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\textcolor{white}{\textbf{Sobre Funciones Generalizadas En Física}}\\
\textcolor{yellow}{Nimrod Rodríguez}\\
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 En los primeros ejercicios del libro que se usaba como texto en un curso de Física, aparecían estos problemas de cálculo:\\
Dada la siguiente definición:\\
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{
\[
\delta (x) =
\left\{
\begin{array}{rl}
0, & si\  x \neq 0\\
\infty, & si\  x = 0\\
\end{array}
\hspace{0.3cm} con: \hspace{0.3cm} \displaystyle\int_{-\infty}^{+\infty} \!\!\delta (x)\ \mathrm{d}x = 1
\right.
\]
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Evaluar las siguientes integrales:
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\begin{itemize}
\item $\displaystyle\int_{-3}^{+1} \!\!(x^3-3x^2+2x-1)\ \delta(x+2)\mathrm{d}x$
\item $\displaystyle\int_{0}^{\infty} \!\![\cos(3x)+2]\ \delta(x-\pi)\mathrm{d}x$
\item $\displaystyle\int_{-1}^{+1} \!\! e^{|x|+3} \ \delta(x-2)\mathrm{d}x$
\end{itemize}
};
Hasta aquí todo iba bien, pero cuando se llegó al ejercicio:\\
Demostrar que:\\
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node[fill=black,text width=9cm]
{
$$\delta (x)\ =\frac{1}{2\pi}\displaystyle\int_{-\infty}^{+\infty} \!\! e^{ikx} \ \mathrm{d}k$$
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Aquí el asunto era ya de un análisis más cuidadoso.
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La figura muestra la sugerencia en la que se aproxima la funcion delta de Dirac, mediante funciones rectangulares normalizadas ($A_n=2\epsilon_n\ ×\frac{1}{2\epsilon_n} =1$), y asi lograr
$\displaystyle\int_{-\infty}^{+\infty} \!\!\delta (x)\ \mathrm{d}x = 1$
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