Limit of distributions of principal value

Question

What is the limit in $D'(\mathbb{R})$ (i.e. in the distribution sense) of \begin{equation} \lim_{t \rightarrow +\infty}\frac{e^{ixt}}{x+i0} \end{equation} where $x+i0=p.v.(\frac{1}{x})-i\pi\delta(x)$ and $p.v.(\frac{1}{x})$ is the principal value function of $\frac{1}{x}.$ Thank you.

Answer 1

As clarified in the comment above, we need to calculate $$ \lim_{t\to\infty}\lim_{\epsilon\to 0} \int_{|x|>\epsilon}\frac{e^{itx}}{x}\varphi(x) dx = \lim_{t\to\infty} \lim_{\epsilon\to 0}\int_{|x|>\epsilon} \frac{\cos tx}{x}\varphi(x) dx + i \lim_{t\to \infty} \lim_{\epsilon\to 0}\int_{|x|>\epsilon} \frac{\sin tx}{x}\varphi(x) dx. $$ The sinus integral converges to $\pi \varphi(0)$ as the integrand is continous at $0$. Using the substitution $tx\mapsto y$ and $\lim_{\epsilon\to 0}\int_{|x|>\epsilon} \frac{f(x)}{x} dx = \int_0^\infty \frac{f(x) - f(-x)}{x} dx$ the cosinus integral becomes $$ \int_0^\infty \frac{\cos y}{y} (\varphi(\frac{y}{t}) - \varphi(-\frac{y}{t})) dy. $$ Splitting this integral into $\int_0^1 ...$ and $\int_1^\infty$, the latter converges to $0$ for $t\to\infty$ by the dominated convergence theorem. Applying the mean value theorem, the first integral becomes $$ \int_0^1\frac{\cos y}{y}\frac{2y}{t} \varphi'(\xi_t) dy $$ for some $\xi_t\in [-t,t]$. As $\varphi'$ is bounded, for $t\to\infty$ this integral converges to $0$ too.

To summarize everything: $$ \lim_{t\to\infty} \frac{e^{ixt}}{x+i0} = 0. $$