Divergence of $\int_{0}^{1}\frac{\sin^{2}x}{x^{2}}dx$

Question

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 $Si(2)$?

Answer 1

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.

Answer 2

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.

Answer 3

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: $$ \int_{[0,1]}f < \infty $$