I have this limit: $\displaystyle \lim\limits_{x\to \infty}\dfrac{1}{x}\int_{0}^x \dfrac{1}{2+\cos(\mathrm t)}\, \mathrm{dt}$

I said that

Edit: $-1\leq \cos\mathrm{t}\leq 1 \Rightarrow \dfrac{1}{2+\cos\mathrm{t}} > 0 \text{ and because that expression has no limit as x -> inf}\\\Rightarrow \displaystyle \int_{0}^x \dfrac{1}{2+\cos\mathrm{t}}\, dt \to \infty \\ \\ \Rightarrow \lim\limits_{x\to \infty} \dfrac{\displaystyle\int_{0}^x \dfrac{1}{2+\cos\mathrm{t}}\,\mathrm{dt}}{x} = \lim\limits_{x\to \infty} \dfrac{\Big( \displaystyle\int_{0}^x \dfrac{1}{2+\cos \mathrm{t}}\, \mathrm{dt}\Big)'}{x'} = \lim\limits_{x\to \infty} \dfrac{\dfrac{1}{2+\cos \mathrm{x}}}{1} = \lim\limits_{x\to \infty} \dfrac{1}{2+\cos \mathrm{x}}$

But here I got stuck because that limit does not exist, because cos x has no limit when x goes to infinity.

But the correct answer is $\dfrac{1}{\sqrt 3}$, what did I do wrong?

  • $\begingroup$ You did nothing wrong (yet), you just got stuck ;) L'Hôpital's rule is not applicable here. Use $2\pi$-periodicity of the integrand to see that the limit is equal to the expression obtained by setting $x=2\pi$. $\endgroup$ – metamorphy Jun 10 '19 at 19:13
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    $\begingroup$ $f(t)>0$ does not imply $\int_0^\infty f(t)\,dt=\infty$. For example, take $f(t)=e^{-t}$. $\endgroup$ – MathIsFun7225 Jun 10 '19 at 19:14
  • $\begingroup$ @MathIsFun Yes, But because 2/(1+cost) does not have limit as x -> inf or x -> -inf it means the integral -> inf. $\endgroup$ – Dan Avrămescu Jun 10 '19 at 19:35
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    $\begingroup$ @metamorphy You are right. Replacing x with 2pi the limit goes 1/sqrt(3). $\endgroup$ – Dan Avrămescu Jun 10 '19 at 19:48
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    $\begingroup$ Your problem is that L'Hopital's rule does not apply. Specifically, L'Hopital says $\lim f/g = \lim f'/g'$ if $\lim f'/g'$ exists, which is not the case here. $\endgroup$ – Quang Hoang Jun 10 '19 at 19:54

Note that \begin{align} \int^{2\pi n}_0 \frac{dt}{2+\cos t} = n \int^{2\pi}_0 \frac{dt}{2+\cos t} \end{align} which means \begin{align} \frac{1}{2\pi n}\int^{2\pi n}_0 \frac{dt}{2+\cos t} = \frac{1}{2\pi}\int^{2\pi }_0 \frac{dt}{2+\cos t}. \end{align}

Hence \begin{align} \lim_{n\rightarrow \infty} \frac{1}{2\pi n}\int^{2\pi n}_0 \frac{dt}{2+\cos t}= \frac{1}{2\pi}\int^{2\pi }_0 \frac{dt}{2+\cos t}= \frac{1}{\sqrt{3}}. \end{align}

This suggests that \begin{align} \lim_{x\rightarrow \infty} \frac{1}{x}\int^x_0 \frac{dt}{2+\cos t} = \frac{1}{\sqrt{3}} \end{align} if the limit exists.

To show that the limit exists, observe we have that \begin{align} \left|\frac{1}{x}\int^{x}_0 \frac{dt}{2+\cos t}- \frac{1}{2\pi} \int^{2\pi}_0 \frac{dt}{2+\cos t}\right| =&\ \left|\frac{1}{2\pi n+r}\int^{2\pi n+r}_0 \frac{dt}{2+\cos t}- \frac{1}{2\pi} \int^{2\pi}_0 \frac{dt}{2+\cos t}\right|\\ =&\ \left|\frac{n}{2\pi n+r}-\frac{1}{2\pi}\right|\int^{2\pi}_0\frac{dt}{2+\cos t}+ \frac{1}{2\pi n+r}\int^{r}_{0} \frac{dt}{2+\cos t} \end{align} where $0\le r< 2\pi$. Then as $n\rightarrow \infty$ we get the desired result.

  • $\begingroup$ Thank you! Now I understand. $\endgroup$ – Dan Avrămescu Jun 10 '19 at 20:04

Use the fact that $\cos x$ is periodic of period $2\pi$ (as per a comment).

Thus, we are just getting the average area over an interval of length $2\pi$.

Using Wolfram alpha, say, we get $\int_0^{2\pi}\dfrac 1{2+\cos x}\operatorname dx=\dfrac {2\pi}{\sqrt3}$.


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