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.
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 $\Re(z) > 0.$