10.2: B- Evaluación de la Integral Gaussiana
- Page ID
- 128910
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)El integral
\[ \int_{-\infty}^{+\infty} e^{-x^{2}} d x\]
llamada la integral gaussiana, no recae en ninguno de los métodos de ataque que aprendiste en el cálculo elemental. Pero se puede evaluar de manera bastante simple usando el siguiente truco.
Definir el valor de la integral para que sea A. Entonces
\[ A^{2}=\int_{-\infty}^{+\infty} e^{-x^{2}} d x \int_{-\infty}^{+\infty} e^{-y^{2}} d y=\int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} d x d y e^{-\left(x^{2}+y^{2}\right)}.\]
En el último paso hemos escrito A 2 como una integral de dos variables sobre todo el plano. Esto parece perverso, porque la mayoría de las veces trabajamos arduamente para reducir las integrales bidimensionales a integrales unidimensionales, mientras que aquí vamos a la inversa. Pero mira de nuevo al integrand. Cuando se considera como una integral en el plano, es claro que podemos considerar x 2 + y 2 como solo r 2, y esto sugiere que debemos convertir la integral de coordenadas cartesianas (x, y) a polares (r, θ):
\[ A^{2}=\int_{0}^{\infty} d r \int_{0}^{2 \pi} r d \theta e^{-r^{2}}=2 \pi \int_{0}^{\infty} r e^{-r^{2}} d r\]
La última integral sugiere inmediatamente la sustitución u = r 2, dando
\[ A^{2}=\pi \int_{0}^{\infty} e^{-u} d u=-\pi\left.e^{-u}\right|_{0} ^{\infty}=\pi.\]
Concluimos que
\[ \int_{-\infty}^{+\infty} e^{-x^{2}} d x=\sqrt{\pi}.\]
B.1 (I) Problema: Otra integral
Demostrar que
\[ \int_{0}^{\infty} \frac{e^{-x}}{\sqrt{x}} d x=\sqrt{\pi}.\]
(Clue: Usa la sustitución\(y = \sqrt{x}\).