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The expression I'd like to simplify is $\frac{\Gamma{[1+q\cdot x]}}{\Gamma{[1+x]^{q}}}$, If somebody could give me a hint I will appreciate it.

Thanks in advance

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  • $\begingroup$ Is $x$ real and $q$ integer ? $\endgroup$
    – C. Dubussy
    Mar 30, 2017 at 11:06
  • $\begingroup$ Not very much. The "simplifications" I can see are not better than the actual form. $\endgroup$
    – Masacroso
    Mar 30, 2017 at 11:06
  • $\begingroup$ In my case, x represent a natural number and q is an integer. Thank you $\endgroup$ Mar 30, 2017 at 14:58

2 Answers 2

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I don't know how helpful this will be, but you could use $\Gamma\left( 1+z \right)=z\Gamma\left( z \right),\,B\left( a,\,b\right) =\frac{\Gamma\left( a \right)\Gamma\left( b \right)}{\Gamma\left( a+b \right)}$ to write $$\frac{\Gamma\left( 1+qx \right)}{\Gamma^q\left( 1+x \right)}=\frac{qx^{1-q}}{\prod_{j=1}^{q-1}B\left( jx,\,x\right)}.$$

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  • $\begingroup$ Thanks, using the relation $\Gamma[1+z]=z\Gamma[z]$ and the property $\Gamma[z]=z^{z-1/2}e^{-z}\sqrt(2\pi)$ I have simplified the relation in a convenient form for my problem: $\frac{\Gamma[1+qz]}{\Gamma[1+z]^{q}}=z^{\frac{1}{2}(1-q)}q^{qz-\frac{1}{2}}(2\pi)^{\frac{1}{2}(1-q)}$ $\endgroup$ Mar 30, 2017 at 15:16
  • $\begingroup$ @N.Sobrino That's of course only approximate, but for large $z$ it'll give you a suiutable asymptotic form. I hadn't realised that was the context you sought. $\endgroup$
    – J.G.
    Mar 30, 2017 at 15:36
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Do you know a simplification for the special case $$\frac{(3n)!}{n!^3}\quad?$$ Actually, I think that is probably the simplest way to write it.

You could say: in $(a+b+c)^{3n}$, take the coefficient of the term $a^nb^nc^n$. You can try to do something similar in your general case.

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