Solution verification of the improper integral $\int_1^{+\infty}\frac{\cos^2{t}}{t}\,dt$

Question

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.

Answer 1

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.

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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.

Answer 2

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.