La transformada de Fourier de $\operatorname{erfc}^2\left|x\right|$

Question

Podría usted por favor me ayude a encontrar la transformada de Fourier de $$f(x)=\operatorname{erfc}^2\left|x\right|,$$ donde $\operatorname{erfc}z$ denota la función complementaria de error.

Answer 1

Tenemos

$$ \int_{-\infty}^{\infty} \operatorname{erfc}^{2}(|x|) e^{-i\xi x} \, dx = \frac{4}{\xi}e^{-\xi^{2}/4} \left\{ \operatorname{erfi}\left( \frac{\xi}{2} \right) - \operatorname{erfi}\left( \frac{\xi}{2\sqrt{2}} \right) \right\}, \tag{1} $$

donde $\operatorname{erfi}$ es la imaginaria de la función de error definida por

$$\operatorname{erfi}(x) = \frac{2}{\sqrt{\pi}} \int_{0}^{x} e^{t^{2}} \, dt. $$

Así que aquí va mi solución. Observe que podemos escribir

$$ \operatorname{erfc}(|t|) = \frac{2}{\sqrt{\pi}} \int_{1}^{\infty} \left| t \right| e^{-t^{2}x^{2}} \, dx. $$

Con esto, podemos escribir

\begin{align*} \int_{-\infty}^{\infty} \operatorname{erfc}^{2}(|t|) e^{-i\xi t} \, dt &= \frac{4}{\pi} \int_{-\infty}^{\infty} \int_{1}^{\infty} \int_{1}^{\infty} t^{2} e^{-(x^{2}+y^{2})t^{2}} e^{-i\xi t} \, dxdydt \\ {\scriptsize(\because \text{ Fubini})} &= \frac{4}{\pi} \int_{1}^{\infty} \int_{1}^{\infty} \left( \int_{-\infty}^{\infty} t^{2} e^{-r^{2}t^{2}} e^{-i\xi t} \, dt \right) \, dxdy. \tag{2} \end{align*}

El uso de algún tipo de norma compleja técnica de análisis, se puede demostrar que

$$ \int_{-\infty}^{\infty} t^{2} e^{-r^{2}t^{2}} e^{-i\xi t} \, dt = \frac{\sqrt{\pi}}{4} \frac{2r^{2} - \xi^{2}}{r^{5}} e^{-\xi^{2} / 4r^{2}}. \tag{3} $$

De hecho, hemos

\begin{align*} \int_{-\infty}^{\infty} t^{2} e^{-r^{2}t^{2}} e^{-i\xi t} \, dt &= e^{-\xi^{2}/4r^{2}} \int_{-\infty}^{\infty} t^{2} \exp \left\{ - r^{2} \left( t + \frac{i\xi}{2r^{2}} \right)^{2} \right\} \, dt \\ {\scriptsize(\because \text{ contour shift})} &= e^{-\xi^{2}/4r^{2}} \int_{-\infty}^{\infty} \left( t - \frac{i\xi}{2r^{2}} \right)^{2} e^{-r^{2}t^{2}} \, dt \\ &= e^{-\xi^{2}/4r^{2}} \int_{0}^{\infty} \frac{4r^{4}t^{2} - \xi^{2}}{2r^{4}} e^{-r^{2}t^{2}} \, dt, \end{align*}

lo que de inmediato los rendimientos $\text{(3)}$ por la explotación de la función gamma. Conectar $\text{(3)}$ vuelta a nuestro cálculo de $\text{(2)}$, obtenemos

\begin{align*} \int_{-\infty}^{\infty} \operatorname{erfc}^{2}(|t|) e^{-i\xi t} \, dt &= \frac{1}{\sqrt{\pi}} \int_{1}^{\infty} \int_{1}^{\infty} \frac{2r^{2} - \xi^{2}}{r^{5}} e^{-\xi^{2} / 4r^{2}} \, dxdy \\ {\scriptsize(\because \text{ symmetry})} &= \frac{2}{\sqrt{\pi}} \iint_{1\leq y\leq x} \frac{2r^{2} - \xi^{2}}{r^{5}} e^{-\xi^{2} / 4r^{2}} \, dxdy \\ {\scriptsize(\because \text{ polar coordinate})} &= \frac{2}{\sqrt{\pi}} \int_{0}^{\frac{\pi}{4}} \int_{\csc\theta}^{\infty} \frac{2r^{2} - \xi^{2}}{r^{4}} e^{-\xi^{2} / 4r^{2}} \, drd\theta. \end{align*}

Mediante la sustitución de $u = \xi / 2r$, obtenemos

\begin{align*} \int_{-\infty}^{\infty} \operatorname{erfc}^{2}(|t|) e^{-i\xi t} \, dt &= \frac{8}{\xi\sqrt{\pi}} \int_{0}^{\frac{\pi}{4}} \int_{0}^{\frac{\xi}{2}\sin\theta} (1 - 2u^{2}) e^{-u^{2}} \, dud\theta \\ &= \frac{8}{\xi\sqrt{\pi}} \int_{0}^{\frac{\pi}{4}} \left[ u e^{-u^{2}} \right]_{0}^{\frac{\xi}{2}\sin\theta} \, d\theta \\ &= \frac{4}{\sqrt{\pi}} \int_{0}^{\frac{\pi}{4}} \sin \theta \exp\left( -\frac{\xi^{2}}{4} \sin^{2}\theta \right) \, d\theta. \end{align*}

Finalmente, mediante la sustitución de $v = \frac{1}{2}\xi \cos\theta$, obtenemos

\begin{align*} \int_{-\infty}^{\infty} \operatorname{erfc}^{2}(|t|) e^{-i\xi t} \, dt &= \frac{8}{\xi \sqrt{\pi}} e^{-\frac{1}{4}\xi^{2}} \int_{\xi/2\sqrt{2}}^{\xi/2} e^{v^{2}} \, dv = \frac{4}{\xi} e^{-\frac{1}{4}\xi^{2}} \left[ \operatorname{erfi}(v) \right]_{\xi/2\sqrt{2}}^{\xi/2}. \end{align*}

Esto demuestra $\text{(1)}$ como se desee.

Answer 2

$\newcommand{\+}{^{\daga}} \newcommand{\ángulos}[1]{\left\langle #1 \right\rangle} \newcommand{\llaves}[1]{\left\lbrace #1 \right\rbrace} \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\equalby}[1]{{#1 \cima {= \cima \vphantom{\enorme}}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\piso}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\mitad}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\a la derecha\vert\,} \newcommand{\cy}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left( #1 \right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\parcial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\raíz}[2][]{\,\sqrt[#1]{\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\wt}[1]{\widetilde{#1}} \newcommand{\verts}[1]{\left\vert\, nº 1 \,\right\vert}$ $$ \mbox{Vamos a}\quad \phi_{1}\pars{k} \equiv \int_{-\infty}^{\infty}{\rm erfc}\pars{\verts{x}}\expo{-\ic kx}\,\dd x $$ tal que \begin{align} \phi_{2}\pars{k}&= \int_{-\infty}^{\infty}{\rm erfc}^{2}\pars{\verts{x}}\expo{-\ic kx}\,\dd x =\int_{-\infty}^{\infty} \int_{-\infty}^{\infty}\phi_{1}\pars{p}\expo{\ic px}\,{\dd p \over 2\pi} \int_{-\infty}^{\infty}\phi_{1}^{*}\pars{q}\expo{-\ic qx}\,{\dd q \over 2\pi}\, \expo{-\ic kx}\,\dd x \\[3mm]&= {1 \over 2\pi}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\dd p\,\dd q\, \phi_{1}\pars{p}\phi_{1}^{*}\pars{q}\ \overbrace{\int_{-\infty}^{\infty}\expo{\ic\pars{p - q - k}x}\,{\dd x \over 2\pi}} ^{\ds{\delta\pars{p - q - k}}} \end{align}

$$ \phi_{2}\pars{k} = {1 \over 2\pi}\int_{-\infty}^{\infty} \phi_{1}\pars{p}\phi_{1}^{*}\pars{p k}\,\dd p $$

\begin{align} \phi_{1}\pars{k}&=\int_{-\infty}^{\infty}{2 \over \root{\pi}} \int_{\verts{x}}^{\infty}\dd t\,\expo{-t^{2}}\expo{-\ic kx}\,\dd x ={2 \over \root{\pi}} \int_{-\infty}^{\infty}\dd t\,\expo{-t^{2}} \int_{-\infty}^{\infty}\Theta\pars{t - \verts{x}}\expo{-\ic kx}\,\dd x \\[3mm]&={2 \over \root{\pi}} \int_{-\infty}^{\infty}\dd t\,\expo{-t^{2}}\Theta\pars{t} \int_{-t}^{t}\expo{-\ic kx}\,\dd x ={2 \over \root{\pi}} \int_{0}^{\infty}\dd t\,\expo{-t^{2}}{\sin\pars{kt} \over k} ={2 \over \root{\pi}}\,{{\rm F}\pars{k/2} \over k} \end{align} donde $\ds{{\rm F}\pars{z}}$ es el Dawson Integral.

$$\color{#00f}{\large% \phi_{2}\pars{k} = {2 \\pi^{2}} \int_{-\infty}^{\infty}{{\rm F}\pars{p/2}{\rm F}\pars{\bracks{p k}/2} \sobre k\pars{p k}}\,\dd p} $$