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Study $$\int_1^{+\infty}\frac{\cos^2{t}}{t}\,dt$$ I have thought the following:

  1. I consider the fact $\frac{\cos^2{t}}{t}\leq \frac{1}{t}$ this is not useful since $\int_1^{+\infty}\frac{1}{t}\, dt$ is divergent but this is not helpful to affirm that $\int_1^{+\infty}\frac{\cos^2{t}}{t}\,dt$ is convergent.

  2. I want to try to prove that the integral is divergent. To do this I have thought:

$$\int_1^{+\infty}\frac{\cos^2{t}}{t}\,dt=\int_1^{\pi}\frac{\cos^2{t}}{t}\,dt+\sum_{k=1}^{\infty}\int_{k\pi}^{(k+1)\pi} \frac{\cos^2{t}}{t}\,dt$$ From this for the reason why $\frac{\cos^2{t}}{t}\geq 0$ in $[1,\pi]$ then $\int_1^{k\pi}\frac{\cos^2{t}}{t}\,dt\geq 0$ and it is not an improper integral, so: $$\int_1^{\pi}\frac{\cos^2{t}}{t}\,dt+\sum_{k=1}^{\infty}\int_{k\pi}^{(k+1)\pi} \frac{\cos^2{t}}{t}\,dt\geq \sum_{k=1}^{\infty}\int_{k\pi}^{(k+1)\pi} \frac{\cos^2{t}}{(k+1)\pi}\,dt= \sum_{k=1}^{\infty}\frac{\pi}{2}\frac{1}{(k+1)\pi}\,dt$$

Now since $ \frac{1}{k+1}\sim \frac{1}{k}$ then the series is divergent and so the integral is divergent.

$\textbf{Request:}$ I would be so grateful if you can tell me if my work (so not only the conclusion on the fact that the integral is divergent), described at 2), is right or there are some mistakes and in this case how can I correct them? Thanks a lot in advance.

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  • $\begingroup$ Seems correct to me. $\endgroup$
    – user65203
    Apr 27, 2021 at 10:32

2 Answers 2

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You have some nice arguments. However, note that you are saying that $$[1, \infty) \overset?= [1,\pi-\epsilon]\cup[\pi - \epsilon, \pi + \epsilon]\cup[2\pi - \epsilon, 2\pi + \epsilon]\cup\dots = \bigcup_{k=1}^\infty[k\pi - \epsilon, k\pi + \epsilon]$$ while it is quite problematic since $\pi+\epsilon \ne 2\pi - \epsilon$ and therefore the segments $[\pi - \epsilon, \pi + \epsilon]$ and $[2\pi - \epsilon, 2\pi + \epsilon]$ are not actually joint for any $\epsilon$ small.


For this problem, I would use the fact that $$\frac{\cos^2t}{t} = \frac{1}{2t} + \frac{\cos 2t}{2t}$$ now since the first summand is divergent and the second is convergent due to the Dirichlet's test, the overall integral is divergent.

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  • $\begingroup$ Thanks I agree! I have thought to change the interval into $[k\pi, (k+1)\pi]$ where $\cos^2{t}\geq c\geq 0$, what do you think now? is it right? $\endgroup$
    – pawel
    Apr 27, 2021 at 9:35
  • $\begingroup$ @pawel I still doubt though. How do you ensure that $c \not \equiv 0$? $\endgroup$
    – VIVID
    Apr 27, 2021 at 9:59
  • $\begingroup$ Yes you're right I am still failling! Now I hope I have found out the right via! I don't use $c$ since I can't gaurantee that it is $c>0$, but I solve directly the integral of $cos^{2}{x}$. i have edited the question. Sorry for bore...I hope you can help me in revising now my idea! $\endgroup$
    – pawel
    Apr 27, 2021 at 10:17
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Hint:

As $\cos^2t\ge\frac12$ in half of every period, you have a lower bound by integrating the minimum of $\dfrac1t$ in these intervals. You will obtain an equivalent of the Harmonic series.

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  • $\begingroup$ Do you mean that my idea it is right but I can also use as interval $[-\frac{\pi}{4}+k\pi, \frac{\pi}{4}+k\pi]$ since here $cos^2{t}\geq \frac{1}{2}$? $\endgroup$
    – pawel
    Apr 27, 2021 at 10:42
  • $\begingroup$ @pawel: any "rectangle" under the curve, with constant width and height $\propto t^{-1}$ will do. $\endgroup$
    – user65203
    Apr 27, 2021 at 10:43
  • $\begingroup$ mm.. I have not understood sorry... $\endgroup$
    – pawel
    Apr 27, 2021 at 10:45
  • $\begingroup$ $\int_{1}^{\infty} f(t)\,dt\geq \sum_{k=1}^{\infty} \int_{-\pi /4+k\pi}^{\pi/4+k\pi}f(t)dt\geq \sum_{k=1}^{\infty} \frac{1}{2}\int_{-\pi/4+k\pi}^{\pi/4+k\pi}\frac{1}{\pi/4+k\pi}\,dt$ This is what do you mean? $\endgroup$
    – pawel
    Apr 27, 2021 at 10:50
  • $\begingroup$ is the previuos comment correct? $\endgroup$
    – pawel
    Apr 27, 2021 at 12:26

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