Upper bound on ratio of incomplete Gamma function and Gamma function $\frac{ \Gamma \left( x; a\right)}{\Gamma(x)}$

Question

I am trying to find a tight upper bound the following expression \begin{align} \frac{ \Gamma \left( x; a\right)}{\Gamma(x)} \end{align} where $\Gamma \left( x; a\right)$ is incomplet Gamma function \begin{align} \Gamma \left( x; a\right)= \int_{a}^\infty t^{x-1} e^{-t} dt. \end{align}

Also, assume $x \ge 1/2$. Clearly a trivial upper bound is \begin{align} \frac{ \Gamma \left( x; a\right)}{\Gamma(x)} \le 1 \end{align}

However, I suspect the following bound should be true

\begin{align} \frac{ \Gamma \left( x; a\right)}{\Gamma(x)} \le O( e^{-a} ) \end{align}

Answer 1

$$\int_{a}^{+\infty}t^{x-1}e^{-t}\,dt = e^{-a}\int_{0}^{+\infty}(t+a)^{x-1}e^{-t}\,dt\leq e^{-a}a^{x-1}\int_{0}^{+\infty}\exp\left(\frac{t(x-1)}{a}-t\right)\,dt$$ hence a simple upper bound is: $$ \Gamma(x;a)\leq \frac{a^x e^{-a}}{a+1-x}.$$ That approximation can be refined, producing a continued fraction representation for the incomplete $\Gamma$ function: $$ \Gamma(x;a) = \frac{a^x e^{-a}}{a+\frac{1-x}{1+\frac{1}{a+\frac{2-x}{\ldots}}}}$$ Have also a look at this paper.

Answer 2

Your final inequality seems to be false. For instance, $\dfrac{\Gamma(2;a)}{\Gamma(2)} = (a+1)\mathrm{e}^{-a}$ and $\dfrac{\Gamma(3;a)}{\Gamma(3)} = \dfrac{1}{2}(a^2 + 2a + 2)\mathrm{e}^{-a}$. The bound $\dfrac{\Gamma(x;a)}{\Gamma(x)} \leq O(a^{x-1}\mathrm{e}^{-a})$ seems possible, but I haven't taken the time to find a proof.