I've been trying to work this out but am having little luck. $n$ here is an integer. Ive found that $$\Gamma(n+1/2) = \frac{1}{2}\Gamma(n-1/2)$$ But I am unsure how to handle $\Gamma$ when it holds a negative value.
What is $\Gamma(1/2-n)\Gamma(1/2+n)$?
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$\begingroup$
complex-analysis
gamma-function
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1We always have $\Gamma(x+1) = x\Gamma(x)$, so $\Gamma(n+1/2) = (n-1/2)\Gamma(n-1/2)$. – 2017-02-24
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0@JohnPage The title says $\Gamma(1/2 - n)$ but the body says $\Gamma(n - 1/2)$. Which one do you mean? – 2017-02-24
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0So, you think that $\Gamma(11.5)=\frac12\Gamma(10.5)$? Hmmm... – 2017-02-24
1 Answers
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Euler's reflection formula write $$\Gamma(z)\,\Gamma(1-z)=\frac \pi {\sin(\pi z)}$$ Making $z=\frac 12+n$ then gives $$\Gamma\left(\frac 12+n\right)\,\Gamma\left(\frac 12-n\right)=\frac \pi {\sin\left(\pi (\frac 12+n)\right)}=\pi \sec (\pi n)$$