# Improper integral over an open interval using the comparison test

I'm getting confused with improper integral over a closed interval it seems to me like the opposite of improper integrals when x tends to infinity.

Example: $\int_0^1 \frac{sin^2x}{x^2}$

I can see that the integral is an improper integral when $x \rightarrow 0^+$ because the function tends to infinity. Usually what I would have done is using the comparison test and take the function $\frac{1}{x^2}$ which its improper integral is convergent and $\frac{sin^2x}{x^2} <= \frac{1}{x^2}$ and for that reason $\int_0^1 \frac{sin^2x}{x^2}$ should be convergent too but it doesn't work like that as I said it seems to me like the opposite.

can someone clarify to me please what is going on? I think i'm lacking understanding of it..

• $\lim_{x\to 0} f(x)$ is $1$, so actually at $0$ the integral is proper. – R. Singh Apr 28 '17 at 13:02
• Are you sure? pay attention it's not $\frac{sinx^2}{x^2}$ – user21312 Apr 28 '17 at 13:04
• $\lim_{x\to 0} f(x)$ = $\lim_{x\to 0}\frac{sin^2x}{x^2}$= $1$, because $\lim_{x\to 0}\frac{sinx}{x}$ is $1$. You can always write $\lim_{x\to 0}\frac{sin^2x}{x^2}$=$\{\lim_{x\to 0}\frac{sinx}{x}\}^2$ – R. Singh Apr 28 '17 at 13:08

The is nothing improper in the integral $\int_{0}^{1}\frac{\sin x}{x}\,dx$, since the function $f(x)=\frac{\sin x}{x}$ has a removable discontinuity at $x=0$, $\lim_{x\to 0}\frac{\sin x}{x}\,dx=1$.
Additionally, $\frac{\sin x}{x}$ is an entire function, hence
$$\int_{0}^{1}\frac{\sin x}{x}\,dx = \int_{0}^{1}\sum_{n\geq 0}\frac{(-1)^n x^{2n}}{(2n+1)!}\,dx = \sum_{n\geq 0}\frac{(-1)^n}{(2n+1)(2n+1)!}$$ the integral is given by a fast-convergent series with alternating signs.
Essentially the same argument applies to $\left(\frac{\sin x}{x}\right)^2$:
$$\int_{0}^{1}\left(\frac{\sin x}{x}\right)^2\,dx \stackrel{IBP}{=}-\sin^2(1)+\int_{0}^{1}\frac{\sin(2x)}{x}\,dx.$$