How to solve $\Gamma(-n)$? Which formula is good to get answer for the above question? I tried of various formulae to solve it but I couldn't make it out. What is the right way to approach the above question?
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$\Gamma(-n)$ has simple poles for integers $n \ge 0$. That is, $\Gamma(z) \to (-1)^n\infty$ as $z\to -n$, but $(z+n)\Gamma(z)$ is bounded. – Paul Sinclair May 31 '23 at 03:24
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Simply have a look at the graphical representation of $\Gamma$ – Jean Marie May 31 '23 at 22:43
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Euler's reflection formula for $\Gamma(z)$ is what you want. $$ \Gamma(-z) = \frac{-\pi}{\sin(\pi z)\Gamma(1+z)} $$ From this formula it is clear to see that for even integers $\Gamma(n)$ is $\infty$ or undefined and for odd integers it is $\frac{\pm\pi}{\Gamma(n+1)}$. Hope this helps.
Edit: As some have pointed out im an idiot. Eulers reflection formula is undefined for integers.
Aidan R.S.
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2The even/odd distinction you describe is incorrect, as $\sin\pi n=0$ for all $n\in\Bbb Z$. – J.G. May 30 '23 at 08:10
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1The Gamma function's analytic continuation doesn't (cannot) incloude the negative integers... – DonAntonio May 30 '23 at 08:28
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Since sinπn = 0, can we say Gamma ( -n ) value will be equal to ( - π ) / 0 ? In that case the answer will be ? – ASHIKA karikkalan May 30 '23 at 13:29