Limit of distributions of principal value

What is the limit in $D'(\mathbb{R})$ (i.e. in the distribution sense) of $$\lim_{t \rightarrow +\infty}\frac{e^{ixt}}{x+i0}$$ 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.

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If $f_t = \frac{e^{ixt}}{x+i0}$, what do you mean with $f_t(\varphi)$ for a test function $\varphi$? –  Vobo Jan 30 '13 at 18:42
It is just $f_t(\phi)=\int \frac{e^{ixt}}{x+i0} \phi(x)dx$. –  Frank Zermelo Jan 30 '13 at 21:04
Do you mean $f_t(\varphi) = -i\pi \varphi(0) + \lim_{\epsilon\to 0} \int_{|x|>\epsilon} \frac{e^{ixt}}{x}\varphi(x) dx$? –  Vobo Jan 30 '13 at 21:27
Yes, this is what I mean. –  Frank Zermelo Jan 30 '13 at 23:12
Is there a reason, why you do not comment the answer? –  Vobo Feb 3 '13 at 19:42

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: $$\lim_{t\to\infty} \frac{e^{ixt}}{x+i0} = 0.$$
Well, I didn't comment your answer, because my busy schedule didn't allow me to. I apologize for the delay. In principle, it is the same proof I gave. Some few remarks however: $\frac{sin x}{x}$ is indeed continuous at 0, but it is not integrable, so the convergence of the first integral is not that obvious (at least, not for me). To justify the convergence, my suggestion would be to develop in the Taylor series, and to analyze the convergence of the first terms separately. Secondly, $\xi$ depends on $t$ and on $y$ as well, and actually $\xi \in [-y/t,y,t]$, but it doesn't matter a lot as it –  Frank Zermelo Feb 6 '13 at 10:12
$\phi'$ is bounded as you mentioned.I will post my solution too these days, not because I do not like your solution, but just to have more proofs for this nice fact. In the end, thank you for your proof and for contributing to this post. –  Frank Zermelo Feb 6 '13 at 10:13