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Let $x>1$ and $0<\varphi<\frac{\pi}{2}$ be fixed. I would like to show that for any $s>0$, the following inequality holds: $$ \left| H_{\frac{is}{e^{i\varphi } \cos \varphi}}^{\left( 1 \right)} \left( {\frac{is}{e^{i\varphi } \cos \varphi }x} \right) \right| \le \left| H_{is}^{\left( 1 \right)} \left( isx \right) \right| = iH_{is}^{\left( 1 \right)} \left( isx \right), $$ where $H_\nu^{(1)}(z)$ is the Hankel function. It would be enough to show that for any fixed $v>0$, $x>1$, $$ [0,+\infty) \ni u \mapsto \left| {H_{u + iv}^{\left( 1 \right)} \left( {\left( {u + iv} \right)x} \right)} \right| $$ is a decreasing function. From Debye's asymptotic formula, we have $$ \mathop {\lim }\limits_{s \to + \infty } \left| \frac{H_{\frac{is}{e^{i\varphi } \cos \varphi }}^{\left( 1 \right)} \left( {\frac{is}{e^{i\varphi } \cos \varphi}x} \right)}{H_{is}^{\left( 1 \right)} \left( isx \right)} \right| = \sqrt {\cos \varphi } < 1. $$ So for large $s$ the inequality is valid. However, I could not find any suitable representation of the (absolute value of the) Hankel function to show the inequality for every $s>0$.

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Does this fail for $x < 1$? –  Suvrit Aug 6 '13 at 4:52
Numerical computations suggest that it does. –  Gary Aug 6 '13 at 10:37

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