2
$\begingroup$

The question asks whether the following converges or diverges.

$$ \int_{0}^{\infty } {\left\vert\,\sin\left(\,x\,\right)\,\right\vert \over x^2}\,{\rm d}x $$

Now I think there might be a trick with the domain of sine function but I couldn't make up my mind on this.

I tried to compare it with $1/x^{2}$, $\sin\left(\,x\,\right)/x$, and $\sin\left(\,x\,\right)$.

I actually expected that something good would come from $1/x^{2}$, but as the lower limit of the integral is zero, it ended up with infinity on $\left(\,0,\infty\,\right)$.

Since $1/x^{2}$ is greater than the given function, and is divergent on the given interval, it doesn't help at all.

So I'm wondering what is the right track on this problem ?.

$\endgroup$
  • 1
    $\begingroup$ Hint: near $0$, $\sin x\approx x$. More precisely, $\lim_{x\rightarrow0}{\sin x\over x}=1$. $\endgroup$ – David Mitra Jan 3 '15 at 16:12
  • $\begingroup$ @DavidMitra How should I use that with this question? $\endgroup$ – user2694307 Jan 3 '15 at 16:20
  • $\begingroup$ Or perhaps more simply, for $x$ near $0$, $\frac{\sin x}{x} > \frac{1}{2}$. $\endgroup$ – John Hughes Jan 3 '15 at 16:21
  • $\begingroup$ As john suggests... $\endgroup$ – David Mitra Jan 3 '15 at 16:23
2
$\begingroup$

$\frac{x}{2}<|sin(x)|$ on the interval of $(0,\pi /2)$. Therefore $\frac{|sin(x)|}{x^2} >\frac{\frac{x}{2}}{x^2}=\frac{1}{2x}$.

since $\int_0^{\pi /2} \frac{1}{2x}$ doesn't converge to a positive real number, $\int_0^{\pi /2} \frac{|sin(x)|}{x^2} $ won't converge to a real number.

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ In the question the upper limit goes to infinity, but it is not a problem as the small part of the function is already divergent, is it? $\endgroup$ – user2694307 Jan 3 '15 at 16:46
  • 1
    $\begingroup$ Exactly. The rest is convergent, since $|sin(x)|<1$ and $\int_{\pi/2} ^{\infty} 1/x^2$ is convergent. $\endgroup$ – questioner Jan 3 '15 at 17:11
2
$\begingroup$

Hint. A potential problem is near $0$, recall that, by the Taylor expansion near $0$, you have $$ \sin x =x+\mathcal{O}(x^3) $$ hence $$ \frac{\sin |x|}{x^2} =\frac{1}{x}+\mathcal{O}(x),\quad x \,\, \text{near} \, 0^+, $$ and your integral is divergent as is $\displaystyle \int_0^a \frac{1}{x} dx$, $a>0$.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ It looks simple and clean, thanks for that, but we haven't seen series yet, so I basically don't know how you wrote that sin(x) equality at the first place. $\endgroup$ – user2694307 Jan 3 '15 at 16:34
  • 1
    $\begingroup$ @user2694307 It is not about 'series', but just the use of the Taylor expansion:$$ f(x)=f(0)+xf'(0)+\frac{x^2}{2!}f''(0)+O(x^3)$$ for any $f$ sufficiently smooth near $0$. Here you take $f=\sin$. $\endgroup$ – Olivier Oloa Jan 3 '15 at 16:36

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.