I am working on a question that involves me proving that the above integral converges. It tells me that I can't use the limit comparison test by observing that $|\frac{\sin x}{x}|\leq\frac{1}{x}$ on $[1,\infty)$. However, this doesn't make sense to me as the conditions to use this theorem seem to be satisfied?
For example, we previously concluded that because $\frac{|\sin x|}{x^2}\leq\frac{1}{x^2}$ on $[1,\infty)$, then by the limit comparison test, $\int_1^\infty|\frac{\sin x}{x^2}|dx$ converges, and so by another theorem, $\int_1^\infty\frac{\sin x}{x^2}dx$ converges. I don't see why my problem is different than this.
The conditions to apply the limit comparison test are: Let f and g be integrable on [a, t] for all t>a. If $\int_a^\infty g$ converges, and $0\leq f(x)\leq g(x)$ for all x in $[a,\infty)$, then $\int_a^\infty f$ converges.
Looking at the graphs $|\frac{\sin x}{x}|$ and $\frac{1}{x}$, I see no issue with them being integrable on the interval from 0 to infinity…
The second theorem I am not permitted to use states: Let f be defined on $[a,\infty)$ and assume that $\int_a^\infty |f|$ converges. Then $\int_a^\infty f$ converges and $|\int_a^\infty f|\leq\int_a^\infty |f|$.
$\frac{\sin x}{x}$ is defined on $[1,\infty)$, and the area underneath the graph, aka $\int_a^\infty|\frac{\sin x}{x}| \, dx$, seems to be converging on 0.
Why is it that I can't apply these tests?
I know how to prove that the integral converges because I am given an outline, I am just wondering why I can't prove it using these tests.
Best Answer
Take the improper integral of $\frac{1}{x}$
$$\int_1^\infty \frac{1}{x}\:dx = \lim_{b\to\infty} \ln b \to\infty$$
We are saying then that the integral we want is less than a diverging integral, which isn't saying much of anything at all.