# Evaluate $\lim _{x\to \infty }\frac{1}{x}\int _0^x\cos\left(t\right)dt\:$

I think that $\lim\limits_{x\to \infty }\frac{1}{x}\int _0^x\cos\left(t\right)dt\:$ is divergent, I can prove with taylor series?

• Why Taylor when you can integrate? It is in fact convergent to $0$. – user230734 Apr 19 '15 at 14:05
• The function $$f(x)=\int_0^x\cos(t)dt$$ oscillates between $-1$ and $1$. Divide by $x$, and let $x$ increase. – Arthur Apr 19 '15 at 14:08
• @Arthur with Taylor's series: $cos\left(x\right)=1-\frac{x^2}{2}+...$ , $\int _0^x\:f\left(t\right)dt\ge \int _0^x\left(1-\frac{t^2}{2}\right)dt=x-\frac{x^3}{6}\:\:\rightarrow \:\infty$ which is divergent, so where I was wrong? – Lucas Apr 19 '15 at 14:16
• What you've shown is that the integral in question is greater than $-\infty$ because the $x^3$ term dominates. – Michael Burr Apr 19 '15 at 14:20

It can't be, since $\lvert \cos{t} \rvert<1$, so the integral is bounded between $x$ and $-x$. It could oscillate, but not diverge. However, $$\frac{1}{x}\int_{0}^{x} \cos{x} \, dx = \frac{1}{x} (\sin{x}-\sin{0}) = \frac{\sin{x}}{x},$$ which tends to $0$ since it is absolutely bounded by $1/x$, which tends to $0$.

• with Taylor's series: $cos\left(x\right)=1-\frac{x^2}{2}+...$ , $\int _0^x\:f\left(t\right)dt\ge \int _0^x\left(1-\frac{t^2}{2}\right)dt=x-\frac{x^3}{6}\:\:\rightarrow \:\infty$ which is divergent, so where I was wrong? – Lucas Apr 19 '15 at 14:13
• @Lucas For higher values of $x$ you cannot cut away the higher degree terms like that. Cutting away higher degree terms only works for $x$ close to $0$. – Arthur Apr 19 '15 at 14:17
• @Arthur can you give me an example, to understand... please – Lucas Apr 19 '15 at 14:20
• Lucas: graph $\cos(x)$, and graph $1-x^2/2$. Is the second one really always bigger than the first? – Unit Apr 19 '15 at 14:21
• No, the issue here is that your function is $1-x^2/6$, which actually tends to $-\infty$. What you've proved is that the expression is bigger than something that tends to $-\infty$, which doesn't say much. Taking the next term, i.e. $1-x^2/2+x^4/4!$, shows that the expression is smaller than something that tends to $+\infty$, and so on. – Chappers Apr 19 '15 at 14:23

Note that $$\int_0^x \cos(t)dt=\sin(x)$$ Hence, we have $$\lim_{x \to \infty} \dfrac\int_0^x \cos(t)dt}x = \lim_{x \to \infty} \dfrac{\sin(x)}x =$$ since $\dfrac{\sin(x)}x \in \left[-\dfrac1x,\dfrac1x\right]$.

• with Taylor's series: $cos\left(x\right)=1-\frac{x^2}{2}+...$ , $\int _0^x\:f\left(t\right)dt\ge \int _0^x\left(1-\frac{t^2}{2}\right)dt=x-\frac{x^3}{6}\:\:\rightarrow \:\infty$ which is divergent, so where I was wrong? – Lucas Apr 19 '15 at 14:16
• @Lucas You are using a Taylor expansion around $0$, to "prove" divergence at $\infty$. This is wrong (how can you truncate terms that are unbounded?). The limit is also wrong anyway: when $x\to+\infty$, $x-x^3/6$ does not tend toward $+\infty$, but $-\infty$. – Jean-Claude Arbaut Apr 19 '15 at 14:46

Such limit is zero, because $x\in\pi\mathbb{Z}$ implies $\int_{0}^{x}\cos y\,dy=0$ and for every $x\in\mathbb{R}$ we have: $$\int_{x}^{x+\pi}|\cos x\,|\,dx \leq 2,\tag{1}$$ hence: $$\left|\frac{1}{x}\int_{0}^{x}\cos y\,dy\right|\leq\frac{2}{x}\to 0.\tag{2}$$

Note that, $\int_0^x \cot (t)dt = \sin(x)$. Now, $\lim_{x \to \infty} \sin(x)/x = 0$. Therefore, $\frac1x \int_0^x \cot (t)dt = 0.$

Note- Sorry for typing it in a hurry, I was a bit busy while I typed the solution so thats why I couldn't type in latex. I'll update it as soon as possible.

• I have latexified your answer for you. n.b. doing so is not an endorsement of its correctness ;) – Emily Apr 19 '15 at 15:04
• Sorry but I was in a hurry. By the way, what is n.b.? – homersimpson Apr 19 '15 at 15:17
• en.wikipedia.org/wiki/Nota_bene – Emily Apr 19 '15 at 19:42