Evaluating first term of Integration by Parts

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.

• Remember that $\sin^2x$ is bounded and hence $\lim_{x\to\infty}\frac{\sin^2x}x=0$. Jun 6 '15 at 12:57
• 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 Jun 6 '15 at 13:01
• 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 $$\lim_{x \to \infty} 1 = \lim_{x \to \infty}x \cdot \lim_{x \to \infty} \frac{1}{x}$$
– user14972
Jun 6 '15 at 15:23
• @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)
– user14972
Jun 6 '15 at 15:25
• 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 Jun 6 '15 at 18:00

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.
• Do you know why $1/x$ tends to $0$? Jun 6 '15 at 13:18
• 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 $\sin^{2}(x)/x$. Jun 6 '15 at 14:10
We may write that, as $x \to +\infty$, $$\left|\frac{\sin^2 x}{x}\right|\leq \frac{\left|\sin^2 x\right|}{x}\leq \frac1{x} \to 0.$$