How is the function $\int^{\infty}_x e^{ipx'} dx'$related to the dirac delta function? The function in question is defined to be $\int^{\infty}_x e^{ipx'} dx'$ It must be a function of $x$ and $p$ and related to the delta function. But I cannot figure out how the $x$ influneces the function....Could anyone please help me?
 A: The integral does not converge in ordinary sense, but it can be understood as distribution. Its behavior is easier to track if we choose a suitable regularization. Indeed, for $\epsilon > 0$
\begin{align*}
\int_{x}^{\infty} e^{ipx'}e^{-\epsilon x'}\,dx'
&= \frac{1}{\epsilon - ip} e^{(ip-\epsilon)x} \\
&= \underbrace{ \frac{\epsilon}{\epsilon^2 + p^2} e^{(ip-\epsilon)x} }_{(1)} + \underbrace{ \frac{ip}{\epsilon^2 + p^2} e^{(ip-\epsilon)x} }_{(2)}.
\end{align*}
We can show that its distribution limit as $\epsilon \downarrow 0$ is
$$ ``\int_{x}^{\infty} e^{ipx'} \,dx' \,\text{''} = \pi \delta(p) + \frac{i}{p}e^{ipx}. $$
More precisely, we can prove that

Claim. For each test function $\varphi\in C_c^{\infty}(\mathbb{R})$ we have
$$ \lim_{\epsilon\downarrow 0}\int_{\mathbb{R}}\left(\int_{x}^{\infty} e^{ipx'}e^{-\epsilon x'}\,dx'\right) \varphi(p) \, dp
= \pi \varphi(0) + \mathrm{PV}\int_{\mathbb{R}}\frac{i}{p}e^{ipx}\varphi(p) \, dp. $$

Indeed, for $\text{(1)}$ we have
\begin{align*}
\int_{\mathbb{R}} \left( \frac{\epsilon}{\epsilon^2 + p^2} e^{(ip-\epsilon)x} \right) \varphi(p) \, dp
&\stackrel{p\mapsto \epsilon p}{=} \int_{\mathbb{R}} \frac{1}{1+p^2} e^{\epsilon(ip-1)x} \varphi(\epsilon p) \, dp \\
&\xrightarrow[\epsilon\downarrow0]{} \int_{\mathbb{R}} \frac{1}{1+p^2} \varphi(0) \, dp
= \pi \varphi(0)
\end{align*}
by the dominated convergence theorem with the dominating function $\frac{1}{1+p^2}e^{-x}\|\varphi\|_{\infty}$. For $\text{(2)}$, we have
\begin{align*}
\int_{\mathbb{R}} \left( \frac{ip}{\epsilon^2 + p^2} e^{(ip-\epsilon)x} \right) \varphi(p) \, dp
&= e^{-\epsilon x} \int_{\mathbb{R}} \frac{ip}{\epsilon^2 + p^2} \left( \frac{e^{ipx} \varphi(p) - e^{-ipx} \varphi(-p)}{2} \right) \, dp \\
&\xrightarrow[\epsilon\downarrow0]{} \int_{\mathbb{R}} \frac{i}{p} \left( \frac{e^{ipx} \varphi(p) - e^{-ipx} \varphi(-p)}{2} \right) \, dp \\
&= \mathrm{PV} \int_{\mathbb{R}} \frac{i}{p} e^{ipx} \varphi(p) \, dp,
\end{align*}
again the convergence follows from the dominated convergence theorem with the dominating function $\frac{1}{2p}\left| e^{ipx} \varphi(p) - e^{-ipx} \varphi(-p) \right|$, which is a compactly-supported continuous function thanks to removable singularity. (This hints why we take only odd part of $e^{ipx} \varphi(p)$ before taking limit.)
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
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Another way to introduce the suggested @Sangchul Lee regularization is to use a Fourier Representation of the Heaviside Step Function
  $\ds{\mrm{H}: \mathbb{R}\setminus\braces{0} \to \mathbb{R}}$. 
  
   Namely, $\ds{\mrm{H}\pars{x} = \int_{-\infty}^{\infty}{\expo{\ic kx} \over k - \ic 0^{+}}\,{\dd k \over 2\pi\ic}}$.

\begin{align}
\int_{x}^{\infty}\expo{\ic px'}\dd x' & =
\int_{-\infty}^{\infty}\mrm{H}\pars{x' - x}\expo{\ic px'}\dd x' =
\int_{-\infty}^{\infty}\bracks{\int_{-\infty}^{\infty}{\expo{\ic k\pars{x' - x}} \over k - \ic 0^{+}}\,{\dd k \over 2\pi\ic}}\expo{\ic px'}\dd x'
\\[5mm] &=
-\ic\int_{-\infty}^{\infty}{\expo{-\ic kx} \over k - \ic 0^{+}}
\int_{-\infty}^{\infty}\expo{\ic\pars{k + p}x'}{\dd x' \over 2\pi}\,\dd k =
-\ic\int_{-\infty}^{\infty}{\expo{-\ic kx} \over k - \ic 0^{+}}\,
\delta\pars{k + p}\,\dd k
\\[5mm] & =
-\ic\,{\expo{\ic px} \over -p - \ic 0^{+}} =
\ic\expo{\ic px}\bracks{\mrm{P.V.}{1 \over p} -\ic\pi\,\delta\pars{p}} =
\bbx{\ic\,\mrm{P.V.}{\expo{\ic px} \over p} + \pi\,\delta\pars{p}}
\end{align}
which means
$$
\bbx{\int_{-\infty}^{\infty}\varphi\pars{p}
\pars{\int_{x}^{\infty}\expo{\ic px'}\dd x'}\dd p =
\ic\,\mrm{P.V.}\int_{-\infty}^{\infty}{\varphi\pars{p} \over p}
\,\expo{\ic px}\dd p + \pi\varphi\pars{0}}
$$
as fully explained in the above cited link.
