Consider the problem where the following integral converges or not:

$$\int_{1}^{\infty} \frac{\sin(x+2)}{x^2} \,dx $$

I tried to solve it in two different ways but the results conflict. I am not sure why.

First Solution:

Using comparison criterion we can prove that it converges because

$$ \frac {\sin(x+2)}{x^2} \leq \frac {1}{x^2} $$

and $$\int_{1}^{\infty} \frac{1}{x^2} dx < + \infty$$ converges as a p-intergral with $p=2 > 1$

Second Solution $$ \frac {\sin(x+2)}{x^2} \leq \frac {x+2}{x^2} = \frac {1}{x} + \frac {2}{x^2} $$

Where this converges $$\int_{1}^{\infty} \frac{2}{x^2} dx$$ but this diverges $$ \int_{1}^{\infty} \frac{1}{x} dx$$

Thus the initial integral also diverges because one part of its sum diverges

The solutions conflict and I know that something is wrong with the second solution. But I cannot spot what went wrong. Any ideas?

  • 3
    $\begingroup$ They do not conflict, the upper bound in your second solution is not sharp enough to be an integrable function. $\endgroup$ – nicomezi Mar 4 '20 at 10:31
  • $\begingroup$ @nicomezi What do you mean "sharp enough" ? $\endgroup$ – Dimitris Mar 4 '20 at 10:43
  • 1
    $\begingroup$ I could say that $0 \le \frac 1 x$, and, as you said, $\frac 1 x$ is not integrable. Yet this does not imply that the zero function is not integrable. (and this is basically the reasonning you have done in the second solution). $\endgroup$ – nicomezi Mar 4 '20 at 10:46
  • $\begingroup$ True, thank you! $\endgroup$ – Dimitris Mar 4 '20 at 10:52
  • $\begingroup$ It is not proper to keep editing the question long after answers appear. $\endgroup$ – Kavi Rama Murthy Mar 4 '20 at 11:41

The problem is integrabilty near $0$. Since $\frac {\sin x} x \to 1$ and $ \int_0^{1} \frac 1 x dx$ does not converge, the given integral does not converge.

It is also not true that $\int_0^{\infty} \frac 1 {x^{2}} d x<\infty$.

Answer for the edited version: The integral is convergent because $-\frac 1 {x^{2}} \leq \frac {\sin x} {x^{2}} \leq \frac 1 {x^{2}}$.

[I do not understand why you are considering $\sin (x+2)$].

(The question was edited again after I posted this answer).

  • $\begingroup$ What if the edges of the intergral where $$\int_{1}^{\infty} \frac{sin(x)}{x^2} dx $$ ? I edited my question @Kavi $\endgroup$ – Dimitris Mar 4 '20 at 10:28
  • $\begingroup$ I have edited my answer too! $\endgroup$ – Kavi Rama Murthy Mar 4 '20 at 10:31
  • $\begingroup$ I did a typo again, I am sorry. The initial integral meant to be with x + 2 $$\int_{0}^{\infty} \frac{sin(x+2)}{x^2} dx $$ @Kavi Rama Murthy $\endgroup$ – Dimitris Mar 4 '20 at 10:38
  • $\begingroup$ @nicomezi fixed that, thank you $\endgroup$ – Dimitris Mar 4 '20 at 10:40
  • $\begingroup$ This doesn't say why the second solution is wrong, does it ? $\endgroup$ – Yves Daoust Mar 4 '20 at 11:06

You are writing

$$\int f(x)dx\le \int g(x)dx$$

and conclude that if $$ \int g(x)dx$$ diverges, so does $$\int f(x)dx.$$

This is wrong.


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