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Is there a simple upper bound of the following fraction of gamma functions for any $a,b\geq1/2$: $$\left(\frac{\Gamma(a+b)}{a\Gamma(a)\Gamma(b)}\right)^{1/a}$$ An upper bound in the following form is enough: there exists a constant $C>0$ and a function $f(a,b)$ such that: $$\left(\frac{\Gamma(a+b)}{a\Gamma(a)\Gamma(b)}\right)^{1/a}\leq C\cdot f(a,b), \forall a,b\geq1/2$$

I have a guess of $f$, which is $\frac{a+b}{a}$. Is it true?

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$\newcommand{\Ga}{\Gamma}$

The inequality in question trivially holds if e.g. $C=1$ and $f(a,b)$ equals the left-hand side of the inequality.

More informatively, the bound with $f(a,b)=\frac{a+b}a$ indeed holds: \begin{equation} \frac{\Ga(a+b)}{a\Ga(a)\Ga(b)}\le C^a\Big(\frac{a+b}a\Big)^a \end{equation} for some real $C>0$ and all $a,b\ge1/2$. Indeed, by Stirling's formula for the gamma function, \begin{equation} \Ga(x)\asymp\frac1{\sqrt x}\,\Big(\frac xe\Big)^x \end{equation} for $x\ge1/2$, where $\asymp$ means "equals up to a universal positive constant factor". So, \begin{multline} \frac{\Ga(a+b)}{a\Ga(a)\Ga(b)}\asymp\frac1a\,\sqrt{\frac{ab}{a+b}}\frac{(a+b)^{a+b}}{a^a b^b} \ll \Big(\frac{a+b}a\Big)^a \Big(\frac{a+b}b\Big)^b \\ = \Big(\frac{a+b}a\Big)^a \Big(1+\frac ab\Big)^b < e^a\Big(\frac{a+b}a\Big)^a \end{multline} for $a,b\ge1/2$.

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  • $\begingroup$ Shouldn't $\Gamma(x)\asymp\sqrt{x}(x/e)^x$? $\endgroup$ – neverevernever Oct 18 '18 at 16:21
  • $\begingroup$ @neverevernever : No, it is as I wrote. You may be confusing it with the Stirling formula for $n!=\Gamma(n+1)$. I have added the reference to Stirling's formula; see the last displayed formula in Section "Stirling's formula for the gamma function" there. $\endgroup$ – Iosif Pinelis Oct 18 '18 at 21:10

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