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My attempt: $$\int_{1}^{\infty}\frac{\cos(x) dx}{1+x^2} \le \int_{1}^{\infty} \frac{dx}{1+x^2}$$ and the latter integral converges, so $\int_{1}^{\infty}\frac{\cos(x) dx}{1+x^2}$ also converges. And similar argument follows for $\int_{1}^{\infty}\frac{\sin(x) dx}{1+x^2}$.

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Is this correct ? Then, is it true that all integrals bounded from above (strictly or non-strictly) by a finite value converge?

Sine of the Time
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Physicsa
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    Your approach is not correct. You have to take absolute value of $\cos x$ and $\sin x.$ – Ryszard Szwarc Dec 08 '23 at 16:56
  • @R.J.Mathar integral in the link is different due a a different lower limit – Physicsa Dec 08 '23 at 17:05
  • You can use the Dirichlet's test for this problem. https://math.stackexchange.com/questions/141048/dirichlets-test-for-convergence-of-improper-integrals#:~:text=Theorem.,af%CF%95%20is%20convergent. –  Dec 08 '23 at 17:06

1 Answers1

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As mentioned in the comments, your solution is wrong. Comparison tests work only for non-negative (or similarly non positive) functions. You cannot use this for functions like $\sin$ or $\cos$. However, you can say that $\frac{|\cos(x)|}{1+x^2}\leq\frac{1}{1+x^2}$, and so by the comparison test $\int_1^{\infty}\frac{|\cos(x)|}{1+x^2}dx<\infty$. Thus the integral $\int_1^{\infty}\frac{\cos(x)}{1+x^2}dx$ converges absolutely, and so converges.

You can also prove the convergence of the original integral directly if you are familiar with Dirichlet's test. But that solution would be kind of a waste, as proving absolute convergence is proving a stronger statement. (and it's also easier)

Mark
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