5
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ For (1) subdivide the integral in two integral:$$\int\limits_0^\infty=\int\limits_0^1+\int\limits_1^\infty$$ . For (2) read here a simple extension to Riemann Integral of the monotone convergence theorem: math.ucv.ro/~niculescu/articles/2011/… $\endgroup$
    – DonAntonio
    Commented Oct 18, 2013 at 6:03
  • $\begingroup$ I am trying to prove convergence subdividing that integral into those very same integrals, and proving the second one converges is not that hard, is the one from $0$ to $1$ that's giving me a hard time. While I appreciate the reference, the problems are based on the lectures and we didn't get even close to discussing extensions of the MCT for R-integrals. While a proof based on it may be mathematically correct, I think I am expected to solve this using basic measure-theory results and Lebesgue's theorem. $\endgroup$
    – alonso s
    Commented Oct 18, 2013 at 6:12

1 Answer 1

Reset to default
3
$\begingroup$

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 $x\le n$.

$\endgroup$
3
  • 1
    $\begingroup$ Have you covered the Dominated Convergence Theorem? $\endgroup$
    – robjohn ♦
    Commented Oct 18, 2013 at 6:30
  • $\begingroup$ I got the first part, thank you for the tip. How does one use your hint for the second part? From what I can see, your hint shows that se sequence of integrals is monotone non-decreasing, and is bounded above by the Gamma integral, so the limit exists and is less than or equal to the Gamma function. I don't know if that's what you had in mind. $\endgroup$
    – alonso s
    Commented Oct 18, 2013 at 7:09
  • $\begingroup$ @user2770617: Let $f_n(x)=[x\lt n]\left(1-\frac{x}{n}\right)^nx^{\alpha-1}$. Then the hint shows that $f_n(x)\le e^{-x}x^{\alpha-1}$ and $\lim\limits_{n\to\infty}f_n(x)=e^{-x}x^{\alpha-1}$. Dominated Convergence then gives $$ \lim_{n\to\infty}\int_0^n\left(1-\frac{x}{n}\right)^nx^{\alpha-1}\,\mathrm{d}x =\int_0^\infty e^{-x}x^{\alpha-1}\,\mathrm{d}x $$ $\endgroup$
    – robjohn ♦
    Commented Oct 18, 2013 at 7:56

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .