Answer to Question #94478 in Calculus for Olaitan Olajide

Question

Evaluate Gamma (-frac{3}{2})

Answer 1

$\Gamma(-\frac{3}{2})=\int\limits_0^\infty x^{-\frac{3}{2}-1}e^{-x}dx$

It is known that

$\Gamma(1-z)\Gamma(z)=\frac{\pi}{sin\pi z}$

Hence for z=1/2

$\Gamma^2(\frac{1}{2})=\pi$

$\Gamma(\frac{1}{2})=\sqrt{\pi}$

It is also known that $\Gamma(z+1)=z\Gamma(z)$

Therefore

$\Gamma(\frac{1}{2})=-\frac{1}{2}\Gamma(-\frac{1}{2})=\frac{3}{4}\Gamma(-\frac{3}{2})$

$\Gamma(-\frac{3}{2})=\frac{4}{3}\Gamma(\frac{1}{2})=\frac{4\sqrt{\pi}}{3}$