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From Havil & Dyson, "Gamma: Exploring Euler's Constant", section 6.1 I can't prove the following Euler's theorem :

... on 13 October 1729, Euler had already proposed to Goldbach the definition $$\Gamma (x) = \lim_{r\ \to \ \infty} \Gamma_r (x) $$ where $$\Gamma_r (x) = \dfrac{r! r^x}{x(x+1)(x+2) \dots (x+r)}$$ [which is valid whenever $x$ is not zero or negative integer].

After 1 hours and 40 minutes of attempt I can't turn $\lim_{r\ \to \ \infty} \Gamma_r (x)$ to the following definition of $\Gamma (x)$ that I know, i.e. $$\Gamma (x) = \int_{0}^{\infty} t^{x-1} e^{-t} dt, \ \ \ x>0. $$

So why $$\lim_{r\ \to \ \infty} \dfrac{r^x}{x(1+ \frac{x}{1})(1+\frac{x}{2}) \dots (1+\frac{x}{r})} = \int_{0}^{\infty} t^{x-1} e^{-t} dt , \ \ \ x>0 \ \text{?} $$

EDIT : For the special case when $x$ is a positive integer we will have the problem of showing $$\lim_{r\ \to \ \infty} \dfrac{r!r^n}{(n+r)!} =1 $$ to be valid.

  • What about: "for non-negative $x$, not necessarily an integer"? Though you want $x$ not to be an integer, have you looked at integer $x$ an integer, since then it may be easier to check out? – jim Apr 13 '16 at 10:01
  • @jim - the book has consider $x\in \mathbb {R} - (\mathbb {Z} \cup {{0}})$ which is reasonable for both $\Gamma_r$ and $\Gamma$. For positive integer case I'll upload an edit. –  Apr 13 '16 at 10:07
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    The usual approach for this is to first show that $\Gamma(1)=1$, $\Gamma(x+1)=x\Gamma(x)$ and that $\log\Gamma(x)$ is convex on $(0, \infty)$ from the integral representation of $\Gamma(x)$. Then one has to show that these three properties define a unique function $f$ for $x>0$ such that $\left[\log f(x)-\log \Gamma_r(x)\right]_{r\rightarrow \infty}=0$. – vnd Apr 13 '16 at 10:41
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    @vnd - If the mentioned three conditions hold for some function $f(x)$ then $f(x)$ must be the Gamma function (i.e. no other function will do) according to The Bohr-Mollerup Theorem. But I don't know if I can understand its proof but I will try! –  Apr 13 '16 at 10:46
  • Another way to start it off, it is by using the Cauchy Integral Formula, using the integrand $s^{z-1}\cdot \exp(-s) ds/(s_0-z)/(s_1-z)\cdots /(s_r-z)$, with a multiple keyhole countour around all the negative integers and ending at $s_0=0$ and with $s_1=-1$, $s_2=-2$, etc. Each $s_k$ will contribute an appropriate $Res_f@(s_k=z)$, etc... –  Apr 13 '16 at 23:08
  • @vnd, I read the proof on wiki. It doesn't answer the question, it implies that if have the Euler's definition then it assures us that $\Gamma (x)$ is the expected function we were looking for. –  Apr 14 '16 at 09:04

1 Answers1

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The gamma function is well-defined by

$$\Gamma(z) = \lim_{r \to \infty} \Gamma_r(z),$$

as it can be shown that the limit exists as a convergent infinite product when $z$ is not zero or a negative integer.

We then can derive the integral form of the gamma function by first relating Euler's definition to the beta function.

The beta integral is defined as

$$B(x,y) = \int_0^1 t^{x-1}(1-t)^{y-1} \, dt.$$

This satisfies

$$B(x,y) - B(x+1,y)= B(x,y+1) = \int_0^1t^{x-1}(1-t)^y \, dt.$$

Integrating by parts, we get

$$B(x,y+1) = \frac{y}{x}B(x+1,y)$$

Hence,

$$B(x,y) = B(x,y+1) + B(x+1,y) = B(x,y+1) + \frac{x}{y}B(x,y+1) \\ \implies B(x,y) = \frac{x+y}{y}B(x,y+1).$$

Now using this recurrence relation $r$ times we find

$$B(x,y) = \frac{(x+y)(x+y+1) \ldots (x + y + r)}{y(y+1) \ldots (y +r)}B(x,y+r+1) \\ =\frac{\Gamma_r(y)}{\Gamma_r(x+y)}r^{x}B(x,y+r+1).$$

Note that

$$r^{x}B(x,y+r+1) = r^{x}\int_0^1t^{x-1}(1-t)^{y+r} \, dt.$$

Changing variables with $s = rt$ we get

$$r^{x}B(x,y+r) = \int_0^rs^{x-1}\left(1-\frac{s}{r}\right)^{y+r} \, ds = \int_0^rs^{x-1}\left(1-\frac{s}{r}\right)^{r} \left(1-\frac{s}{r}\right)^{y}\, ds. $$

Hence,

$$B(x,y) = \frac{\Gamma_r(y)}{\Gamma_r(x+y)}\int_0^rs^{x-1}\left(1-\frac{s}{r}\right)^{r} \left(1-\frac{s}{r}\right)^{y}\, ds.$$

We can progress towards representing the gamma function as an integral over $[0, \infty),$ by taking the limit of both sides of this equation as $r \to \infty$. The LHS will remain unchanged since there is no explicit dependence on $r$.

First, note that

$$\int_0^rs^{x-1}\left(1-\frac{s}{r}\right)^{r} \left(1-\frac{s}{r}\right)^{y}\, ds = \int_0^{\infty}s^{x-1}\left(1-\frac{s}{r}\right)^{r} \left(1-\frac{s}{r}\right)^{y}1_{[0,r]}\, ds , $$

and by the Lebesgue dominated convergence theorem

$$\lim_{r \to \infty}\int_0^rs^{x-1}\left(1-\frac{s}{r}\right)^{r} \left(1-\frac{s}{r}\right)^{y}\, ds \\ = \int_0^{\infty} \lim_{r \to \infty}s^{x-1}\left(1-\frac{s}{r}\right)^{r} \left(1-\frac{s}{r}\right)^{y}1_{[0,r]}\, ds \\ = \int_0^\infty s^{x-1}e^{-s}\, ds.$$

Since $\lim_{r \to \infty} \Gamma_r(z) = \Gamma(z)$, we find

$$B(x,y) = \frac{\Gamma(y)}{\Gamma(x+y)}\int_0^\infty s^{x-1}e^{-s}\, ds.$$

Finally, with $y = 1$ and using $\Gamma(1) = 1$ and $\Gamma(x+1) = x\Gamma(x),$ we find

$$\begin{align} \int_0^\infty s^{x-1}e^{-s}\, ds &= \frac{\Gamma(x+1)}{\Gamma(1)}B(x,1) \\ &= \frac{x\Gamma(x)}{1}\int_0^1t^{x-1} \, dt \\ &= \,\,\,\Gamma(x) \end{align}$$

RRL
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  • Thank you so much. This is a brilliant answer. Just two remarks: a. $B(x,y) = B(x,y+r+1)(x+y)(x+y+1) \ldots (x + y + r)/\big(y(y+1) \ldots (y +r)\big)$ i.e. $B(x,y+r+1)$ to be replaced by $B(x,y+r)$. Doing so, it makes no difference after the limit is done. [...] –  Apr 14 '16 at 09:49
  • [...] b. I don't none of "Lebesgue dominated convergence theorem", what is $1_{[0,1]}$, why entering $1_{[0,1]}$ in the integral change the upper-bound from $r$ to $\infty$ and how limit can pass through the integral. It's very disappointing not to know these many but (unless you do a great favor for me to explain them) I have to accept them for now till I learn them. Thank you again :-) –  Apr 14 '16 at 09:49
  • @AgnetaGauck: You're welcome. For (a) you are indeed correct. It should be $B(x,y+r+1)$. I corrected above. The rest of the argument is unchanged. – RRL Apr 14 '16 at 15:13
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    For (b), the symbol $1_{[0,r]}$ is the indicator function which in this case equals $1$ for $0 \leqslant s \leqslant r$ and $0$ otherwise. That just allows us to replace the upper limit of integration with $\infty$, since the integrand is taken to be $0$ beyond $s = r$. The key step here is switching the limit with integral. This can be justified here in several ways. With LDCT we use the fact that $|(1 - s/r)^r(1 - s/r)^y 1_{[0,r]}| \leqslant e^{-s}$ with the latter being an integrable function. We also know this converges to $e^{-s}$. – RRL Apr 14 '16 at 15:29
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    The other way is to use the monotone convergence theorem,applicable here because this integrand sequence is increasing in $r$. You most likely will learn such things at some point. – RRL Apr 14 '16 at 15:33
  • Thank You :-) I've already started the excellent book "Real Analysis" by H. L. Royden and P. M. Fitzpatrick, and so luckily it includes LDCT later in Ch.4. –  Apr 14 '16 at 15:59
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    Also, as you are particularly interested in Euler's constant it appears in a related gamma product: http://math.stackexchange.com/a/1614105/148510 – RRL Apr 15 '16 at 04:14
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    You might also find this interesting: http://math.stackexchange.com/a/844075/148510 – RRL Apr 15 '16 at 04:15