8
$\begingroup$

I have this problem

$\displaystyle \int_{0}^{\infty}\frac{\sin^{2}x}{x}\ dx$

I think it diverges, but I cannot prove it. (I'm in my first calculus course, so I don't know of any more advanced methods than

  • u-substitution (including trig subs)
  • integration by parts
$\endgroup$

3 Answers 3

4
$\begingroup$

Consider the periodic function $f:\Bbb{R}\to\Bbb{R}$ with period $2\pi$ defined as $$ f(x)=\begin{cases} 1/2 & \text{if }\quad \pi/4\le x\le 3\pi/4\\ 0 & \text{if }\quad 0 \le x<\pi/4\quad \text{or}\quad 3\pi/4<x\le 2\pi \end{cases} $$ then $\sin^2 x\ge f(x)$ so we get $$\int_0^\infty \frac{\sin^2x}{x}dx\ge\int_0^\infty \frac{f(x)}{x}dx=\sum_{n=0}^\infty\int_{(1/4+n)\pi}^{(3/4+n)\pi} \frac{dx}{2x}$$

and $$\sum_{n=0}^\infty\int_{(1/4+n)\pi}^{(3/4+n)\pi} \frac{dx}{2x}=\frac{1}{2}\sum_{n=0}^\infty\ln\frac{3+4n}{1+4n}=\frac{1}{2}\sum_{n=0}^\infty\ln\left( 1+\frac{2}{4n+1}\right).$$ You can easily check that the series of the right side diverges.

$\endgroup$
2
$\begingroup$

Since $(\sin^2 x)/x$ is non-negative on $(0, \frac{\pi}{2})$, $$ \int_{\frac{\pi}{2}}^\infty \frac{\sin^2 x}{x}\, dx \leq \int_0^\infty \frac{\sin^2 x}{x}\, dx. $$ Further, $\cos^2(x + \frac{\pi}{2}) = \sin^2 x$ for all real $x$, so substitution ($x = u + \frac{\pi}{2}$) gives $$ \int_{\frac{\pi}{2}}^\infty \frac{\cos^2 x}{x}\, dx = \int_0^\infty \frac{\sin^2 x}{x + \frac{\pi}{2}}\, dx \leq \int_0^\infty \frac{\sin^2 x}{x}\, dx. $$ Adding these inequalities, $$ \int_{\frac{\pi}{2}}^\infty \frac{dx}{x} = \int_{\frac{\pi}{2}}^\infty \frac{\cos^2 x}{x}\, dx + \int_{\frac{\pi}{2}}^\infty \frac{\sin^2 x}{x}\, dx \leq 2\int_0^\infty \frac{\sin^2 x}{x}\, dx, $$ and the lower bound is clearly divergent.

$\endgroup$
1
$\begingroup$

It diverges. Notice that $\sin^2(\frac\pi2+k\pi)=1$. By continuity you can find $\delta>0$ such that $\sin^2x >\frac12$ for $|x-(\frac\pi2 + k\pi)| < \delta$. Then you can write this integral as $$\int_{0}^{\infty}\frac{\sin^{2}x}{x}\ dx = \sum_{n=0}^\infty\int_{n\pi}^{(n+1)\pi}\frac{\sin^{2}x}{x}dx \geq \sum_{n=0}^\infty\int_{n\pi+\frac\pi2-\delta}^{n\pi+\frac\pi2+\delta}\frac{\sin^{2}x}{x}dx$$ now we can see that $$\int_{n\pi+\frac\pi2-\delta}^{n\pi+\frac\pi2+\delta}\frac{\sin^{2}x}{x}dx \geq \int_{n\pi+\frac\pi2-\delta}^{n\pi+\frac\pi2+\delta}\frac{1}{2x}dx \geq \frac{\delta}{2n\pi + \pi + 2\delta}$$ Series $$\sum_{n=0}^\infty\frac{\delta}{2n\pi + \pi + 2\delta}$$ is divergent, so $\int_{0}^{\infty}\frac{\sin^{2}x}{x}\ dx$ also diverges

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .