# About a limit with Euler $\Gamma$ function.

For all $x \in \Bbb R_+^*$, we put: $$f(x)=\frac{1}{\Gamma(x)}\int_x^{+\infty}t^{x-1}e^{-t}dt.$$ Can we compute the limit : $\displaystyle\lim_{x \to +\infty} f(x)$?

• Consider the limit of a product of two functions of $x$. – Mhenni Benghorbal Dec 29 '14 at 3:55
• This should be $1/2$, and the intuition is that $f(x) = \Bbb{P}(X \geq \Bbb{E}X)$ for $X \sim \mathrm{Gamma}(x, 1)$. Since the gamma distribution arises when summing i.i.d. exponential distribution, exploiting CLT allows us to guess that the limit should be $1/2$. – Sangchul Lee Dec 29 '14 at 4:01
• @Mhenni: if you mean the two fuctions $F=\frac 1{\Gamma}$ and $G:x \mapsto \int_x^{+\infty} t^{x-1}e^{-t} dt$ , it is known that $\displaystyle \lim_{+\infty} F = 0$ and i think we can prove that: $\displaystyle \lim_{+\infty} G=+\infty$ ... – Mohamed Dec 29 '14 at 4:17
• @Mohamed: the limit of G should be $0$. – Mhenni Benghorbal Dec 29 '14 at 4:31
• @Mhenni:Thank you, i well verify this. – Mohamed Dec 29 '14 at 4:58

## 2 Answers

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,{\rm Li}_{#1}} \newcommand{\norm}[1]{\left\vert\left\vert\, #1\,\right\vert\right\vert} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $\ds{\fermi\pars{x}\equiv{1 \over \Gamma\pars{x}} \int_{x}^{\infty}t^{x - 1}\expo{-t}\,\dd t:\ {\large ?}}$

First, we have to derive an asymptotic behavior of the integral when $\ds{x \gg 1}$. The derivation is similar to the one which leads to the Gamma function asymptotic behavior ( namely, the Stirling approximation ):

\begin{align}&\left.\dsc{\int_{x}^{\infty}t^{x - 1}\expo{-t}\,\dd t}\, \right\vert_{\,x\ \gg\ 1} =\int_{x}^{\infty}\exp\pars{\bracks{x - 1}\ln\pars{t} - t}\,\dd t \\[5mm]&\sim\int_{x}^{\infty} \exp\pars{\bracks{x - 1}\ln\pars{x - 1} - \bracks{x - 1} - {\bracks{t - x + 1}^{2} \over 2\bracks{x - 1}}}\,\dd t \\[5mm]&=\pars{x - 1}^{x - 1}\expo{-\pars{x - 1}} \int_{1}^{\infty}\exp\pars{-t^{2} \over 2\bracks{x - 1}}\,\dd t \\[5mm]&=\bracks{\pars{x - 1}^{x - 1}\expo{-\pars{x - 1}}}\root{2\pars{x - 1}} \int_{1/\root{2\pars{x - 1}}}^{\infty}\exp\pars{-t^{2}}\,\dd t \\[5mm]&\sim\pars{x - 1}^{x - 1/2}\expo{-\pars{x - 1}}\root{2}\ \overbrace{\int_{0}^{\infty}\exp\pars{-t^{2}}\,\dd t}^{\dsc{\root{\pi} \over 2}} \\[5mm]&=\dsc{\half}\, \bracks{\root{2\pi}\pars{x - 1}^{x - 1/2}\expo{-\pars{x - 1}}} \end{align}

Then

\begin{align}&\color{#66f}{\large% \left.{\int_{x}^{\infty}t^{x - 1}\expo{-t}\,\dd t \over \Gamma\pars{x}}\, \right\vert_{\,x\ \gg\ 1}} \sim{\pars{\dsc{1/2}}\bracks{% \root{2\pi}\pars{x - 1}^{x - 1/2}\expo{-\pars{x - 1}}}\over \root{2\pi}\pars{x - 1}^{x - 1/2}\expo{-\pars{x - 1}}} \color{#66f}{\large\to\ \half} \end{align}

The $\ds{\Gamma\pars{s,z}}$ asymptotic behavior as given in this link is misleading because $\ds{\tt\mbox{it should be valid for fixed}}$ $\ds{s}$. In another words, $\ds{\lim_{x\ \to\ \infty}\lim_{s\ \to\ x}\Gamma\pars{s,x}}$ can not be evaluated with such expansion which was the one I used blindly in my previous answer.

Thanks to @Mohamed who call my attention to this point and to this paper and thanks to @anorton who was worried about the whole procedure. Thanks to both of them.

• Isn't computing $\lim_{x\to\infty}\Gamma(x, x)$ still pretty difficult, even with that nice asymptotic form? (Or am I just not seeing something?) – apnorton Dec 29 '14 at 4:12
• I find this method pretty easy. We do not have to 'reinvent the wheel'. Otherwise we are forced to rederive a well known result. Thanks. – Felix Marin Dec 29 '14 at 4:24
• Hmm... Ok. I guess my limit-taking abilities aren't that great with sums like that. I certainly agree that reducing to a known problem is much easier, I just can't see how the sum reduces to the fraction you've listed... I'll work on it more in the morning, I guess. – apnorton Dec 29 '14 at 4:42
• Why this paper gives $\frac 12$? See pages 468 and 469. cs.purdue.edu/homes/wxg/selected_works/section_02/068.pdf – Mohamed Dec 29 '14 at 5:14
• @Mohamed I'll check it. Thanks. – Felix Marin Dec 29 '14 at 5:51

Does this help?

$\int_{0}^{\infty}t^{x-1}e^{-t}dt=\int_{0}^{x}t^{x-1}e^{-t}dt+\int_{x}^{\infty}t^{x-1}e^{-t}dt$

so

$\int_{x}^{\infty}t^{x-1}e^{-t}dt=\Gamma\left(x\right)-\int_{0}^{x}t^{x-1}e^{-t}dt$

In taking the limit we get zero: $\lim_{x\rightarrow\infty}\frac{\Gamma\left(x\right)-\int_{0}^{x}t^{x-1}e^{-t}dt}{\Gamma\left(x\right)}=1-1=0$

• You've only asserted that the limit in the last line goes to zero. How do we know that? – apnorton Dec 29 '14 at 14:00