1
$\begingroup$

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?

$\endgroup$
5
  • 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
3
$\begingroup$

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).

$\endgroup$
9
  • $\begingroup$ 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
5
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.