Let $a >0, b >0$, and $r \in \mathbb{R}$. I am trying to find a lower bound for the integral $$\int_a^\infty y^{-r} \exp\left( - b(y-a)^2\right) \,\mathrm dy.$$ After consulting the Wikipedia page for the incomplete gamma function and a couple of other websites I still have not been able to find a reference for a lower bound.

Do there exist a standard or well-known lower bounds for the incomplete gamma function?

  • $\begingroup$ Lower or upper? $\endgroup$
    – Pedro
    Apr 8, 2012 at 1:03
  • $\begingroup$ Upper notice $a >0$ is the type of integral I am interested in $\endgroup$
    – user7980
    Apr 8, 2012 at 1:08
  • $\begingroup$ Maybe this can help. $\endgroup$
    – Pedro
    Apr 8, 2012 at 1:17
  • $\begingroup$ Not sure if it'll help, but in this answer I proved that the quantity $$\Bigl(e^a/(e^a-1)\Bigr)^x \,\Bigl(\Gamma(x)-\Gamma(x,a)\Bigr)$$ is nondecreasing in $a$ when $a > 0$ for fixed $x \geq 1$ and strictly increasing in $a$ when $a > 0$ for fixed $x > 1$. Depending on what you need, you could vary $a$ to get different bounds. $\endgroup$ Jul 28, 2012 at 23:07

2 Answers 2


NB: This answer has two parts. The first part addresses the integral in the question and the second provides some simple lower bounds on $\Gamma(s,x)$.

Part I: Bounds on the integral in the question.

Here we provide some tight bounds on the integral $$ \newcommand{\rd}{\,\mathrm d} I(a) := \int_a^\infty y^{-r} e^{-b(y-a)^2}\rd y $$ and, in particular, give an asymptotic equivalence.

Claim: For $a > 0$, $b > 0$, and $r > 0$, $$ \sqrt{\frac{\pi}{4b}}\, \Big(a+\frac{1}{\sqrt{b\pi}} \Big)^{-r} \leq I(a) \leq \sqrt{\frac{\pi}{4b}} \,a^{-r} \>. $$ In particular, we have that $a^r I(a) \uparrow \sqrt{\pi/4b}$ as $a \to \infty$.

Proof. (Lower bound.) Use the substitution $z = \sqrt{2b}(y-a)$ to rewrite the integral in the "canonical form" $$ \sqrt{\frac{\pi}{4 b}} \int_0^\infty (a+z/\sqrt{2b})^{-r} \frac{2 e^{-\frac{1}{2}z^2}}{\sqrt{2\pi}} \rd z = \sqrt{\frac{\pi}{4 b}} \mathbb E (a+Z/\sqrt{2b})^{-r}\>, $$ where $Z$ is a standard half-normal random variable. The function $(a+z/c)^{-r}$ is convex in $z$ and so Jensen's inequality yields the stated lower bound by noting that $\mathbb E Z = \sqrt{2/\pi}$.

(Upper bound). Trivially $(a+z/\sqrt{2b})^{-r} \leq a^{-r}$ and so monotonicity of expectation yields the stated upper bound.

(Asymptotic result). That $I(a) \sim \sqrt{\pi/4b} \,a^{-r}$ as $a \to \infty$ follows directly from the upper and lower bounds. That $a^r I(a) \uparrow \sqrt{\pi/4b}$ follows from the fact that $(1+z/a)^{-r}$ is monotone increasing in $a$ and from monotonicity of expectation.

Part II: Lower bounds on $\Gamma(s,x)$.

Below, we assume real $s$ and $x > 0$. We use similar tricks to those in the previous part. Generally, the bounds we obtain below are asymptotically of the form $g(s) e^{-x} x^{s-1}$ as $x \to \infty$.

First of all, note that $$ \Gamma(s,x) = \int_x^\infty t^{s-1} e^{-t} \rd t = e^{-x} \int_0^\infty (t+x)^{s-1} e^{-t} \rd t = e^{-x} \mathbb E(T+x)^{s-1} \>, $$ where $T \sim \mathrm{Exp}(1)$.

Claim: For $s \leq 1$ and $s \geq 2$, we have $\Gamma(s,x) \geq e^{-x} (1+x)^{s-1}$.

Proof. $(t+x)^{s-1}$ is convex in $t$ for the corresponding $s$. Applying Jensen's inequality to the right-hand side of the above display equation and noting $\mathbb E T = 1$ yields the result.

Claim: For $s \geq 1$, we have $\Gamma(s,x) \geq e^{-x} \Gamma(s) (1+x/s)^{s-1}$. The inequality reverses for $0 < s < 1$.

Proof: The function $f(t) = (1+x/t)^{s-1}$ is convex for $s \geq 1$ and concave for $s < 1$. Apply Jensen's inequality to the right-hand side of $$ \Gamma(s,x) = \int_x^\infty t^{s-1} e^{-t} \rd t = e^{-x} \int_0^\infty (1+x/t)^{s-1} t^{s-1} e^{-t} \rd t = e^{-x} \Gamma(s) \mathbb E(1+x/G)^{s-1} \>, $$ where $G$ is a gamma random variable with shape parameter $s$ and scale parameter 1, and hence mean $\mathbb E G = s$. (For the reverse inequality when $s < 1$, the restriction to $s > 0$ is necessary to maintain the probabilistic interpretation and, hence, the application of Jensen's inequality.)

  • 1
    $\begingroup$ I know this post is very old but would like to ask you for a reference link of the results you've stated here, particularly part II. $\endgroup$
    – SL_MathGuy
    Aug 14, 2020 at 5:47
  • $\begingroup$ @SL_MathGuy No reference that I'm aware of. I just developed them from basic convexity considerations. $\endgroup$
    – cardinal
    Aug 22, 2020 at 5:59

I've been looking at bound and other formulas for $\Gamma(a,x)$ for a few months now. There are two works of reference for this:

  1. "The Incomplete Gamma Functions Since Tricomi" by Gautschi, 1998

  2. "Uniform Bounds for the Complementary Incomplete Gamma Function" by Borwein and Chan, 2009.

The first one is a pretty comprehensive survey of pre-1998 results on the incomplete gamma function, including upper and lower bounds.

The second one is a refinement of existing lower and upper bound to complex numbers. The problem is the bounds only work for a > 1, some of them even need a > 2 or a > 3, which has been useless for me :(

  • 1
    $\begingroup$ Gautschi's work is always a good source for gamma-related results. $\endgroup$
    – cardinal
    Jul 29, 2012 at 0:33
  • $\begingroup$ @cardinal : Good lower bounds, the part for 0 < s < 1 was very useful for me. Is there anything for -2 < s < 0 ? Also, is any of this published anywhere? $\endgroup$ Jul 29, 2012 at 7:40
  • 1
    $\begingroup$ The same bound works for the stated integral whenever $s < 1$ and $x > 0$ with no change in the proof argument. Note that, like in the first part of my answer, we can get a "trivial" (but, useful) upper bound for $s < 1$ via the inequality $(t+x)^{s-1} \leq x^{s-1}$. That is, $\Gamma(s,x) \leq e^{-x} x^{s-1}$. I don't know about published versions of these; but, I have to imagine they exist. $\endgroup$
    – cardinal
    Jul 29, 2012 at 11:37
  • $\begingroup$ Do you maybe think you could add these two (upper and lower) bounds on the incomplete gamma function to the Wikipedia page? I have been on the lookout for ages for such bounds, and I know I'm not the only one. $\endgroup$
    – HdM
    Nov 5, 2015 at 15:52

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