Calcule
B\left(\frac{9}{2},-\frac{1}{2}\right)

Passo 1

O enunciado pede para calcular o valor da função Beta
B\Bigl(\tfrac{9}{2},-\tfrac{1}{2}\Bigr)
Usaremos a relação entre Beta e Gamma:
B(x,y)=\frac{\Gamma(x)\,\Gamma(y)}{\Gamma(x+y)}.

Passo 2

Aplicando a fórmula com x=\tfrac{9}{2} e y=-\tfrac{1}{2}, temos:
B\Bigl(\tfrac{9}{2},-\tfrac{1}{2}\Bigr) =\frac{\Gamma\bigl(\tfrac{9}{2}\bigr)\,\Gamma\bigl(-\tfrac{1}{2}\bigr)}{\Gamma\bigl(\tfrac{9}{2}-\tfrac{1}{2}\bigr)} =\frac{\Gamma\bigl(\tfrac{9}{2}\bigr)\,\Gamma\bigl(-\tfrac{1}{2}\bigr)}{\Gamma(4)}.

Passo 3

Calcular \Gamma\bigl(\tfrac{9}{2}\bigr) usando a fórmula para semi-inteiros:
\Gamma\Bigl(\tfrac{9}{2}\Bigr) =\frac{7}{2}\cdot\frac{5}{2}\cdot\frac{3}{2}\cdot\frac{1}{2}\,\sqrt{\pi} =\frac{105}{16}\,\sqrt{\pi}.

Passo 4

Calcular \Gamma\bigl(-\tfrac{1}{2}\bigr) a partir de \Gamma(\tfrac{1}{2})=\sqrt{\pi} e da relação \Gamma(z+1)=z\,\Gamma(z):
\Gamma\Bigl(-\tfrac{1}{2}\Bigr) =\frac{\Gamma(\tfrac{1}{2})}{-\tfrac{1}{2}} =-2\,\sqrt{\pi}.

Passo 5

Calcular \Gamma(4):
\Gamma(4)=3!=6.

Passo 6

Substituindo tudo em
B\Bigl(\tfrac{9}{2},-\tfrac{1}{2}\Bigr) =\frac{\frac{105}{16}\sqrt{\pi}\,\bigl(-2\sqrt{\pi}\bigr)}{6} =-\frac{210}{96}\,\pi =-\frac{35}{16}\,\pi.

Resposta

-\frac{35}{16}\,\pi

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