I've learned factorial. But today I saw a question which I don't know how to start with:
$$\left(-\frac{1}{2}\right)!$$
Can anyone explain how to solve it? Thanks
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If you take $a!$ to mean $\Gamma(a+1)$, then this is $\Gamma(1/2) = \sqrt \pi$. ${}\qquad{}$ – Michael Hardy Mar 29 '15 at 12:21
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You'll have to use the extension of the factorial which is the Gamma function : $$\left (-\frac{1}{2}\right )!:=\Gamma\left (\frac{1}{2}\right )=\sqrt{\pi}.$$ – Workaholic Mar 29 '15 at 12:22
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As suggested by Workaholic, $$I=\left(-\frac{1}{2}\right)!=\Gamma\left(\frac{1}{2}\right) = \int_{0}^{+\infty}x^{-1/2}e^{-x}\,dx = 2\int_{0}^{+\infty}e^{-y^2}\,dy = \int_{\mathbb{R}}e^{-y^2}\,dy $$ and by Fubini's theorem: $$ I^2 = \int_{\mathbb{R}^2}e^{-(y^2+z^2)}\,dy\,dz = \int_{0}^{2\pi}\int_{0}^{+\infty}\rho\, e^{-\rho^2}\,d\rho\,d\theta = \pi \int_{0}^{+\infty}2\rho\, e^{-\rho^2}\,d\rho = \color{red}{\pi} $$ so:
$$ \left(-\frac{1}{2}\right)! = \color{red}{\sqrt{\pi}}. $$
Jack D'Aurizio
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