Prove the limit of this Lebesgue integral in connection with gamma function I want to prove that
$$\lim _{n \rightarrow \infty} \int_{0}^{1} x^{\alpha-1} e^{-n \beta x}(1-x)^{n} \mathrm{~d} x=(\beta+1)^{-\alpha} \Gamma(\alpha)$$ when $\alpha > 0$ and $\beta > -1.$
First I set $y=nx$, then it can be rewrited as
$$\int_{0}^{n} n^{-\alpha}y^{\alpha-1} e^{- \beta y}(1-\frac{y}{n})^{n} \mathrm{~d} y$$
I can prove that $$\lim _{n \rightarrow \infty} \int_{0}^{n}\left(1-\frac{x}{n}\right)^{n} x^{\alpha-1} d x=\Gamma(\alpha)$$, but it is different from above. How can I overcome this obstacle?
 A: What you actually want to prove is
$$\lim_{n\to\infty}\:\color{red}{n^\alpha}\int_0^1x^{\alpha-1}\,\mathrm e^{-n\beta x}(1-x)^n\,\mathrm dx=(\beta+1)^{-\alpha}\Gamma(\alpha).$$
Assume $|\beta|<1$. Let $n\in\Bbb N$. Then
\begin{align*}
n^\alpha\int_0^1x^{\alpha-1}\,\mathrm e^{-n\beta x}(1-x)^n\,\mathrm dx
&=n^\alpha\int_0^1\sum_{k=0}^\infty\frac{(-n\beta)^k}{k!}\,x^{\alpha+k-1}(1-x)^n\,\mathrm d x\\[.4em]
&=\sum_{k=0}^\infty\frac{(-\beta)^k}{k!}\,n^{\alpha+k}\,\mathrm B(\alpha+k,n+1)\tag{$\star$}
\end{align*}
because the series of functions $\sum_kx\mapsto\frac{(-n\beta)^k}{k!}\,x^{\alpha+k-1}(1-x)^n$ is normally convergent on $[0,1]$.
Next we note the limit $n^{\alpha+k}\,\mathrm B(\alpha+k,n+1)\xrightarrow[n\to\infty]{}\Gamma(\alpha+k)$, with
$$0\le n^{\alpha+k}\,\mathrm B(\alpha+k,n+1)=\Gamma(\alpha+k)\,\frac{n!\,n^{\alpha+k}}{\Gamma(\alpha+k+n+1)}\le\Gamma(\alpha+k),$$
so the summand in $(\star)$ is bounded (uniformly in $n$) by $\frac{\beta^k}{k!}\Gamma(\alpha+k)$, which is summable.
The dominated convergence theorem allows us to take the limit $n\to\infty$:
$$L:=\lim_{n\to\infty}n^\alpha\int_0^1x^{\alpha-1}\,\mathrm e^{-n\beta x}(1-x)^n\,\mathrm dx=\sum_{k=0}^\infty\frac{\Gamma(\alpha+k)}{k!}(-\beta)^k.
$$
Recalling the power series
$$(1-x)^{-\alpha}=\sum_{k=0}^\infty\frac{\Gamma(\alpha+k)}{\Gamma(\alpha)k!}x^k,\quad |x|<1$$
we find $L=\Gamma(\alpha)(1+\beta)^{-\alpha}$ if $|\beta|<1$. This should extend to all $\beta\in\mathbb C,\,\Re{\beta}>{–1},$ by analytic continuation.
