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I would like to bound from above the expression

$$ \frac{\Gamma(\alpha,x)-\Gamma(\alpha,y)}{\Gamma(\beta,x)-\Gamma(\beta,y)} $$ for $x>y>0$. By plotting the above expression I have found that it should hold $$ \frac{\Gamma(\alpha,x)-\Gamma(\alpha,y)}{\Gamma(\beta,x)-\Gamma(\beta,y)}< y^{\alpha-\beta} $$ for $\alpha < \beta$ and $$ \frac{\Gamma(\alpha,x)-\Gamma(\alpha,y)}{\Gamma(\beta,x)-\Gamma(\beta,y)}< x^{\alpha-\beta} $$ for $\alpha > \beta$. To prove the first estimate my guess was to show that a function $$ f(\alpha)=y^{-\alpha}(\Gamma(\alpha,x)-\Gamma(\alpha,y)) $$ is increasing in $\alpha$. It seemed natural to me to compute $\partial_\alpha f(\alpha)$ but things started to get rough because $\partial_\alpha \Gamma(\alpha,x)$ has to be expressed as Meijer G-function. So the question is is there a more elementary proof of the presented estimates? Are there "better" estimates of which my estimates are special cases?

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  • $\begingroup$ By writing the first ratio as a ratio of integrals, what do you get if you maximize the numerator and minimize the denominator? That should yield a trivial upper bound? Is it easy to do this for $\frac{\int_y^x t^{\alpha-1}e^{-t}}{\int_y^x t^{\beta-1}e^{-t}}$? $\endgroup$
    – Suvrit
    Commented Sep 29, 2013 at 17:19
  • $\begingroup$ @suvrit How exactly do you plan to minimize the denominator? By taking $x\to y$ or vice versa? Anyway, this would result in denominator equal to zero. I tried maximizing/minimizing the term $e^{-t}$ which for numerator is $e^{-y}$ and for denominator is $e^{-x}$ but this gives an estimate $\dots<e^{x-y}\frac{\beta}{\alpha}\frac{x^\alpha-y^\alpha}{x^\beta-y^\beta}$ which is far from optimal due to $e^{x-y}$ term. $\endgroup$
    – pwl
    Commented Sep 30, 2013 at 12:45

2 Answers 2

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In fact it seems to be a consequence of the Cauchy theorem from calculus. Really, by it and a formula for derivative of incomplete gamma function (cf. Wiki for example) we evaluate

$\frac{f(x)-f(y)}{g(x)-g(y)}=\frac{f'(c)}{g'(c)}$ with some intermediate c, $0<y<c<x$ and then by the above mentioned formula $\Gamma^{'}_{x}(s,x)=-\frac{x^{s-1}}{e^x}$ it follows for your expression the exact formula $$ \frac{\Gamma(a,x)-\Gamma(a,y)}{\Gamma(b,x)-\Gamma(b,y)}=c^{a-b} $$ and everything is proved due to $0<y<c<x$.

And even more:

  1. In fact we proved estimates from both sides, upper and lower ones;

  2. They both are sharp for $x\to y$.

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As @suvrit suggests we have $$ \frac{\Gamma(\alpha,x)-\Gamma(\alpha,y)}{\Gamma(\beta,x)-\Gamma(\beta,y)}=\frac{\int_y^x t^{\alpha-1}e^{-t}dt}{\int_y^x t^{\beta-1}e^{-t}dt}=\frac{\int_y^x (t^{\alpha-\beta})t^{\beta-1}e^{-t}dt}{\int_y^x t^{\beta-1}e^{-t}dt}. $$ We can now maximize $t^{\alpha-\beta}$ depending on the sign of $\alpha-\beta$. For any $t\in[y,x]$ we have $$ t^{\alpha-\beta}\le\begin{cases}x^{\alpha-\beta}\quad &\text{ for } \alpha>\beta\\y^{\alpha-\beta}\quad &\text{ for } \alpha<\beta\\\end{cases} $$ which gives the requested estimate.

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