I'm using the following criteria (now I wonder if it's accurate): "if $f(x)\geq 0,\quad \int_{0}^{+\infty}f(x) dx \; \text{ converges } \iff \forall \epsilon \; \exists x_\epsilon \; \forall A,B \;\; B \geq A \geq x_\epsilon \;\; |\int_{A}^{B}f(x) dx|<\epsilon.$"

$$\int_{A}^{B}\frac{|\mathrm{sin}(x)|}{x} dx\leq\int_{A}^{B}\frac{{1}}{x} dx=\mathrm{log}(x)\Big|^B_A=\mathrm{log}\Bigl(\frac{B}{A}\Bigr)\to 0, \; (A\to+\infty),$$

so the integral converges. Can somebody spot the horrible blunder that I may have made? Thanks for helping me.

  • $\begingroup$ $\displaystyle\lim_{B\to \infty,A\to 0+}\ln\bigg(\frac{B}{A}\bigg)=??$ $\endgroup$ – Sumanta Sep 8 '20 at 12:16
  • $\begingroup$ I don't understand, A goes to infinity $\endgroup$ – stew.ita Sep 8 '20 at 12:18
  • $\begingroup$ Keeping $B$ fix let $A\to 0+\implies \frac{B}{A}\to \infty\implies \ln\big(\frac{B}{A}\big)\to \infty$. $\endgroup$ – Sumanta Sep 8 '20 at 12:20
  • $\begingroup$ Similarly, keeping $A$ fix and letting $B\to \infty$ we have $\ln\big(\frac{B}{A}\big)\to \infty$. $\endgroup$ – Sumanta Sep 8 '20 at 12:21
  • $\begingroup$ $A$ just needs to be bigger than a certain $x_\epsilon$, so it doesn't mind if there are problems close to 0 $\endgroup$ – stew.ita Sep 8 '20 at 12:23

There's a couple of things here wrong with the argument. First of all,

$${\log\left(\frac{B}{A}\right) \rightarrow -\infty \text{ as } A\rightarrow \infty}$$

Also, you maybe misunderstanding the criteria. What it's basically saying is that the integral converges iff you can find some point ${x_{\epsilon}}$ that makes the "remaining area" after that point as small as you want. So it's basically saying:

$${\lim_{s\rightarrow \infty}\int_{s}^{\infty}f(x)dx=0}$$

With ${\frac{1}{x}}$, this doesn't happen. If you take the upper bound ($B$) large enough in comparison to the lower bound (which you can do since the criteria says ${\forall A,B}$ such that ${B\geq A}$) - the area will (very slowly, just like the harmonic series) get larger. So this argument doesn't work.

  • $\begingroup$ Ok, got it. First part is wrong, because B is always larger than A, so it can't tend to $-\infty$. But then you're right; if I fix an A there's always a B large enough. Thank you :) $\endgroup$ – stew.ita Sep 8 '20 at 13:14
  • $\begingroup$ @stew.ita no problem. And yeah that is true, but without also stating the condition ${B>A}$ that limit alone is ${-\infty}$ :) $\endgroup$ – Riemann'sPointyNose Sep 8 '20 at 13:39

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