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Determine whether the integral $$\int_0^\infty \frac{\sin^2x}{x^2}~\mathrm dx$$ converges.

I know it converges, since in general we can use complex analysis, but I'd like to know if there is a simpler method that doesn't involve complex numbers. But I cannot come up with a function that I could compare the integral with.

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  • $\begingroup$ Hint: Split the single integral into two integrals $\endgroup$ – gammatester Aug 24 '17 at 15:14
  • $\begingroup$ @FlybyNight No need to make it red. $\endgroup$ – Simply Beautiful Art Aug 24 '17 at 15:31
  • $\begingroup$ @FlybyNight He probably means to split it over $(0,1)$ and $(1,\infty)$. $\endgroup$ – Simply Beautiful Art Aug 24 '17 at 15:32
  • $\begingroup$ @SimplyBeautifulArt Good point. $\endgroup$ – Fly by Night Aug 24 '17 at 15:32
  • $\begingroup$ "since in general we can use complex analysis": what justification is that ?? $\endgroup$ – Yves Daoust Nov 1 '17 at 9:35
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Hint:$$x>1\implies0\le\frac{\sin^2(x)}{x^2}\le\frac1{x^2}\\0<x<1\implies1\le\frac{\sin^2(x)}{x^2}\le\frac1{\cos^2(x)}\le\frac1{\cos^2(1)}$$The second of the two coming from the proof of the derivative of $\sin$ using squeeze theorem.

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For $x>1$ $$\frac{\sin^2x}{x^2}\le \frac{1}{x^2}$$ and For $x<1$ $$ |\sin x|\le |x|\implies \frac{\sin^2 x}{x^2}\le 1.$$

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Using integration by part $$\int_0^\infty \frac{\sin^2(x)}{x^2}dx=\\ -\frac{1}{x}\sin^2(x) |^b_a-\int_{0}^{\infty}-\frac{1}{x}2\sin(x)\cos(x)dx$$note that $\lim_{x\to 0}-\frac{1}{x}\sin^2(x)\to 0$and also $\lim_{x\to \infty}-\frac{1}{x}\sin^2(x)\to 0$ so $$-\frac{1}{x}\sin^2(x) |^b_a-\int_{0}^{\infty}-\frac{1}{x}2\sin(x)\cos(x)dx=\\0-(-)\int_{0}^{\infty}\frac{1}{x}2\sin(x)\cos(x)dx=\\ \int_{0}^{\infty}\frac{\sin(2x)}{x}dx=\\\frac{\pi}{2}$$ It is well known $\int_{0}^{\infty}\frac{\sin(ax)}{x}dx=\frac{\pi}{2}$

$\bf{Remark}$: let:

$$I(t)=\int_{0}^{\infty} \frac{\sin x}{x} e^{-tx} dx$$

Note:

$$\frac{\partial}{\partial t} \frac{\sin x}{x} e^{-tx}=\frac{\sin x}{x} e^{-tx}(-x)$$

So by differentiation of parameter,we have

$$I'(t)=-\int_{0}^{\infty} e^{-tx} \sin x dx$$

And through integration by parts twice we have:

$$I'(t)=-\frac{1}{t^2+1}$$

Hence,

$$I(t)=\int -\frac{1}{t^2+1} dt$$

$$I(t)=-\arctan (t) +c$$

when $t \to \infty$, $I(t) \to 0$ so :

$$I(t)=\frac{\pi}{2}-\arctan t$$

Let $t \to 0^+$:

$$\int_{0}^{\infty} \frac{\sin x}{x} dx=\frac{\pi}{2}$$and general form is $\int_{0}^{\infty} \frac{\sin (ax)}{x} dx=\frac{\pi}{2}$

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  • $\begingroup$ It's not a problem here, but $\int_0^{\infty} \frac{\sin(ax)}{x}~dx=\frac{\pi}{2}$ is only true if $a>0$. $\endgroup$ – projectilemotion Aug 25 '17 at 14:38
  • $\begingroup$ It is not completely satisfactory to resort to arguments like "it is well known that..." and cite an equally sophisticated result, because then it is much shorter to say "it is well known that this integral converges". $\endgroup$ – Yves Daoust Nov 1 '17 at 9:38

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