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I'm trying to investigate the convergence of the following:

$$ \int_{1}^{\infty} (1-cos(\dfrac{1}{x})) \,dx $$

Initially, its easy to see that the limit of $cos(\dfrac{1}{x})$ when $x\rightarrow \infty$ is $1$, therefor the the $(1-cos(\dfrac{1}{x}))$ goes to $0$ as $x\rightarrow \infty$, thus allowing me to deduce that the improper integral indeed converges.

Yet I'm trying to prove this using the comparative/dirichlet/absolute convergence methods.

I've tried playing around with trigonometric identities, substituting $1- cos(\dfrac{1}{x})$ with $2sin^2(\dfrac{1}{2x}))$ yet that didn't get me anywhere.

Any assistance would be indeed helpful.

On a different note- I'm having similar issues with $\int_{0}^{\infty}\dfrac{e^{2x}}{1+x^2}dx $, I've proved that $\int_{-\infty}^{1}\dfrac{e^{2x}}{1+x^2}dx $ converges but then our class lecturer decided that you cant use comparison tests with integrals in form of $\int_{-\infty}^{a}$ where $a\in \mathbb{R}$.

Thanks in advance!

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  • $\begingroup$ You cannot say $\int_1^{\infty}\left(1-\cos(1/x)\right)\mathrm{d}x$ converges simply because $1-\cos(1/x)$ approaches $0$ as $x\rightarrow +\infty$. Notice $1/x\rightarrow 0$ as $x \rightarrow \infty$ too but $\int_1^{\infty}\frac{\mathrm{d}x}{x}$ diverges. To show your improper integral converges, consider $\lim_{x\rightarrow \infty}\frac{1-\cos(1/x)}{1/x^2}$. $\endgroup$
    – Matthew H.
    May 1, 2022 at 15:06
  • $\begingroup$ can you elaborate on how you made the connection between what I'm searching for and $\dfrac{1-cos(\dfrac{1}{x})}{\dfrac{1}{x^2}}$? I didnt say it's converging according to what I saw, the integrand slowly approaches $0$ therefor the area under the graph of $1-cos(\dfrac{1}{x})$ converges, not an official proof tho, only direction of thought $\endgroup$
    – Aishgadol
    May 1, 2022 at 15:10

1 Answer 1

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Hint: Using the inequality $\sin(t)\le t$ for $t\ge 0$, we have $$0\le 2\sin^2\left(\frac{1}{2x}\right)\le \frac{1}{2x^2} $$ for $x\ge 1$.

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  • $\begingroup$ Exactly what I was looking for, thank you! Regarding the second part of my question, I'm rather stuck. trying to bound it by $\dfrac{1}{1+x^2}$ didn't work so well. Kind of at a thought standstill about this :( $\endgroup$
    – Aishgadol
    May 1, 2022 at 15:38
  • $\begingroup$ @Aishgadol Note that the policy on MSE is to ask a single question per post. That said, the second integral is clearly divergent because $\lim_{x\to\infty}e^{2x}/(1+x^2)=\infty$ by the properties of the exponential function. Or you can show that $e^{2x}/(1+x^2)\ge 1$ for $x\ge 0$. $\endgroup$
    – bjorn93
    May 1, 2022 at 15:46
  • $\begingroup$ I see, showing that $e^{2x}\ge (1+x^2)$ for $x\ge 0$ is sufficient in order to claim that the integral is divergent? $\endgroup$
    – Aishgadol
    May 1, 2022 at 15:51

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