## $f(x) = \frac{\int_{\alpha x}^{x} e^{-t} t^{b+1}\ dt}{x \int_{\alpha x}^{x} e^{-t} t^b\ dt}:\ ]\ 0,+\infty\ [\ \to \mathbb{R}$

where $\ 0<\alpha<1\ $ and $\ b>0$

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## $f(x) = \frac{\int_{\alpha x}^{x} e^{-t} t^{b+1}\ dt}{x \int_{\alpha x}^{x} e^{-t} t^b\ dt}:\ ]\ 0,+\infty\ [\ \to \mathbb{R}$

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where $\ 0<\alpha<1\ $ and $\ b>0$

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First perform a change of variables in the integration using $t = sx$. This gives

$$ f(x) = \frac{\int_a^1 e^{-xs} s^{b+1}~ \mathrm{d}s}{\int_a^1 e^{-xs} s^{b} ~\mathrm{d}s} = - \frac{d}{dx} \ln \int_a^1 e^{-xs} s^b ~\mathrm{d}s $$

Define $$ g(x) = \int_a^1 e^{-xs} s^b ~\mathrm{d}s $$ you have $$ f'(x) \leq 0 \iff \ln g(x) \text{ is convex} \iff \ln g(\frac{\alpha + \beta}{2}) \leq \frac12 (\ln g(\alpha) + \ln g(\beta)) $$ this in turn $$ \iff g(\frac{\alpha+\beta}{2}) \leq \sqrt{g(\alpha)g(\beta)} $$

This final line holds by Cauchy-Schwarz applied to $e^{-\alpha s / 2} e^{-\beta s/2}$ integrating against the measure $s^b ~\mathrm{d}s$.