I was reading the derivation of Euler's reflection formula for Gamma function on Wikipedia. It's an easy enough proof that does not require anything sophisticated or technical as other proofs. Then I realized I don't know how to justify the obvious looking representation of the gamma function
$$\lim_{n\to \infty} \int_0^n t^{x-1} \left(1-\frac{t}{n}\right) ^n dt = \Gamma(x) := \int_0^\infty t^{x-1}e^{-t}dt,$$ where $x>0$. It seems like it's obvious, but stuff like this doesn't hold automatically. Take e.g. $$f_n(x)= \begin{cases} \frac{1}{n} & 0\leq x\leq n,\\ 0 & x>n, \end{cases} $$ then $f_n \to 0$ (even uniformly), but $\int_0^n f_n = 1$.
If we don't want to just believe it, we need to show that $$\int_0^n t^{x-1}\left|e^{-t}-\left(1-\frac{t}{n}\right)^n\right| dt \to 0.$$
It would suffice if we had for arbitrary $\varepsilon>0,m>-1$ an estimation $$\left|e^{-t}-\left(1-\frac{t}{n}\right) ^n\right|\leq \frac{\varepsilon}{nt^m}$$ for $t\in [0,n]$ for large enough $n$'s. But my numerical experiments say this isn't actually true; the approximation of the exponential function is not that good.
Seems like something more subtle is needed. But this is where I'm essentially stuck. Any ideas? Quite possibly I'm missing something straightforward.
Let $$f_n(t) := t^{x-1} (1-t/n)^n \mathbf{1}_{[0, n]}(t),$$ $$f(t) := t^{x-1} e^{-t} \, dt.$$
One can show that $(1-t/n)^n \le e^{-t}$ for all $n$ by checking that $(1-t/n)^n$ is increasing in $n$ and converges to $e^{-t}$.
So, $0 \le f_n(t) \le f(t)$ for $t \ge 0$. Since $f$ is integrable, we can use the dominated convergence theorem to conclude $\int_0^\infty f_n(t) \, dt \to \int_0^\infty f(t) \, dt$ from the pointwise convergence $f_n \to f$.
Side note: the conditions for the dominated convergence theorem fail to hold for your example $f_n(t) = \frac{1}{n} \mathbf{1}_{[0,n]}(t)$ because $\sup_{t \ge 0} |f_n(t)|$ decays like $1/t$ which is not integrable.