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Could you please provide or point me to a proof of inequality 5.6.8 found at this site? That is,

$\left|\frac{\Gamma(z+a)}{\Gamma(z+b)}\right| \leq \frac{1}{|z|^{b-a}}$

for $z\in \mathbb{C}$, $a,b\in\mathbb{R}$, and $a≥0, b≥a+1, Re(z)>0$.

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At that link it has additional assumptions which you left out: $a \ge 0$, $b \ge a + 1$, $\text{Re}(z) > 0$. – Robert Israel May 4 '12 at 4:03
The (i) link to the right of the inequality gives a reference to Paris and Kaminski (2001): – Robert Israel May 4 '12 at 4:08
Also you got the left side wrong: it should be $\left| \dfrac{\Gamma(z+a)}{\Gamma(z+b)}\right|$ – Robert Israel May 4 '12 at 4:31
Ah thanks for that! – yep May 4 '12 at 5:11
The book is available at Google Books, but p. 34, to which the citation points, isn't shown (for me). – joriki May 4 '12 at 5:37
up vote 2 down vote accepted

Thanks to joriki's link to Google Books, here is the passage in question.

enter image description here

enter image description here

R. B. Paris and D. Kaminski, Asymptotics and Mellin-Barnes Integrals, pp. 33-34.

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For real $x>0$, the strict log-convexity of $\Gamma$ (see the end of this answer) implies that for $a\ge0$, $b\ge1$ and $b\ge a$, $$ \frac{\log\Gamma(x+1)-\log\Gamma(x)}{1}\le\frac{\log\Gamma(x+b)-\log\Gamma(x+a)}{b-a}\tag{1} $$ Which translates to $$ \frac{\Gamma(x+a)}{\Gamma(x+b)}\le\frac{1}{x^{b-a}}\tag{2} $$ Simpler, but not as general.

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Thanks - I think this is useful to have as well. – yep May 6 '12 at 3:04

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