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I am facing some troubles related to the Gamma function. As I do not want to add to much complex elements to my questions, imagine that in a part of my equation I have this: $\frac{\Gamma(-m-1/2)}{2}+\Gamma(m+1/2)\cdot 2$

I found online that it considers only one part or the other one according to the following condition: if $Rm>-1/2$ then the solution is $\Gamma(m+1/2)\cdot 2$ otherwise is $\frac{\Gamma(-m-1/2)}{2}$.

Could you kindly explain why? What happens if m isn't a complex number? I do not understand why the $\Gamma(x)$ when x is non-positive should be deleted and considered equal to zero. I cannot find any literature about that.

EDIT: I got it from here: Asymptotics of this HyperGeometric Function In point 2, you can obtain: $x^{2m}\left(x^1\frac{\Gamma(3/2)\Gamma(-m-1/2)}{\Gamma(-m)} + x^{-2m} \frac{\Gamma(1/2+m)}{2(\Gamma(3/2+m))}\right)$ I cannot find how they reach the result in point 2.

It seems that the solution provided in the link above is wrong according to the comments. May I ask you how can I work with the previous expression and how can I solve the limit when x tends to zero? (as in the link)

Silvia F.
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    This is not an equation. An equation has an equals sign. If you would mean that you would like to simplify this expression, then it does not simplify to the result you wrote. The source you got this from is wrong. – Snaw Dec 02 '21 at 08:47
  • (Gamma has no zeroes so such a cancellation would not be possible.) – Snaw Dec 02 '21 at 08:54
  • well if you are interested in negative values of gamma function: Then it doesn't exist for negative integers however, for fractions does. – Darshan P. Dec 02 '21 at 08:55
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    "I found online" means nothing. There are all kinds of people online. Please add a link. – Taladris Dec 02 '21 at 08:55
  • This screenshot is from "Hypergeometric Summation", Wolfram Koepf – Darshan P. Dec 02 '21 at 09:10
  • @I_don't_know_maths_ I found the same screenshot as you. May I ask you if you can help me related to the main problem? How can you solve the limit of the expression above? I also provided the link where I found the approach explained in my question. – Silvia F. Dec 02 '21 at 09:30
  • This seems to be an XY problem. Could you please write down explicitly what is the original problem you are considering? – Gary Dec 03 '21 at 06:38

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