I am taking a somewhat hard measure theory course and I was asked to prove this:
a) Let $\alpha > 0$ be a real number. Prove that
$$\Gamma(\alpha):=\int_0^\infty e^{-x}x^{\alpha-1}dx$$
exists. (We are studying the relationship between being Lebesgue-integrable and Riemann-integrable, so I am not quite sure what kind of integral should I prove exists, but I suspect it's the latter).
b)Prove that:
$$\lim_{n\to\infty}\int_0^n \left( 1-\frac{x}{n}\right)^n x^{\alpha-1}dx=\Gamma(\alpha).$$
Hint: $(1-\frac{x}{n})^n\le (1-\frac{x}{n+1})^{n+1}$ whenever $\frac{x}{n}<1$.
This reeks of some limit theorem such as Monotone Convergence, Fatou or Lebesgue's Dominated Convergence, but those apply to Lebesgue-integrable functions. I the inequality stated in the hints makes me think I should use Monotone Convergence.
Any insight would be greatly appreciated. We are using Bartle's Measure Theory Book and we are more or less around the $\mathcal{L}_p$ spaces part.
Thank you in advance.
Best Answer
1) Hint: For $x\ge0$ and $n\ge1$, $e^x=\sum\limits_{k=0}^\infty\frac{x^k}{k!}\ge1+\frac{x^n}{n!}$, thus $e^{-x}\le\frac{n!}{n!+x^n}$
2) Hint: For $m\ge n$, $$ \int_0^m\left(1-\frac xm\right)^m x^{\alpha-1}\,\mathrm{d}x \ge\int_0^n\left(1-\frac xm\right)^m x^{\alpha-1}\,\mathrm{d}x \ge\int_0^n\left(1-\frac xn\right)^n x^{\alpha-1}\,\mathrm{d}x $$ and $\left(1-\frac xn\right)^n\le e^{-x}$ for $x\le n$.