How to prove $\lim_{t\rightarrow \infty} \frac{e^{i x t}}{x- {i\mkern1mu} 0^+} =2 \pi {i\mkern1mu} \delta(x)$ Recently I meet following two equations in physics. Their proof is like a magic and I can't understand.
$$\lim_{t\rightarrow -\infty} \frac{e^{i x t}}{x- {i\mkern1mu} 0^+} =0$$
$$\lim_{t\rightarrow \infty} \frac{e^{i x t}}{x- {i\mkern1mu} 0^+} =2 \pi {i\mkern1mu} \delta(x)$$
I'm very confused about in what sense these equations are hold? Could you recommend some literatures about these part of math? According to my knowledge(e.g. mathematical analysis or calculus) there should be no limit. 
PS: I think the $0^+$ here means
$$\lim_{\epsilon\rightarrow 0^{+}}\lim_{t\rightarrow -\infty}\frac{e^{i x t}}{x- {i\mkern1mu} \epsilon} =0$$
$$\lim_{\epsilon\rightarrow 0^{+}}\lim_{t\rightarrow \infty} \frac{e^{i x t}}{x- {i\mkern1mu} \epsilon} =2 \pi {i\mkern1mu} \delta(x)$$
Actually I don't know which order of taking limit is correct: 
$\lim_{\epsilon\rightarrow 0^{+}}\lim_{t\rightarrow -\infty}$ or $\lim_{t\rightarrow -\infty}\lim_{\epsilon\rightarrow 0^{+}}$ or the order doesn't matter.
 A: Just analyze the behaviour of this expression.
For any interval $[-a,a]$ with $a>0$ and for any holomorphic function $f(x)$ we have
$$
\int_{-a}^{a} \lim_{t\rightarrow \pm\infty} \, \frac{e^{i x t}}{x- {i\mkern1mu} 0^+} \, f(x) \, {\rm d}x = \lim_{t\rightarrow \pm\infty} \int_{-at}^{at} \frac{e^{iz}}{z-i0^+t} \, f\left(\frac{z}{t}\right) \, {\rm d}z \, .
$$
We can add an integral over the semi-circle $\gamma_r$ in the upper half-plane of radius $r=at$ and estimate
$$
\left| \lim_{r \rightarrow \infty} \int_{\gamma_r}  \frac{e^{iz}}{z-i0^+t} \, f\left(\frac{z}{t}\right) \, {\rm d}z \right| \leq \left| f\left(ae^{i\phi_0}\right) \lim_{r\rightarrow \infty} \int_{\gamma_r}  \frac{e^{iz}}{z-i0^+t} \, {\rm d}z \right| = 0
$$
Thus for $t\rightarrow +\infty$ we have
$$
\lim_{t\rightarrow \infty} \int_{-at}^{at} \frac{e^{iz}}{z-i0^+t} \, f\left(\frac{z}{t}\right) \, {\rm d}z = 2\pi i \lim_{t\rightarrow \infty} e^{-0^+ t} \, f\left(i0^+\right) = 2\pi i \, f(0)
$$
by the residue theorem and we need to assume $$\lim_{t\rightarrow \infty} 0^+ t = 0 \, ,$$
so for instance you could choose $$0^+ = \lim_{t\rightarrow \infty} \frac{1}{t^2} \, .$$
In contrast for $t \rightarrow -\infty$ we still have to close the semi-circle in the upper half-plane, because only that arc vanishes. But for negative $t$ the pole is in the lower half-plane and hence the residue-theorem yields $0$.
A: This is a distributional limit, given by applying a sequence of functionals to a well-behaved test function $\phi$ to get a sequence of complex numbers and then taking the usual limit $L$ of that sequence. The distributional limit is the functional that maps $\phi$ to $L$. The usual limit gives
$$\lim_{\epsilon \downarrow 0} \int_{-\infty}^\infty
 \frac {\phi(x)} {x - i \epsilon} dx =
\operatorname{v.\!p.} \int_{-\infty}^\infty \frac {\phi(x)} x dx +
i \pi \phi(0),$$
which, dropping the test function and working with functionals, can be written as
$$\frac 1 {x - i0} =
\lim_{\epsilon \downarrow 0} \frac 1 {x - i \epsilon} =
\frac 1 x + i \pi \delta(x).$$
Similarly,
$$\lim_{t \to \pm \infty}
 \operatorname {v.\!p.} \int_{-\infty}^\infty
  \frac {e^{i t x}} x \phi(x) dx =
\lim_{t \to \pm \infty} 
 \operatorname{sgn} t \,\operatorname {v.\!p.} \int_{-\infty}^\infty
  \frac {e^{i x}} x \phi \!\left( \frac x t \right) dx =
\pm i \pi \phi(0),$$
or, in the distributional sense,
$$\lim_{t \to \pm \infty} \frac {e^{i t x}} x =
\pm i \pi \delta(x), \\
\lim_{t \to \pm \infty} \frac {e^{i t x}} {x - i0} =
(1 \pm 1 ) i \pi \delta(x).$$
A: METHODOLOGY $1$:
Let $\phi\in C^\infty_C$ be a test function and let $\epsilon>0$.  Then, we have
$$\begin{align}
\int_{-\infty}^\infty \frac{e^{itx}}{x-i\epsilon}\,\phi(x)\,dx&=\int_{-\infty}^\infty \frac{x}{x^2+\epsilon^2}\,e^{itx}\phi(x)\,dx+i\int_{-\infty}^\infty \frac{\epsilon}{x^2+\epsilon^2}\,e^{itx}\phi(x)\,dx\\\\
&=\int_{-\infty}^\infty \frac{e^{itx}}{x}\phi(x)\,dx-(\epsilon-i)\int_{-\infty}^\infty \frac{\epsilon}{x^2+\epsilon^2}e^{itx}\phi(x)\,dx\\\\
&=\text{sgn}(t)\int_{-\infty}^\infty \frac{e^{ix}}{x}\phi(x/t)\,dx-(\epsilon-i)\int_{-\infty}^\infty \frac{\epsilon}{x^2+\epsilon^2}\,e^{itx}\phi(x)\,dx\tag1
\end{align}$$

For the second integral on the right-hand side of $(1)$, the Dominated Convergence Theorem guarantees that 
$$\begin{align}
\lim_{\epsilon\to 0}\int_{-\infty}^\infty \frac{\epsilon}{x^2+\epsilon^2}\,e^{itx}\phi(x)\,dx&=\lim_{\epsilon\to 0}\int_{-\pi/2}^{\pi/2} e^{it\epsilon \tan(x)}\phi(\epsilon \tan(x))\,dx\\\\
&=\int_{-\pi/2}^{\pi/2}\lim_{\epsilon\to 0}\left( e^{it\epsilon \tan(x)}\phi(\epsilon \tan(x))\right)\,dx\\\\
&=\pi\phi(0)
\end{align}$$

For the first integral on the right-hand side of $(1)$, we have 
$$\begin{align}
\lim_{t\to\pm \infty}\int_{-\infty}^\infty \frac{e^{ix}}{x}\phi(x/t)\,dx&=i\pi \phi(0)\end{align}$$
where we used the result in THIS ANSWER.

Putting it together, we have 
$$\lim_{t\to \pm \infty}\lim_{\epsilon\to 0^+}\int_{-\infty}^\infty \frac{e^{itx}}{x-i\epsilon}\,\phi(x)\,dx=i \pi\phi(0)(1\pm 1)$$
whence we can write in distribution that 
$$\lim_{t\to\pm\infty}\lim_{\epsilon\to 0^+}\frac{e^{itx}}{x-i\epsilon}\sim i\pi(1\pm 1)\delta(x)$$

METHODOLOGY $2$:
Note that for any $\epsilon>0$ and $t>0$
$$\begin{align}
\int_{-\infty}^\infty \left(\frac{e^{itx}}{x-i\epsilon}\right)\,dx&=e^{-t\epsilon}\int_{-\infty-i\epsilon}^{\infty-i\epsilon}\frac{e^{itx}}{x}\,dx\\\\
&=i2\pi e^{-t\epsilon}\tag1
\end{align}$$

For any test function $\phi\in C_C^\infty$ and any given number $\nu>0$ we write
$$\begin{align}
\int_{-\infty}^\infty \left(\frac{e^{itx}}{x-i\epsilon}\right)\phi(x)\,dx&=i2\pi e^{-t\epsilon}+\int_{-\infty}^\infty \left(\frac{e^{itx}}{x-i\epsilon}\right)(\phi(x)-\phi(0))\,dx\\\\
&=i2\pi e^{-t\epsilon}+\int_{|x|\le \nu} \left(\frac{e^{itx}}{x-i\epsilon}\right)(\phi(x)-\phi(0)) \,dx\\\\
&+\int_{|x|\ge\nu} \left(\frac{e^{itx}}{x-i\epsilon}\right)(\phi(x)-\phi(0)) \,dx\tag2
\end{align}$$

For the first integral on the right-hand side of $(2)$, application of $(1)$ along with the mean value theorem yields
$$\int_{|x|\le \nu} \left(\frac{e^{itx}}{x-i\epsilon}\right)(\phi(x)-\phi(0)) \,dx = \int_{|x|\le \nu} \left(\frac{e^{itx}x}{x-i\epsilon}\right)\phi'(\xi(x))\,dx\tag3$$
for $0<\xi(x)<x$.

For the integral on the right-hand side of $(3)$, we have the estimate
$$\left|\int_{|x|\le \nu} \left(\frac{e^{itx}x}{x-i\epsilon}\right)\phi'(\xi(x))\,dx\right| \le 2||\phi'||_{\infty}\left(\sqrt{\nu^2+\epsilon^2}-\epsilon\right)\tag4$$
so that $\lim_{\epsilon\to 0}\left|\int_{|x|\le \nu} \left(\frac{e^{itx}x}{x-i\epsilon}\right)\phi'(\xi(x))\,dx\right| \le 2||\phi'||_{\infty}\nu$.

For the second integral on the right-hand side of $(2)$, integration by parts followed by application of the Riemann-Lebesgue Lemma yields 
$$\begin{align}
\lim_{t\to\infty}\lim_{\epsilon\to 0}\int_{|x|\ge\nu} \left(\frac{e^{itx}}{x-i\epsilon}\right)(\phi(x)-\phi(0)) \,dx&=\lim_{t\to\infty}\int_{|x|\ge\nu} \left(\frac{e^{itx}}{x}\right)(\phi(x)-\phi(0)) \,dx\\\\
&=0\tag5
\end{align}$$

Putting it all together, reveals that for any given $\nu>0$
$$\lim_{t\to\infty}\lim_{\epsilon\to0}\left|\int_{-\infty}^\infty \left(\frac{e^{itx}}{x-i\epsilon}\right)\phi(x)\,dx-i2\pi \phi(0)\right|\le 2||\phi'||_{\infty}\nu$$
whence we have established that in distribution 
$$\lim_{t\to\infty}\lim_{\epsilon\to 0^+}\frac{e^{itx}}{x-i\epsilon}\sim2\pi i \delta(x)$$
