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Solve integral $$\int_0^\infty\frac{\sin^2x}{x^2}dx$$ initially used Integration by Parts: $$\begin{align} \int_0^\infty\frac{\sin^2x}{x^2}\,dx&=-\left.\frac{\sin^2x}{x}\right|_0^\infty+\int_0^\infty\frac{2\sin x\cos x}{x}\,dx\\ \end{align}$$ Please show how to evaluate $\frac{\sin^2x}{x}|_0^\infty $ in $$\int_0^\infty\frac{\sin^2x}{x^2}\,dx=-\left.\frac{\sin^2(x)}{x}\right|_0^\infty+\int_0^\infty\frac{2\sin x\cos x}{x}dx$$ Was told could not evaluate $\frac{\sin^2x}{x}|_0^\infty $ at infinity.

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    $\begingroup$ Remember that $\sin^2x$ is bounded and hence $\lim_{x\to\infty}\frac{\sin^2x}x=0$. $\endgroup$
    – xpaul
    Jun 6 '15 at 12:57
  • $\begingroup$ Because the latter term is bounded, $\displaystyle -1 \times lim_{x\to \infty}\dfrac{-\sin(x)}{x} \leq lim_{x\to \infty}\dfrac{-\sin(x)}{x}\times lim_{x\to \infty} \sin(x) \geq 1 \times lim_{x\to \infty}\dfrac{-\sin(x)}{x}$, so it's zero $\endgroup$ Jun 6 '15 at 13:01
  • $\begingroup$ To really drive home that the teacher's argument is in error, it would also imply that $\lim_{x \to \infty} 1$ doesn't exist, because $$\lim_{x \to \infty} 1 = \lim_{x \to \infty}x \cdot \lim_{x \to \infty} \frac{1}{x} $$ $\endgroup$
    – user14972
    Jun 6 '15 at 15:23
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    $\begingroup$ @AnalysisStudent: A good argument can be made along those lines, but not as you've written it. $\lim_{x \to \infty} \sin(x)$ doesn't exist, so any formulas involving it arenonsensical. (although you can change the meaning of the notation to be the multivalued operation that returns the set of limit points) $\endgroup$
    – user14972
    Jun 6 '15 at 15:25
  • $\begingroup$ Of course it's not correct, but he asked how could his teacher's (wrong) notation lead to the correct result, and I showed him how $\endgroup$ Jun 6 '15 at 18:00
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Your teacher said that?! They are very incorrect.
First, the limit: $$0\le\frac{\sin^{2}(x)}{x}\le\frac{1}{x}$$ So the limit is forced to go to $0$.

You CANNOT split up a limit of a product into the product of 2 limits, unless both limits exist. Since $\lim_{x \to \infty}\sin(x)$ doesn't exist, you can't split up this limit.

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  • $\begingroup$ Do you know why $1/x$ tends to $0$? $\endgroup$ Jun 6 '15 at 13:18
  • $\begingroup$ So by choosing $x$ large enough, we can force $0<\frac{1}{x}<\epsilon$ for any $\epsilon$ we choose in advance. This inequality shows we can do the same for $\sin^{2}(x)/x$. $\endgroup$ Jun 6 '15 at 14:10
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We may write that, as $x \to +\infty$, $$ \left|\frac{\sin^2 x}{x}\right|\leq \frac{\left|\sin^2 x\right|}{x}\leq \frac1{x} \to 0. $$

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