Non-asymptotic upper bound of right tail of Gamma function

Question

I'm wondering if there is any non-asymptotic upper bound for the following Gamma function: $$f_a(x)=\int_{x}^{\infty}t^a\exp(-t)dt$$ for $x>0,a>0$? Something like $x^a\exp(-x)$?

Answer 1

In Gabcke thesis, that you can download at http://hdl.handle.net/11858/00-1735-0000-0022-6013-8 page 84 you can find

Theorem: The incomplete gamma function $$\Gamma(a,x)=\int_x^\infty e^{-v}v^{a-1}\,dv,\qquad \text{$x>0$ and $a$ real numbers}$$ satisfies for $a\ge 1$ and $x>a$
$$\Gamma(a,x)\le a e^{-x} x^{a-1}.$$

Answer 2

In terms of the OP, the result by Gabcke cited in the answer by user juan is as follows: \begin{equation*} f_a(x)=\int_{x}^{\infty}t^a e^{-t}\,dt=\Gamma(a+1,x)\le(a+1) x^a e^{-x}=:B_1(a,x) \end{equation*} \begin{equation*} \text{for $a>0$ and $x\ge a+1$.} \end{equation*}

One can obtain the following improvement of this result: \begin{equation*} f_a(x)=\Gamma(a+1,x)\le x^a e^{-x}\,\frac1{1-a/x}=:B_2(a,x) \tag{1} \end{equation*} \begin{equation*} \text{for $a>0$ and $x>a$. } \tag{2} \end{equation*} Indeed, for $a>0$ and $x>a+1$ we have $B_2(a,x)<B_1(a,x)$.

Moreover, by l'Hospital's rule, for each real $a>0$ \begin{equation*} B_2(a,x)\sim f_a(x), \quad B_1(a,x)\sim (a+1)f_a(x) \end{equation*} as $x\to\infty$. So, the bound $B_2(a,x)$ is asymptotically optimal as $x\to\infty$, whereas the bound $B_2(a,x)$ is not.

---

Let us now prove the claim (1) under condition (2). For $t>x[>a>0]$, let $g(t):=a\ln t-t$. Since the function $g$ is concave, we have \begin{equation*} g(t)\le g_x(t):=g(x)+g'(x)(t-x)=a\ln x-x+(a/x-1)(t-x). \end{equation*} So, \begin{equation*} f_a(x)=\int_{x}^{\infty}e^{g(t)}\,dt \le \int_{x}^{\infty}e^{g_x(t)}\,dt =B_1(a,x), \end{equation*} as claimed.