Prove that for any $x>1$, $\lim_{n \to \infty} \int_0^n t^{x-1}(1-\frac{t}{n})^n dt = \Gamma(x)$

Question

Prove that for any $x>1$, $$\lim_{n \to \infty} \int_0^n t^{x-1}(1-\frac{t}{n})^n dt = \Gamma(x)$$

The hint is to use the dominated convergence theorem, so I did let $f(x)=t^{x-1}e^{-t}$ and let

$$f_n(t) = \begin{cases} t^{x-1}(1 - \dfrac{t}{n} )^n & t \in [0,n]\\ 0 & t > n\end{cases}$$

then $f_n$ converges to $f$, but I am struggling to find a function $g$ such that $|f_n(t)| \leq g(t)$ and that $\int_0^\infty g$ is finite.

Please help me solve this problem, thanks in advance.

Answer 1

Hint: Show that $x+\log(1-x)\leq 0$ on $[0,1($. Then replace $x$ by $t/n$, $0\leq t< n$, multiply by $n$, and deduce that $f(t)=t^{x-1}\exp(-t)$ is a dominating function $g$ for the DCT.