Acabo de leer un fascinante prueba del valor de la integral $$ \int_{-\infty}^\infty e^{-ax^2} dx, $$ which proceeds by dimensional analysis, as follows: we know that we can write $$ \int_{-\infty}^\infty e^{-ax^2} dx = f(a) $$ for some $f$. Suppose $x$ represents some length, so that $x$ has dimension $[L]$. The argument of the exponential function must be dimensionless, so $a$ must have dimension $[L]^{-2}$. On the LHS, $e^{-ax^2}$ has dimension $[1]$, and $dx$ has dimension $[L]$, so $f(a)$ must also have dimension $[L]$. Hence, we can write $$ f(a) \sim \frac{1}{\sqrt a} $$ where $\sim$ represents proportionality with respect to a dimensionless constant. Now, we need only invoke the well-known result $$ \int_{-\infty}^\infty e^{-x^2} dx = \sqrt{\pi}, $$ which shows that $f(1) = \sqrt{\pi}$. Thus, we have $f(a) = \frac{\sqrt{\pi}}{\sqrt un}$, and $$ \int_{-\infty}^\infty e^{-ax^2} dx = \frac{\sqrt{\pi}}{\sqrt a}. $$ This approach of evaluating an integral by dimensional analysis is one that I have never seen before, and it is not obvious to me that I should accept its validity. Why should I expect an equation to remain dimensionally correct when I introduce an arbitrary dimensional constraint (in this case, $x$ having dimension $[L]$)? Bajo qué condiciones es un paso válido?
Respuesta
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Parece que con la punta de @anon, he descubierto la respuesta a mi propia pregunta.
La justificación es simple; es decir, lineal sustituciones de trabajo para las integrales. Aquí está la idea: $$\int_{-\infty}^\infty e^{-ax^2} dx = f(a)$$ and make the substitutions $x \a kx$ and $\a \frac{a}{k^2}$. This gives $$k \int_{-\infty}^\infty e^{-ax^2} dx = f\left( \frac{a}{k^2} \right).$$ With a bit of rearrangement, we have the functional equation $$f(a) = \frac{1}{k} f\left( \frac{a}{k^2} \right)$$ whose unique solution on $(0, \infty)$ is $$f(a) = \frac{f(1)}{\sqrt a}.$$ Esto completa la prueba sin hacer uso de cota, y es fácilmente visto como equivalente a las dimensiones método de análisis se demostró anteriormente.
