Question

Prove que

$\int^{\infty}_{-\infty} {e^{\left(\dfrac{-x^2}{2\sigma^2}\right)} \, dx} = \sqrt[2]{2\pi}|\sigma}|$

Respostas com brincadeiras ou incompletas serão desconsideradas

Answer 1

Mostrar que

$\displaystyle\int_{-\infty}^{+\infty}\exp\bigg\{-\frac{x^{2}}{2\sigma^{2}}\bigg\}\,dx=\sqrt{2\pi}|\sigma|$

Resolução:

Considere $a=\displaystyle\int_{-\infty}^{+\infty}\exp\bigg\{-\frac{x^{2}}{2\sigma^{2}}\bigg\}\,dx$, então

$\bigg(\displaystyle\int_{-\infty}^{+\infty}\exp\bigg\{-\frac{x^{2}}{2\sigma^{2}}\bigg\}\,dx\bigg)\cdot\bigg(\int_{-\infty}^{+\infty}\exp\bigg\{-\frac{x^{2}}{2\sigma^{2}}\bigg\}\,dx\bigg)=a\cdot a$

Mudando a variável da segunda integral:

$\displaystyle\bigg(\int_{-\infty}^{+\infty}\exp\bigg\{-\frac{x^{2}}{2\sigma^{2}}\bigg\}\,dx\bigg)\cdot\bigg(\int_{-\infty}^{+\infty}\exp\bigg\{-\frac{y^{2}}{2\sigma^{2}}\bigg\}\,dy\bigg)=a^{2}$

Do cálculo, sabemos que $\iint_{A}f(x)g(x)\,dxdy=\big(\int_{A}f(x)dx\big)\big(\int_{A}g(y)dy\big)$ (para domínio A simples, que é o caso)

Então,

$\displaystyle\iint_{-\infty}^{+\infty}\exp\bigg\{-\frac{x^{2}}{2\sigma^{2}}\bigg\}\cdot\exp\bigg\{-\frac{y^{2}}{2\sigma^{2}}\bigg\}\,dxdy=a^{2}\\\\\\\iint_{\mathbb{R}^{2}}\exp\bigg\{-\frac{x^{2}}{2\sigma^{2}}-\frac{y^{2}}{2\sigma^{2}}\bigg\}\,dxdy=a^{2}\\\\\\\iint_{\mathbb{R}^{2}}\exp\bigg\{-\frac{(x^{2}+y^{2})}{2\sigma^{2}}\bigg\}\,dxdy=a^{2}$

Podemos considerar $\mathbb{R}^{2}$ como um círculo de raio infinito e utilizar a mudança de coordenadas polares. Isto é,

$x=r\cos\theta\\y=r\sin\theta$

Com $r\in(0,\infty),\,\theta\in[0,2\pi]$. Além disso,

$\bullet\,\,x^{2}+y^{2}=r^{2}\\\\\bullet\,\,|\mathbf{J}|=r\,\,\,(\mathsf{jacobiano\,\,\,da\,\,\,transf.})$

Logo,

$a^{2}=\displaystyle\int\limits_{0}^{2\pi}\int\limits_{0}^{+\infty}\exp\{-\frac{r^{2}}{2\sigma^{2}}\}\,|\mathbf{J}|\,drd\theta\\\\\\a^{2}=\int\limits_{0}^{2\pi}\int\limits_{0}^{+\infty}\exp\bigg\{-\frac{r^{2}}{2\sigma^{2}}\bigg\}\,r\,drd\theta\\\\\\a^{2}=\int\limits_{0}^{2\pi}\bigg[\int\limits_{0}^{+\infty}r\,\exp\bigg\{-\frac{r^{2}}{2\sigma^{2}}\bigg\}\,dr\bigg]d\theta$

Sabemos resolver a integral de dentro, já que

$\displaystyle\frac{d}{dr}\bigg[-\sigma^{2}exp\bigg\{-\frac{r^{2}}{2\sigma^{2}}\bigg\}\bigg]=-\sigma^{2}\exp\bigg\{-\frac{r^{2}}{2\sigma^{2}}\bigg\}\cdot\bigg(-\frac{r}{\sigma^{2}}\bigg)=\\\\\\=r\exp\bigg\{-\frac{r^{2}}{2\sigma^{2}}\bigg\}$

Então

$\displaystyle\int\limits_{0}^{+\infty}r\exp\bigg\{-\frac{r^{2}}{2\sigma^{2}}\bigg\}\,dr=\bigg[-\sigma^{2}\exp\bigg\{-\frac{r^{2}}{2\sigma^{2}}\bigg\}\bigg]_{0}^{+\infty}=\\\\\\=\sigma^{2}-\lim\limits_{r\to+\infty}\exp\bigg\{-\frac{r^{2}}{2\sigma^{2}}\bigg\}=\sigma^{2}-0=\sigma^{2}$

Substituindo na integral dupla:

$a^{2}=\displaystyle\int\limits_{0}^{2\pi}\sigma^{2}d\theta\\\\\\a^{2}=\sigma^{2}\int\limits_{0}^{2\pi}d\theta\\\\\\a^{2}=\sigma^{2}(2\pi-0)\\\\\\a^{2}=2\pi\sigma^{2}$

Tirando raiz quadrada dos dois lados da igualdade:

$\sqrt{a^{2}}=\sqrt{2\pi\sigma^{2}}\\\\\\a=\sqrt{2\pi}\sqrt{\sigma^{2}}\\\\\\\boxed{\boxed{a=\sqrt{2\pi}|\sigma|}}$