Question

Função Gama: Calcule Γ(3/2)

Answer 1

Função gamma é dada pela integral:
$\displaystyle \Gamma(n+1)=\int_{0}^{\infty}t^{n-1}\cdot e^{-t}dt$
Para calcular a integral (que observaremos ter uma relação de recorrência):
$\Gamma(n+1)=n\Gamma(n)\implies \Gamma(n+1)=n(n-1)\Gamma(n-1)$

Perceba que se $n \leq 0$ a função Gamma diverge.

Calculando Gama de 3/2 pela integral imprópria

Temos que levar em consideração que 
$\boxed{\Gamma(n)=\int\limits_{0}^{\infty}t^{n-1}\cdot e^{-t}dt}$
ou seja:
$\displaystyle \Gamma\left(\frac{3}{2}\right)=\int\limits_{0}^{\infty}t^{\frac{3}{2}-1}\cdot e^{-t}dt=\int\limits_{0}^{\infty}t^{\frac{1}{2}}\cdot e^{-t}dt$

usando integração por partes encontraremos:
$\displaystyle \Gamma\left(\frac{3}{2}\right)=\int\limits_{0}^{\infty}t^{\frac{1}{2}}\cdot e^{-t}dt\\\\i)~~~~\lim_{\psi\to\infty}\int\limits_{0}^{\psi}t^{\frac{1}{2}}\cdot e^{-t}dt\\\\ii)~~~partes:\\~~~v=-e^{-t}\implies dv=e^{-t}dt\\~~~u=\sqrt{t}\implies du=\frac{1}{2v}\\\\iii)~~\int udv=u\cdot v-\int vdu\\\\iv)~~\int\limits_{0}^{\infty}\sqrt{t}\cdot e^{-t}dt=-\frac{\sqrt{t}}{e^{t}}+\int\limits_{0}^{\infty}\frac{1}{2}t^{-\frac{1}{2}}\cdot e^{-t}dt$
$\displaystyle iv)~~\int\limits_{0}^{\infty}\sqrt{t}\cdot e^{-t}dt=-\frac{\sqrt{t}}{e^{t}}+\int\limits_{0}^{\infty}\frac{1}{2}t^{-\frac{1}{2}}\cdot e^{-t}dt\\\\ v)~~\lim_{\psi\to\infty}\left.-\sqrt{t}\cdot\frac{1}{e^t}\right]\limits_{0}^{\psi}+\boxed{\left(\frac{1}{2}\right)\int\limits_{0}^{\infty}t^{-\frac{1}{2}}\cdot e^{-t}dt}\\\\vi)~\lim_{\psi\to\infty}\left.-\sqrt{t}\cdot\frac{1}{e^t}\right]\limits_{0}^{\psi}=\left(-\sqrt{\psi}\cdot\frac{1}{e^{\psi}}\right)-\left(-\sqrt{0}\cdot\frac{1}{e^0}\right)=0+0$

agora resolvemos a segunda integral encontrada (a destacada em "v")

$\displaystyle \int\limits_{0}^{\infty}t^{-\frac{1}{2}}\cdot e^{-t}dt=\Gamma\left(\frac{1}{2}\right)$
$\displaystyle i)~~~~\int\limits_{0}^{\infty}t^{-\frac{1}{2}}\cdot e^{-t}dt\implies \int\limits_{0}^{\infty}\frac{e^{-t}}{\sqrt{t}}dt\\\\ii)~~u=\sqrt{t}\implies du=\frac{dt}{2u}\implies dt=2udu\\~~~~~~~~~~t=u^2\\\\iii)~~\int\limits_{0}^{\infty}t^{-\frac{1}{2}}\cdot e^{-t}dt=\int\limits_{0}^{\infty}\frac{2\exp(-u^2)u}{u}du=2\int\limits_{0}^{\infty}2\exp(-u^2)du$

levando em conta que:
$\boxed{erf(u)=\frac{2}{\sqrt{u}}\int\limits_{0}^{u}e^{-u^2}du}\implies \boxed{\int\limits_{0}^{u}e^{-u^2}du=\frac{\sqrt u}{2}erf(u)}$
onde erf é função erro de Gauss, teremos:
$\displaystyle 2\int\limits_{0}^{\infty}t^{-\frac{1}{2}}\cdot e^{-t}dt=\sqrt{\pi}$

então encontraremos
$\displaystyle \Gamma\left(\frac{3}{2}\right)=\int\limits_{0}^{\infty}\sqrt{t}\cdot e^{-t}dt=\left(\frac{1}{2}\right)\int\limits_{0}^{\infty}t^{-\frac{1}{2}}\cdot e^{-t}dt=\left(\frac{1}{2}\right)\cdot \sqrt{\pi}=\boxed{\frac{\sqrt{\pi}}{2}}$
que equivale a $\displaystyle \frac{3}{2}!$
olhe:
Se for na calculadora e calcular 3/2 fatorial encontraremos:

$\displaystyle \boxed{\frac{3}{2}!=0,88622692545275801364908374167057}$

a integral nos forneceu:

$\displaystyle \frac{\sqrt{\pi}}{2}=\frac{\sqrt{3,1415926535897932384626433832795}}{2}\\\\\frac{\sqrt{\pi}}{2}=\frac{1,7724538509055160272981674833411}{2}\\\\\boxed{\frac{\sqrt \pi}{2}=0,88622692545275801364908374167057=\frac{3}{2}!}$

encontramos o resultado como esperado :)
precisando de tirar dúvida é só comentar abaixo. Entre por um navegador de computador para visualizar as atividades.