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I have this expression:

Gamma[n - 0.691 + 2.35 I] Gamma[n - 0.691 - 2.35 I]

which returns real values for all $n$. However, neither Simplify nor FullSimplify can reduce it to a form without involving $i$. Is there a way to force these commands to do a better job?

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  • $\begingroup$ Welcome to Mathematica StackExchange! What exactly do you want as a final formula? If you are bothered by + 0. I in the result, then use Chop, for example: Gamma[n - 0.691 + 2.35 I] Gamma[n - 0.691 - 2.35 I] /. n -> 3 // Chop $\endgroup$
    – Domen
    Mar 16 at 16:40
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    $\begingroup$ "Is there a way to force these commands to do a better job?" -- This seems to assume there is a better result that they fail to produce. What is the result you expect? Or identity that they fail to apply? $\endgroup$
    – Michael E2
    Mar 17 at 11:37
  • $\begingroup$ If n being real is a safe assumption, perhaps use Re[Gamma[n - 0.691 + 2.35 I] Gamma[n - 0.691 - 2.35 I]]. $\endgroup$
    – Michael E2
    Mar 17 at 11:40
  • $\begingroup$ Maybe this little helps: $$\Gamma (n-i x) \Gamma (n+i x)=\frac{\Gamma (n)^2}{\prod _{k=1}^{\infty } \left(1+\frac{x^2}{(n-1+k)^2}\right)}$$ $\endgroup$
    – Mariusz Iwaniuk
    yesterday

2 Answers 2

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Using the fact that $\overline{\Gamma(x)}=\Gamma(\overline{x})$, you can rewrite your expression as

Abs[Gamma[n - 0.691 + 2.35 I]]^2

which is manifestly real (even though it still contains an I). Honestly, I would be very surprised if you found a useful expression without any I, so this might be the best you can do.

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  • $\begingroup$ Do you happen to know how to achieve this in Mathematica? FullSimplify[Gamma[z] Gamma[Conjugate[z]] == Abs[Gamma[z]]^2] does return True, yet FullSimplify[Gamma[z] Gamma[Conjugate[z]]] doesn't give the simpler form (which has smaller LeafCount). $\endgroup$
    – Domen
    Mar 17 at 12:32
  • $\begingroup$ @Domen unfortunately I don't... So far, I have had very limited success when trying to get Mathematica to simplify expressions containing Conjugate - it knows the relevant relations if you ask it directly (as you did), but it never seems to use that knowledge to simplify anything $\endgroup$
    – Lukas Lang
    2 days ago
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Check to see that the imaginary part is so small that you are "sure" it is zero. If so return just the real part:

\[Epsilon] = 10^-10;
g[n_] := Gamma[n - 0.691 + 2.35 I] Gamma[n - 0.691 - 2.35 I]
h[n_] := If[Im[g[n]] < \[Epsilon], Re[g[n]], g[n]]

g[55.5] returns 1.04586*10^142 + 0. I

h[55.5] returns 1.04586*10^142

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