# How to prove that the following function is monotonically decreasing?

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

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

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 $s^b ~\mathrm{d}s$.