2
$\begingroup$

In the context of a measure theory and Lebesgue integration course I have to compute the limit $$\lim_{n\to \infty} \int_{1}^{\infty} \exp(-x)(\text{cos}(x))^n dx $$

To me this looks like the application of some convergence theorem, either the dominated or monotone one. I first checked some general properties relating to the convergence theorems:

1) Pointwise convergence: Since $\vert \text{cos}(x) \vert \leq 1$, it holds that $\vert (\text{cos}(x))^n \vert \leq 1 \forall n \in \mathbb{N}$. Hence I know that for the pointwise limit holds that $$-\exp(-x) \leq \lim_{n \to \infty} \exp(-x) (\text{cos}(x))^n \leq exp(-x)$$ But what can I infer about pointwise convergence from this? Since cosine oscillates between 1 and -1 there seems to be no pointwise limit. But there has to be in order to apply either one of the convergence theorems. So what do I miss here?

2) Dominating function: Let's apply the dominated convergence theorem. For this we need a non-negative function $g \in L(\lambda)$ for which it holds that $$\vert f_n(x) \vert \leq g(x) \quad \forall x \in [1,\infty),\quad \forall n \in \mathbb{N}$$. Obviously, a good candidate is $g(x) = \exp(-x)$ since it is non-negative and $$\int_{[1,\infty)} \vert\exp(-x)\vert d\lambda(x) < \infty$$

3) Now we can apply the dominated convergence theorem and write $$ \lim_{n \to \infty} \int_{0}^{\infty} \exp(-x) (\text{cos}(x))^n = \int_{0}^{\infty} \lim_{n \to \infty} \exp(-x) (\text{cos}(x))^n d\lambda(x)$$

Now all I need to do is evaluate the actual limit $$ \lim_{n \to \infty} \exp(-x) (\text{cos}(x))^n$$ but I don't see it since $(\text{cos(x)})^n$ jumps back and forth for $\lim_{n \to \infty}$.

$\endgroup$
  • $\begingroup$ the limit doesn't exist (your last one). Do you know the method of steepest descent? $\endgroup$ – tired Feb 13 '17 at 16:30
  • $\begingroup$ So there is no way to apply a convergence theorem? $\endgroup$ – Taufi Feb 13 '17 at 16:30
  • $\begingroup$ A bit, yes. But only from other courses that do not relate to this one. How does it help here? $\endgroup$ – Taufi Feb 13 '17 at 16:31
  • $\begingroup$ another idea would be to expand $\cos^n$ into a binomial series and integrate termwise. after that is done it should be simple to read of the dominant contribution $\endgroup$ – tired Feb 13 '17 at 16:57
3
$\begingroup$

Your ansatz with the dominated convergence theorem is right. A last step is missing: We know that for $f_n(x)=\exp(-x)(\text{cos}(x))^n$ we have $f_n(x) \to \exp(-x) \cdot 0$ for all $x \in \mathbb{R} \setminus \pi\mathbb{Z}$. But $\pi\mathbb{Z}$ is a set of measure zero in $\mathbb{R}$ (since it is discrete). Hence $f_n(x) \to 0$ almost everywhere. It follows

$$\lim_{n \to \infty} \int_{0}^{\infty} \exp(-x) (\text{cos}(x))^n = \int_{0}^{\infty} \lim_{n \to \infty} \exp(-x) (\text{cos}(x))^n d\lambda(x) $$ $$= \int_{0}^{\infty} 0 d\lambda(x) = 0$$

$\endgroup$
  • 1
    $\begingroup$ well done! (+1) $\endgroup$ – tired Feb 13 '17 at 16:57
  • 1
    $\begingroup$ I'm a bit confused here. $\cos(x)^n$ goes to zero, not $1$, at points which are not integer multiples of $\pi$. So shouldn't the limit be zero? $\endgroup$ – Ian Feb 13 '17 at 17:16
  • $\begingroup$ Oh yes, you are right. I will correct it. Thanks. $\endgroup$ – Muzi Feb 13 '17 at 17:44
  • $\begingroup$ Why don't you take $x \in \mathbb{R} - \frac{\pi}{2}\mathbb{Z}$? Then it would be $f_n(x) \to exp(-x)\cdot 1 \forall x \in \mathbb{R} - \frac{\pi}{2}$ since this is also a set of measure zero? $\endgroup$ – Taufi Feb 13 '17 at 18:53
  • 1
    $\begingroup$ For $x \in \mathbb{R} \setminus \pi\mathbb{Z}$ it holds that $|\cos(x)| < 1 \implies |\cos(x)^n| = |\cos(x)|^n \to 0$ as $n \to \infty$. Hence $\cos(x)^n \to 0$. For $x \in \pi\mathbb{Z}$ we have $\cos(x) = \pm 1$. I.e. $\cos(x)^n \not\to 0$ for $x \in \pi\mathbb{Z}$. But $\pi\mathbb{Z}$ is a set of measure zero. So we can ignore it. $\endgroup$ – Muzi Feb 13 '17 at 19:19

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.