# How to prove asymptotic limit of an incomplete Gamma function

How can I prove: $$\lim_{z\to\infty}{\Gamma(z, x)\over\Gamma(z)} = 1$$ ? Here $\Gamma(z, x)$ is the upper incomplete gamma function and $\Gamma(z)$ is the gamma function.

This must be something trivial, but I can't figure it out. Any ideas?

-
Nice question (+1) – OFFSHARING Sep 21 '12 at 10:23

The integrand $t^{z-1}\mathrm e^{-t}$ is unimodal and has its maximum at $t=z-1$. Thus for $z-1\gt x$ we can bound the integrand on both sides of $x$ by its value at $x$ to obtain

\begin{align} \Gamma(z)-\Gamma(z,x) &= \int_0^xt^{z-1}\mathrm e^{-t}\mathrm dt \\ &\lt\int_0^xx^{z-1}\mathrm e^{-x}\mathrm dt \\ &=xx^{z-1}\mathrm e^{-x} \\ &=\frac x{z-1-x}\int_x^{z-1}x^{z-1}\mathrm e^{-x}\mathrm dt \\ &\lt\frac x{z-1-x}\int_x^{z-1}t^{z-1}\mathrm e^{-t}\mathrm dt \\ &\lt\frac x{z-1-x}\int_0^\infty t^{z-1}\mathrm e^{-t}\mathrm dt \\ &=\frac x{z-1-x}\Gamma(z)\;, \end{align}

so for fixed $x$ the ratio of the difference to $\Gamma(z)$ tends to $0$ for $z\to\infty$.

-
So you proved that $\gamma(z, x) < {x\over z-1-x}\Gamma(z)$ for $z-1 > x > 0$ and from that the limit follows immediately. Nice! – Ondřej Čertík Sep 21 '12 at 18:10
I verified your steps, I think it's all correct. I am accepting your answer as it gives a simple proof. – Ondřej Čertík Sep 21 '12 at 21:15

In this answer I showed that

$$\left(\frac{e^x}{e^x-1}\right)^z \gamma(z,x) \leq \Gamma(z)$$

when $z \geq 1$ and $x > 0$, where

$$\gamma(z,x) = \Gamma(z) - \Gamma(z,x)$$

is the lower incomplete gamma function. Rearranging we get

$$0 \leq 1 - \frac{\Gamma(z,x)}{\Gamma(z)} = \frac{\gamma(z,x)}{\Gamma(z)} \leq \left(\frac{e^x-1}{e^x}\right)^z \longrightarrow 0$$

as $z \to \infty$ when $x$ is fixed.

In fact, this shows that the result is still true if $x = f(z) \geq 0$ and

$$z \log\left(1-e^{-f(z)}\right) \longrightarrow -\infty.$$

In particular, it's still true if

$$\frac{z}{e^{f(z)}} \longrightarrow \infty.$$

-
Thanks! As a matter of fact, I actually wanted to prove that $\gamma(z, x)\over\Gamma(z)$ goes to zero, but thought it'd be easier to do it with the upper incomplete gamma function. – Ondřej Čertík Sep 21 '12 at 18:13