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 $\lim_{n \to \infty}$.

  • $\begingroup$ 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

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

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

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    $\begingroup$ well done! (+1) $\endgroup$ – tired Feb 13 '17 at 16:57
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    $\begingroup$ 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
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    $\begingroup$ 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

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