Is it possible to prove that a ratio of an upper incomplete gamma functions is monotonically decreasing?

Specifically, let $\Gamma(s+1,u) = \int_u^\infty \tau^{s} \mathrm{e}^{-\tau}\mathrm{d}\tau$ be the upper incomplete gamma function. Then, I would like to show that $$\frac{\Gamma(s,u+z)}{\Gamma(s,u)}$$ is monotonically decreasing as $u$ grows, assuming $u$ and $z\in \mathbb{R}_+$ .

That is, for $0 < u_1 \leq u_2 \leq... \leq u_n < \infty$

$$\frac{\Gamma(s,u_1+z)}{\Gamma(s,u_1)}\geq \frac{\Gamma(s,u_2+z)}{\Gamma(s,u_2)}\geq ...\geq \frac{\Gamma(s,u_n+z)}{\Gamma(s,u_n)}$$

I tried the first derivative test, without any success.

I think it may be solved by using L'Hôpital's rule, but I'm not sure how.

Any help and ideas are much appreciated.


1 Answer 1


L'Hopital rule it is. It is enough to show that $$ \frac{\partial }{\partial u}\frac{\Gamma(s,u+z)}{\Gamma(s,u)} = \frac{\frac{\partial }{\partial u}\Gamma(s,u+z)\cdot\Gamma(s,u)-\frac{\partial }{\partial u}\Gamma(s,u)\cdot\Gamma(s,u+z)}{\Gamma(s,u)^2}\leq 0$$ or $$ (u+z)^s e^{-(u+z)}\int_{u}^{+\infty}t^s e^{-t}\,dt-u^s e^{-u}\int_{u+z}^{+\infty} t^s e^{-t}\,dt\geq 0 $$ or $$ \int_{0}^{+\infty}\left[(u+t)^s (u+z)^s-u^s(u+t+z)^s\right] e^{-t}\,dt\geq 0 $$ which is just a consequence of $$ (u+t)(u+z)\geq u(u+t+z)\quad\Longleftrightarrow\quad tz\geq 0.$$

  • 1
    $\begingroup$ Thanks @Jack D'Aurizio ! This is wonderful. $\endgroup$
    – sefi
    Commented Oct 5, 2017 at 9:04

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