Divergence of $\int_{0}^{1}\frac{\sin^{2}x}{x^{2}}dx$ Does $\int_{0}^{1}\frac{\sin^{2}x}{x^{2}}dx$ diverge or converge?
Symbolab says it diverges, and I get why, but I don't get the logic behind this because you can clearly see that the graph is bounded and continuous (also $\underset{x\to0^{+}}{\lim}\frac{\sin^{2}x}{x^{2}}=1$). Wolfram gives a definite answer with $Si(2)$, what is $Si(2)$?
 A: You are correct that the integral exists because the function can be continuously extended to the closed interval $[0, 1]$.
Symbolab writes the integral as the difference of two divergent integrals
$$
\int_{0}^{1}\frac{\sin^{2}x}{x^{2}}\, dx = \int_{0}^{1}\frac{1}{x^{2}}\, dx - \int_{0}^{1}\frac{\cos^{2}x}{x^{2}}\, dx
$$
but of course nothing can be concluded from that.
You can compute the integral with integration by parts:
$$
 \int_{0}^{1}\frac{\sin^{2}x}{x^{2}}\, dx = \Bigl[ -\frac 1x \sin^2(x) \Bigr]_{x=0}^{x=1} + \int_0^1 \frac{2 \sin(x)\cos(x)}{x} \, dx \\
= -\sin^2(1) + \int_0^1 \frac{\sin(2x)}{x} \, dx
= -\sin^2(1) + \int_0^2 \frac{\sin(t)}{t} \, dt \\
= \operatorname{Si}(2) - \sin^2(1)
$$
where
$$
\operatorname{Si}(x) = \int_0^x \frac{\sin(t)}{t} \, dt
$$
is the sine integral function.
A: We have $\frac{\sin^2x}{x^2}\le1\forall \ 0\le x\le1$, which implies that the integral of $\frac{\sin^2x}{x^2}$ from $0$ to $1$ is bounded by $1$. 
In response to @Alex 's comment, consider $g(x)=\sin x-x$. We have $g(0)=0$ and $g'(x)\le0\implies \sin x-x\le0\forall x\in\mathbb{R}$ from which the above follows. 
A: Here's another way: 
Since $\lim_{x \to 0^{+}} f(x) = 1$ and $f$ is continuous on $(0,1]$, so $f$ is continuous on $[0,1]$. If you take the derivative:
$$
f'(x) = \sin 2x -2 x f(x)
$$
which is also continuous on $(0,1]$ and 
$$
\lim_{x \to 0^{+}}f'(x) = 0
$$
so $f'(x)$ is continuous on $[0,1]$. This means $f(x)$ is uniformly continuous on $[0,1]$. Therefore $f$ is Riemann integrable:
$$
\int_{[0,1]}f < \infty
$$ 
