How is the function $\int^{\infty}_x e^{ipx'} dx'$related to the dirac delta function?

Question

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?

Answer 1

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.)

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

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.