Let $C=\int_{-1}^{1}[\sin 1 / x]^{2} d x$. Since the integrand is bounded and continuous on [-1,1] except at the point $x=0,$ the integral exists. Moreover, since the integrand is never negative, $C>0 .$ Put $u=1 / x$ so that $f(x)$ becomes $(\sin u)^{2}$ and $d x=-u^{-2} d u$. When $x=1, u=1,$ and when $x=-1, u=-1$ Thus, $$ C=\int_{-1}^{1}(\sin u)^{2}\left(-u^{-2}\right) d u=-\int_{-1}^{1}\left[\frac{\sin u}{u}\right]^{2} d u $$ The integrand in this integral is continuous on [-1,1] since the discontinuity at $u=0$ is removable; thus, the new integral exists, and is negative!
Is it because we are supposed to integrate over $[-1,1]$, but after performing the change in variables, the interval for $u$ should be $(-\infty,-1] \cup [1,\infty)$, since $u=1/x$, not $[-1,1]$, i.e. an improper integral?
Source:
Buck, Advanced Calculus, pg 183-184
Yes. I think their point is a simple one. You cannot simply plug in the lower and upper limits of integration into $u=1/x$ and put them as the new limits. The new integral is
$$\lim_{a\rightarrow-\infty}\int_a^{-1}\left(\frac{\sin u}{u}\right)^2du+\lim_{b\rightarrow\infty}\int_1^{b}\left(\frac{\sin u}{u}\right)^2du$$
because $x\in[-1,0)$ means $u\in(-\infty,-1]$ and $x\in(0,\infty]$ means $u\in(1, \infty)$.