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I would like to show that $$\frac{\Gamma(1-2x)\Gamma(1+x)}{\Gamma(1-x)}\geq 1$$ for real $x$ such that $|x|\leq \frac12$, where $\Gamma$ is the usual gamma function.

I looked at the derivatives and went a long road to prove this. I though somehow believe there should be a simpler way. I appreciate any hints or comments. Many thanks!

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  • $\begingroup$ You can show that $$\beta(1-2x,1+x)=\frac{\Gamma(2-x)}{\Gamma(1-2x)\Gamma(1+x)}=(1-x)\frac{\Gamma(1-x)}{\Gamma(1-2x)\Gamma(1+x)}\leqslant 1-x$$ which $\beta(a,b)$ is the beta function. $\endgroup$
    – Nosrati
    Jan 4, 2017 at 16:12
  • $\begingroup$ @MyGlasses Your relationships are incorrect. $B(x,y)=\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$. $\endgroup$
    – Mark Viola
    Jan 4, 2017 at 20:43
  • $\begingroup$ @Dr.MV Sure. I left a bad formula in mathematics history. Thanks alot. $\endgroup$
    – Nosrati
    Jan 4, 2017 at 21:11

2 Answers 2

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Note that the cases for $x=0$ and $x=-1/2$ are trivial.

Next, we can write for $0<x<1/2$

$$\begin{align} \frac{\Gamma(1-2x)\Gamma(1+x)}{\Gamma(1-x)}&=x\frac{\Gamma(1-2x)\Gamma(x)}{\Gamma(1-x)}\\\\ &=xB(1-2x,x)\\\\ &=x\int_0^1 t^{-2x}(1-t)^{x-1}\,dt\\\\ &\ge x\int_0^1(1-t)^{x-1}\\\\ &=1 \end{align}$$

as was to be shown!


For $-1/2<x<0$, we let $y=-x$ and write

$$\begin{align} \frac{\Gamma(1-2x)\Gamma(1+x)}{\Gamma(1-x)}&=\frac{\Gamma(1+2y)\Gamma(1-y)}{\Gamma(1+y)}\\\\ &=2yB(2y,1-y)\\\\ &=2y\int_0^1 t^{2y-1}(1-t)^{-y}\,dt\\\\ &\ge 2y\int_0^1 t^{2y-1}\,dt\\\\ &=1 \end{align}$$

as was to be shown!

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  • $\begingroup$ Many thanks +1 for this neat approach. I should have been able to see this, but did not :-/ $\endgroup$
    – Math-fun
    Jan 5, 2017 at 8:08
  • $\begingroup$ You're welcome! My pleasure. -Mark $\endgroup$
    – Mark Viola
    Jan 5, 2017 at 16:05
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Since $$ \Gamma(1+t)=e^{-\gamma t}\prod_{n\geq 1}\left(1+\frac{t}{n}\right)^{-1}e^{t/n} \tag{1}$$ we have: $$ h(x)=\frac{\Gamma(1-2x)\,\Gamma(1+x)}{\Gamma(1-x)}=\prod_{n\geq 1}\frac{\left(1-\frac{x}{n}\right)}{\left(1-\frac{2x}{n}\right)\left(1+\frac{x}{n}\right)}\tag{2}$$ and it is straightforward to check that $h(x)$ is a log-convex function on $I=\left(-\frac{1}{2},\frac{1}{2}\right)$.
Since $h'(0)=0$, $h(0)=1$ is a local minimum for $h(x)$ on $I$.


We may even compute the Taylor series of $\log h(x)$. Since: $$ \sum_{n\geq 2}\frac{\zeta(n)}{n} t^n = -\gamma t+\log\Gamma(1-t)\tag{3} $$ we have: $$ \log h(x) = \sum_{n\geq 2}\color{red}{\frac{\zeta(n)}{n}(2^n+(-1)^n-1)}\, x^n.\tag{4}$$

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  • $\begingroup$ very interesting approach! +1 many thanks Jack. I have learnt a lot from you on this website! $\endgroup$
    – Math-fun
    Jan 5, 2017 at 8:07

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