Kummer's hypergeometric negative integer

Question

It is stated in many texts that when a-b+c=1 that 2F1(a,b;c;-1)=$\frac{\Gamma(\frac{b}2+1)\Gamma(b-a+1)}{\Gamma(\frac{b}2+1-a)\Gamma(b+1)}$ and when b is a negative integer using $\Gamma(z)\Gamma(1\!-\!z)=\frac{\pi}{\sin{\pi z}}$ and taking limit as b approaches a negative integer obtain $2\cos(\frac{\pi b}{2})\frac{\Gamma(|b|)\Gamma(b-a+1)}{\Gamma(\frac{b}2-a+1)\Gamma(\frac{|b|}2)}$ It is stated in more than one place but nowhere as I have yet found is it sufficiently in detail proven and I have tried all I can think of such as $\Gamma(\frac 1 2)=\sqrt{\pi},\Gamma(n+\frac 1 2)=2^{-n}(2n-1)!!,\frac{\Gamma(1+b)}{\Gamma(b)}=2\frac{\Gamma(1+\frac b 2)}{\Gamma(\frac b 2)},\sin(x+\frac\pi 2)=\cos(x),\sqrt{1-\sin(x)^2}=\cos(x)$ etc. Does anyone have any idea how to explicitly and in detail proof ? Sorry scratch that as i have already now answered below.

Answer 1

This is too long for comment so put it here and have to rescind that prior statement "...nothing that remotely...$\Gamma(z)\Gamma(1\!-\!z)$ " because another item may try is using the general expression given that $\Gamma(z)\Gamma(1\!-\!z)\!=\!\frac \pi {\sin(\pi z)}\;$, substitute $\Gamma(\frac b2\!+\!1)\!=\!\frac{\pi}{\sin(\pi(\frac b2+1))\;\Gamma(-\frac b2)}$ and substitute $\Gamma(b\!+\!1)\!=\!\frac{\pi}{\sin(\pi(b+1))\;\Gamma(-b)}$ into the original rhs given in the original question which is $\frac{\Gamma(\frac b2+1)\Gamma(b-a+1)}{\Gamma(b+1)\Gamma(\frac b2-a+1)}$ and after doing that clearly get $\frac{\sin(\pi(b +1))}{\sin(\pi(\frac b2+1))}\frac{\Gamma(|b|)\Gamma(b-a+1)}{\Gamma(\frac{b}2-a+1)\Gamma(\frac{|b|}2)}$ where have replaced $-b,-\frac b2$ by $|b|,\frac{|b|}2\;$ when b is neg. in a couple of places,specifically in $\Gamma(-b),\Gamma(-\frac b2)$ so clearly obtain

$\frac{\sin(\pi(b +1))}{\sin(\pi(\frac b2+1))}\frac{\Gamma(|b|)\Gamma(b-a+1)}{\Gamma(\frac{b}2\!-a+1)\Gamma(\frac{|b|}2)}$. Now the problem is that if $\;b\;$were an even integer the denom. and num. could both go to zero because of the $\frac{\sin(\pi(b +1))}{\sin(\pi(\frac b2+1))}$ factor so need to re-express in a form which would take care of that limiting form issue if $\;b$ approached a negative even integer(if $\;b\;$ were odd integer the ans. is clearly$\;0$). So the remaining issue to obtain the texts answers is the most difficult problem of showing as $\;b\;$approaches negative integer that$\;\frac{\sin(\pi(b +1))}{\sin(\pi (\frac b2+1))}\;$can be replaced by $\;2\cos(\frac{\pi b}{2})\;$. It would seem that we may consider try using expressions for $\:\sin(a1\!+\!a2)\;$that is expressions for sin of a sum of 2 quantities in terms of expressions involving fct's of the 2 quantities separately and/or double angle formulas etc. which actually turns out to be fairly straightforward and not difficult because using sin of sum formula identity, in num. obtain $\;\sin(\pi(b\!+\!1))=\!-\!\sin(\pi b)=\!-\!2\cos(\pi\frac b2)\sin(\frac{\pi b}2))$ and in denom. get $\;\sin(\pi (\frac b2\!+\!1))=\!-\!\sin(\frac{\pi b}2))\;$where have also used$\;\cos(\frac {\pi}2)\!=\!0,\,\cos(\pi)\!=\!-\!1,\sin(\frac{\pi}2)\!=\!1,\,\sin(\pi)\!=\!0\;$. So this clearly shows $\;\frac{\sin(\pi(b +1))}{\sin(\pi (\frac b2+1))}\;$can be replaced by $\;2\cos(\frac{\pi b}{2})\;$ so that derives the answer to the original question.