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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1At that link it has additional assumptions which you left out: $a \ge 0$, $b \ge a + 1$, $\text{Re}(z) > 0$. – Robert Israel May 04 '12 at 04:03
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2The (i) link to the right of the inequality gives a reference to Paris and Kaminski (2001): http://dlmf.nist.gov/bib/P#bib1827 – Robert Israel May 04 '12 at 04:08
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1Also you got the left side wrong: it should be $\left| \dfrac{\Gamma(z+a)}{\Gamma(z+b)}\right|$ – Robert Israel May 04 '12 at 04:31
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Ah thanks for that! – yep May 04 '12 at 05:11
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The book is available at Google Books, but p. 34, to which the citation points, isn't shown (for me). – joriki May 04 '12 at 05:37
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Thanks to joriki's link to Google Books, here is the passage in question.
R. B. Paris and D. Kaminski, Asymptotics and Mellin-Barnes Integrals, pp. 33-34.
Antonio Vargas
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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.
