I need to prove $$\Gamma(\alpha+x)\Gamma(\beta+y)\leq C(\alpha; \beta)\Gamma(x+y),$$ for $x>\alpha$ and $y>\beta$ with $0<\alpha,\beta \leq \frac{1}{2}$ constants, and $C(\alpha, \beta)$ is a constant depending on $\alpha$ and $\beta$. I guess this holds from numerical computations and I need the especific case where $\alpha=\frac{1}{6}$ and $\beta=\frac{1}{2}$. It probably has something to do with the fact that $$\Gamma(x)\Gamma(y)\leq \Gamma(x+y-1),$$ for $x,y\geq1$, which I know is true for naturals, as $$\Gamma(x)\Gamma(y)=(x-1)!(y-1)!\leq(x+y-2)!=\Gamma(x+y-1),$$ which can be proven with induction, but I also can't prove for the general case where $x,y\in\mathbb{R}$. Any help, suggestion or proof to either of the inequalities is of great help. Thanks in advance!


This is not true for any positive $\alpha,\beta$ without some further hypothesis on $x$ and $y$. Fix positive $y<\alpha$, and let $x \rightarrow \infty$. Then the left-hand side is a constant times $\Gamma(x+\alpha)$ and the right-hand side is a constant times $\Gamma(x+y)$; so the desired inequality would assert that $\Gamma(x+\alpha)/\Gamma(x+y)$ is bounded. But in fact (using Stirling's asymptotic formula for $\Gamma(z)$, for example) this ratio $\Gamma(x+\alpha)/\Gamma(x+y)$ is asymptotic to $x^{\alpha-y} \rightarrow \infty$ as $x \rightarrow \infty$.

Instead the inequality $$ \Gamma(x) \, \Gamma(y) \leq \Gamma(x+y-1) $$ can be proved for all real $x,y \geq 1$ as follows. Equality holds when either $x=1$ or $y=1$; I claim that otherwise $\Gamma(x) \, \Gamma(y) < \Gamma(x+y-1)$. Fix $y>1$. It is known that the Gamma function is logarithmically convex upwards. Therefore $\Gamma(x+y-1)/\Gamma(x)$ is an increasing function of $x$. Because it equals $\Gamma(y)$ for $x=1$, it exceeds $\Gamma(y)$ for all $x>1$, and we are done.

  • $\begingroup$ You're right, my bad. It's because in my case I'm interested in values such that $x>\alpha$ and $y>\beta$. In this case your argument doesn't hold, or does it? I'll edit the question. Do you have any idea on how to show the second inequality, that seems to be easier to me? Thanks anyway! $\endgroup$ – Mateus Sampaio Oct 23 '13 at 23:41
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    $\begingroup$ Yes, the second inequality is true; I just added a paragraph giving its proof. $\endgroup$ – Noam D. Elkies Oct 24 '13 at 0:04

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