I stumbled upon this when trying to calculate an infinite sum, but wolframalpha yielded no solutions except approximations. The question is to evaluate $$\lim_{x\to \infty} \frac{\Gamma(x+1,x)}{\Gamma(x+1)}$$ If it helps, x is supposed to be a positive integer. I thought about simply using L'Hoptals rule, but the solution is overly complicated and still $\frac{\infty}{\infty}$, and it so complicated, using L'Hopital again seems impossible (WolrframAlpha can't do it). Also, this relies on the assumption that $$\lim_{x\to \infty}\Gamma(x+1,x)=\infty$$ which seems to be true intuitively and when plugging in large numbers, it gets absurdly large, but I didn't manage to prove it. For Equation 1, I originally assumed to the limit would be $0$, since the lower bound of the integral definition of the incomplete gamma function is getting larger compared to the normal one. But plugging in values as high as $10000000000$ (the computer couldn't go higher), yields approximately $0.5000026596152$. I'm pretty sure at this point the answer is $0.5$, but I need an actual proof for my paper. If anyone does answer this, I'll cite you if that would be okay with you. Thanks in advance!

  • $\begingroup$ For the proof in the case the limit is taken along integer values, see this posting as well. $\endgroup$ Jul 10 '20 at 5:32

Currently I do not have any nice reference to this particular limit (although when the limit variable is restricted along the positive integers, it can be rephrased in terms of the Central Limit Theorem applied to the Poisson or Exponential variables.)

Instead, let me present a quick proof: First we apply the substitution $t=x+u\sqrt{x}$ to write

\begin{align*} \Gamma(x+1,x) &= \int_{x}^{\infty} t^x e^{-t} \, \mathrm{d}t \\ &= x^{x+\frac{1}{2}}e^{-x} \int_{0}^{\infty} \left( 1 + \frac{u}{\sqrt{x}}\right)^x e^{-\sqrt{x}u} \, \mathrm{d}u. \end{align*}

Then by noting that

$$ \lim_{x \to \infty} \left( 1 + \frac{u}{\sqrt{x}}\right)^x e^{-\sqrt{x}u} = e^{-\frac{u^2}{2}}, $$

we obtain1

$$ \lim_{x\to\infty} \int_{0}^{\infty} \left( 1 + \frac{u}{\sqrt{x}}\right)^x e^{-\sqrt{x}u} \, \mathrm{d}u = \int_{0}^{\infty} e^{-\frac{u^2}{2}} \, \mathrm{d}u = \sqrt{\frac{\pi}{2}}. $$

On the other hand, by the Stirling's formula2, we get

$$ \Gamma(x+1) \sim \sqrt{2\pi} \, x^{x+\frac{1}{2}}e^{-x} \quad \text{as} \quad x \to \infty. $$

Combining altogether, we obtain

$$ \lim_{x\to\infty} \frac{\Gamma(x+1,x)}{\Gamma(x+1)} = \frac{1}{2} $$

as expected.


  1. Interchanging the order of integration and limit is not always possible, but in this case, the inequality $\log(1+x) \leq x - \frac{x^2}{2(x+1)}$ for $x \geq 0$ shows that $$ \left( 1 + \frac{u}{\sqrt{x}}\right)^x e^{-\sqrt{x}u} \leq e^{-\frac{u^2}{2(u+1)}} $$ for all $x \geq 1$ and $u \geq 0$. Since this bound is integrable on $[0, \infty)$, the interchange is justified by the Dominated Convergence Theorem.

  2. In fact, the idea outlined in this post can be used to prove the Stirling's formula.


To ensure that we are thinking of the same incomplete gamma function, I recite a definition from the DLMF here (and Gamma, while we're at it): \begin{align*} \Gamma(a) &= \int_0^\infty t^{a-1} \mathrm{e}^{-t} \,\mathrm{d}t \text{ and } \\ \Gamma(a,z) &= \int_z^\infty t^{a-1} \mathrm{e}^{-t} \,\mathrm{d}t \text{.} \end{align*}

The following identity is helpful, for $n \in \Bbb{Z}_{\geq 0}$, $$ \Gamma(n+1,z) = n! \mathrm{e}^{-z} e_n(z) $$ where $$ e_n(z) = \sum_{k=0}^n \frac{z^k}{k!} \text{,} $$ is the truncated Taylor series for the exponential function.

Here's another: $$ \Gamma(n+1) = n! \text{.} $$

Using these two identities, $$ \frac{\Gamma(n+1,n)}{\Gamma(n+1)} = \frac{n! \mathrm{e}^{-n} e_n(n)}{n!} = \mathrm{e}^{-n} e_n(n) \text{.} $$ (If you follow the definition of $Q$ in section 8.2, you see this result in item 8.4.10.)

Then your result is item 8.11.13 with $x = 1$: $$ \lim_{n \rightarrow \infty} \frac{e_n(nx)}{\mathrm{e}^{nx}} = \begin{cases} 0 ,& x > 1 \\ \frac{1}{2} ,& x = 1 \\ 1 ,& 0 \leq x < 1 \end{cases} \text{.} $$ (That item references Abramowitz & Stegun item 6.5.34, which I have just verified is equivalent (under the substitution $x \mapsto \alpha$) in my copy (the ninth Dover printing).)

  • $\begingroup$ Interestingly enough, I stiumbled upon this the other way around. My problem included $\mathrm{e}^{-n} e_n(n) $ and I changed it to this gamma function problem ,since I didn't know how to evaluate the sum and the gamma function seemed more "clean" $\endgroup$ Jul 9 '20 at 22:34

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