# Investigating improper integral convergence

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 $$a\in \mathbb{R}$$.

• 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 $\lim_{x\rightarrow \infty}\frac{1-\cos(1/x)}{1/x^2}$. May 1, 2022 at 15:06
• 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 May 1, 2022 at 15:10
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 $$x\ge 1$$.
• 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 :( May 1, 2022 at 15:38
• @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 $x\ge 0$. May 1, 2022 at 15:46
• 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? May 1, 2022 at 15:51