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I'm trying to work through Gauss' derivation of the Gamma function from:

$$ \Gamma : \mathbb{R^+}\mapsto\mathbb{R} \;\;,\;\;\Gamma(n) = \int_0^\infty y^{n-1}e^{-y}dy\,. $$

I know the limit expression for $e^{-y}$ is

$$ e^{-y} = \lim_{m\rightarrow\infty}\bigg(1-\frac{y}{m}\bigg)^m. $$

If I then have

$$ \Gamma : \mathbb{R^+}\mapsto\mathbb{R} \;\;,\;\;\Gamma(n) = \int_0^\infty y^{n-1}\lim_{m\rightarrow\infty}\bigg(1-\frac{y}{m}\bigg)^mdy\,. $$

Am I then able to take out the limit from the integral? What are the conditions/justification of being able to do so? Is it possible to extend this to a complex domain? Would it be possible if the domain was the reals without negative integers?

Thanks.

JL53
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  • May be you ask if $\lim_{m \to\infty} \int_0^\infty y^{n-1}\left(1-\frac{y}{m}\right)^mdy = \Gamma(n)?$ – kmitov Feb 27 '17 at 19:10
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    You should look into the dominated (bounded) convergence theorem, and monotone convergence theorem. They give conditions when you can exchange a limit and integral. – David Feb 27 '17 at 19:11

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The dominated convergence theorem can be used here to justify switching the limit and the integral but some care with the integration limits is required.

Note that the integral of $y^{n-1}(1 - y/m)^m$ over $[0,\infty)$ is divergent and the result you are looking for is

$$\lim_{m \to \infty}\int_0^m y^{n-1} \left(1 - \frac{y}{m} \right)^m \, dy = \int_0^\infty y^{n-1} e^{-y} \, dy. $$

We can justify this first by putting the upper limit on the left-hand side to $\infty$ after multiplying the integrand by the indicator function $\chi_{[0,m]}(y)$ which is defined as

$$\chi_{[0,m]}(y) = \begin{cases}1, \,\,\,0 \leqslant y \leqslant m\\ 0, \,\,\,y > m \end{cases}$$

This gives us

$$\begin{align}\lim_{m \to \infty}\int_0^m y^{n-1} \left(1 - \frac{y}{m} \right)^m \, dy &= \lim_{m \to \infty}\int_0^\infty y^{n-1} \left(1 - \frac{y}{m} \right)^m \chi_{[0,m]}(y) \, dy \\ &= \int_0^\infty \lim_{m \to \infty}y^{n-1} \left(1 - \frac{y}{m} \right)^m \chi_{[0,m]}(y) \, dy \\ &= \int_0^\infty y^{n-1} e^{-y} \, dy \end{align}.$$

The dominated convergence theorem has been applied to justify the interchange for an integrand that is both integrable over $[0,\infty)$ and dominated by the integrable function to which it converges. Note that as $m \to \infty$ we have

$$\left(1 - \frac{y}{m} \right)^m \chi_{[0,m]}(y) \uparrow e^{-y},$$

since the LHS increases monotonically with respect to $m$. An application of Bernoulli's inequality verifies this.

Since $m + 1 > m$ we have

$$\left(1 - \frac{y}{m+1} \right)^{(m+1)/m} \geqslant 1 - \frac{y}{m+1}\frac{m+1}{m} = 1 - \frac{y}{m} \\ \implies \left(1 - \frac{y}{m+1} \right)^{m+1} \geqslant \left(1 - \frac{y}{m} \right)^{m}.$$

The integral representation of $\Gamma(z)$ where $z \in \mathbb{C}$ is convergent only for $\Re(z) > 0.$

RRL
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  • Don't you need some comment about the limit function being Riemann integrable, in order for to apply Lebesgue integral theorems to Riemann integrals (as I expect the OP intended)? e.g., does bounded convergence theorem hold for Riemann integral. – adfriedman Feb 28 '17 at 03:26
  • Right -- this is an improper Riemann integral -- so a form of dominated convergence is applicable. The integrand is of course dominated by the same integrable function. – RRL Feb 28 '17 at 03:43
  • Of course Lebesgue and convergent improper Riemann integrals are equivalent for nonnegative functions. Since the Riemann and Lebesgue integrals are the same on bounded intervals, $\int_0^m f(x) , dx = \int_{[0,m]} f = \int_0^\infty f \chi_{[0,m]}$ and $\lim_{m \to \infty} \int_0^m f(x) , dx = \int_0^\infty f$ where convergence on the right is a consequence of monotone convergence. Given my experience with this type of question asked on this site, most of this effort is not warranted. – RRL Feb 28 '17 at 04:25
  • @RRL I am struggling to accept that \int_0^\infty y^{n-1} e^{-y} , dy = \int_0^\infty \lim_{m \to \infty}y^{n-1} \left(1 - \frac{y}{m} \right)^m \chi_{[0,m]}(y) , dy – JL53 Feb 28 '17 at 11:38
  • I'm sure you know that $\lim_{m \to \infty} (1 - y/m)^m = e^{-y}$. If not than you can find this proved many times on this site. Hence $\lim_{m \to \infty} y^{n-1}(1 -y/m)^m1_{[0,m]}(y) = y^{n-1} e^{-y} 1_{[0,\infty]}(y) = y^{n-1}e^{-y}$. – RRL Feb 28 '17 at 17:08
  • ... just consider fixed $y$ then for all sufficiently large $m$, $y < m$ and $1_{[0,m]}(y) = 1.$ – RRL Feb 28 '17 at 17:10