Questão 2 - Calcule \Gamma\left(\frac{3}{2}\right), \Gamma\left(\frac{5}{2}\right), \Gamma\left(\frac{7}{2}\right), \Gamma\left(\frac{-1}{2}\right), \Gamma\left(\frac{-3}{2}\right), \Gamma\left(\frac{-5}{2}\right).

Passo 1

O enunciado pede para calcular valores da função Gama em argumentos seminteiros, a saber \displaystyle \Gamma\bigl(\tfrac32\bigr), \Gamma\bigl(\tfrac52\bigr), \Gamma\bigl(\tfrac72\bigr), \Gamma\bigl(-\tfrac12\bigr), \Gamma\bigl(-\tfrac32\bigr) e \Gamma\bigl(-\tfrac52\bigr).
O conteúdo necessário inclui a relação de recorrência
\Gamma(x+1)=x\,\Gamma(x)
e o valor conhecido
\Gamma\bigl(\tfrac12\bigr)=\sqrt{\pi}.

Passo 2

Para começar, usamos \Gamma\bigl(\tfrac12\bigr)=\sqrt{\pi} como base e aplicamos a relação de recorrência subindo em incrementos de 1 até \tfrac32, \tfrac52 e \tfrac72.

Passo 3

Cálculo para os argumentos positivos seminteiros:
\Gamma\bigl(\tfrac32\bigr) =\tfrac12\,\Gamma\bigl(\tfrac12\bigr) =\tfrac12\,\sqrt{\pi} =\frac{\sqrt{\pi}}{2},
\Gamma\bigl(\tfrac52\bigr) =\tfrac32\,\Gamma\bigl(\tfrac32\bigr) =\tfrac32\cdot\frac{\sqrt{\pi}}{2} =\frac{3\sqrt{\pi}}{4},
\Gamma\bigl(\tfrac72\bigr) =\tfrac52\,\Gamma\bigl(\tfrac52\bigr) =\tfrac52\cdot\frac{3\sqrt{\pi}}{4} =\frac{15\sqrt{\pi}}{8}.

Passo 4

Para os argumentos negativos seminteiros, usamos a mesma recorrência na forma
\Gamma(x)=\frac{\Gamma(x+1)}{x},
indo de cima para baixo:
\Gamma\bigl(-\tfrac12\bigr) =\frac{\Gamma\bigl(\tfrac12\bigr)}{-\tfrac12} =\frac{\sqrt{\pi}}{-\tfrac12} =-2\sqrt{\pi},
\Gamma\bigl(-\tfrac32\bigr) =\frac{\Gamma\bigl(-\tfrac12\bigr)}{-\tfrac32} =\frac{-2\sqrt{\pi}}{-\tfrac32} =\frac{-2\sqrt{\pi}\times(-2)}{3} =\frac{4\sqrt{\pi}}{3},
\Gamma\bigl(-\tfrac52\bigr) =\frac{\Gamma\bigl(-\tfrac32\bigr)}{-\tfrac52} =\frac{\tfrac{4\sqrt{\pi}}{3}}{-\tfrac52} =\frac{4\sqrt{\pi}}{3}\cdot\frac{-2}{5} =-\frac{8\sqrt{\pi}}{15}.

Resposta

\displaystyle \Gamma\bigl(\tfrac32\bigr)=\frac{\sqrt{\pi}}{2}, \Gamma\bigl(\tfrac52\bigr)=\frac{3\sqrt{\pi}}{4}, \Gamma\bigl(\tfrac72\bigr)=\frac{15\sqrt{\pi}}{8}, \Gamma\bigl(-\tfrac12\bigr)=-2\sqrt{\pi}, \Gamma\bigl(-\tfrac32\bigr)=\frac{4\sqrt{\pi}}{3}, \Gamma\bigl(-\tfrac52\bigr)=-\frac{8\sqrt{\pi}}{15}.

Exercícios relacionados