# Simplification of Gamma functions

I am having some trouble simplifying some Gamma functions. I have a large expression in which some combinations of Gamma functions appear, that can be simplified, but applying FullSimplify won't give the desired result.

As an example, this combination appears:

(111 Gamma[5/4]^3)/(-96 Gamma[9/4]^3 + 40 Gamma[5/4]^2 Gamma[13/4])


after applying FullSimplify,

FullSimplify[(111 Gamma[5/4]^3)/(-96 Gamma[9/4]^3 + 40 Gamma[5/4]^2 Gamma[13/4])]

(* (111 Gamma[5/4]^3)/(-96 Gamma[9/4]^3 + 40 Gamma[5/4]^2 Gamma[13/4]) *)


Although, if I consider the inverse expression, the simplification goes well,

FullSimplify[(-96 Gamma[9/4]^3 + 40 Gamma[5/4]^2 Gamma[13/4])/(111 Gamma[5/4]^3)]

(* -(25/37) *)


I guess that this is due to the sum in the numerator, so the fraction can be expanded, then Mathematica recognises the to ratios of Gamma functions and simplify them separatedly.

I have tried with FunctionExpand, Expand, Apart... and didn't get any result. Any advice?

• Something like: FullSimplify[(111 Gamma[5/4]^3)/(-96 Gamma[9/4]^3 + 40 Gamma[5/4]^2 Gamma[13/4])] //. {Gamma[x_] /; x > 1 -> (x - 1) Gamma[x - 1]}. (Using a conditional rule.) And actually for this example FullSimplify[] is not needed to get the desired answer.
– JimB
Jul 12, 2016 at 16:27
• That totally solved it! Maybe you should wirte it as an answer to close the question. Jul 12, 2016 at 16:35
• Or 1/FullSimplify[1/expr] Jul 12, 2016 at 17:18
• @BobHanlon the problem is that this is not an isolated expression, so I cannot invert it as you suggest. Jul 12, 2016 at 23:45

gamRule = {Gamma[x_] /; x > 1 -> (x - 1) Gamma[x - 1]};

In this particular example, FullSimplify is not needed but the use of //. (ReplaceRepeated) rather than just /. (ReplaceAll) is needed.
• A generalization: (111 Gamma[5/4]^3)/(-96 Gamma[9/4]^3 + 40 Gamma[5/4]^2 Gamma[13/4]) /. Gamma[x_] /; x > 1 :> Product[x - k, {k, IntegerPart[x]}] Gamma[FractionalPart[x]] Jul 13, 2016 at 6:51