4
$\begingroup$

I want to compute that $\lim\limits_{n\to\infty}\int_{[0,1]}\frac{nx^{n-1}}{1+x} \, d\lambda(x)=\frac{1}{2}$, where $\lambda$ is the Lebesgue measure.

I want to use Lebesgue dominated convergence theorem, but I can't find a uniform bound for $|\frac{nx^{n-1}}{1+x}|$, and the limit function tends to be $\begin{cases}0 & x<1 \\\infty & x=1\end{cases}.$ How do I deal with this integral?

$\endgroup$
3
$\begingroup$

Make the substitution of variable $u = x^n$ which implies $du = n x^{n-1}dx$ to get $$\int_0^1 \frac{nx^{n-1}}{1+x} \ dx = \int_0^1 \frac{1}{1+ u^{1/n}} \ du$$

And now, you can apply Lebesgue dominated convergence theorem as $$0 \le \frac{1}{1+ u^{1/n}} \le 1$$ and $\lim\limits_{n \to \infty} \frac{1}{1+ u^{1/n}} = 1/2$ for $u \in (0,1]$.

$\endgroup$
  • $\begingroup$ We didn't officially treat substitutation of variables yet in a measure theoretic version $\endgroup$ – James Oct 22 at 13:59
  • $\begingroup$ OK. How did you define the integral of a function in your course then? $\endgroup$ – mathcounterexamples.net Oct 22 at 14:01
  • $\begingroup$ As the difference between the positive and negative part of the function, then the integral of the positive part is the supremum over the integrals of all step functions which lie under the positive part. $\endgroup$ – James Oct 22 at 14:02
  • $\begingroup$ Won't be easy to use LDC theorem if you use only that. You're not even suppose to know fundamental theorem of calculus? $\endgroup$ – mathcounterexamples.net Oct 22 at 14:11
  • $\begingroup$ We do know about differentiation under the integral sign: If $u:(a,b)\times X\to\mathbb{R}$ is integrable and differentiable with respect to $t$ and $|\partial_tu(t,x)|\leq|w(x)|$ for all $(t,x)$ with $w$ integrable, then I can change differentiation and integration order. $\endgroup$ – James Oct 22 at 14:14
1
$\begingroup$

Dominated convergence allows to interchange limit and integral.The integral of the pointwise limit is zero. However, the limit of the integrals is non-zero. So dominated convergence cannot work for this example.

$\endgroup$
  • $\begingroup$ I agree, what would be a method to compute the integral? $\endgroup$ – James Oct 22 at 13:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.