bf(t)dt\ne 0\Longrightarrow \int_a^be^{itx_n}dt\ne 0 \quad\text{for large }n.$$
Then by fundamental theorem of calculus,
$$x_n=\frac{i(e^{iax_n}-e^{ibx_n})}{\int_a^be^{itx_n}dt}  \quad\text{for large }n.$$
Letting $n\to\infty$, we have
$$\lim_{n\to\infty}x_n=\frac{i(f(a)-f(b))}{\int_a^bf(t)dt}.$$
