# Improper integral comparison test $\frac{\sin(x)}{x^2}$

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 ?.

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

$\frac{x}{2}<|sin(x)|$ 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.
• Exactly. The rest is convergent, since $|sin(x)|<1$ and $\int_{\pi/2} ^{\infty} 1/x^2$ is convergent. – questioner Jan 3 '15 at 17:11
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 $\displaystyle \int_0^a \frac{1}{x} dx$, $a>0$.
• @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 $f=\sin$. – Olivier Oloa Jan 3 '15 at 16:36