Asymptotic Estimation of a function ${\int_{0}^{x} \frac{\sin^2 t}{t^2} dt}$

Question

I'm reading a book "Theory of Approximation of Functions of a Real Variable" by A.F Timan

On the page 14 there is an asymptotic estimation:"... It is an integral function of the first degree, bounded on the real axis and monotonically increasing. Hence $1 - |g(x)| = O(\frac{1}{1+|x|})$, where $g(x) = {\frac{4}{\pi}\int_{0}^{x} \frac{\sin^2 \frac{t}{2}}{t^2} dt}$ "

I don't know exactly how to get this estimation.

I think about $\int \frac{\sin^2 \frac{t}{2}}{x^2} dx=-\frac 12\int \frac { \cos(t)-1}{t^2} dt$, then integrate by parts but it doesn't help.

Answer 1

When $x\geq 1$ we have $$ 1-g(x)=\frac{4}{\pi}\int_x^{+\infty}\frac{\left(\sin(t/2)\right)^2}{t^2}dt\leq \frac{4}{\pi}\int_x^{+\infty}\frac{dt}{t^2}=O(1/x)=O(1/(1+|x|)). $$ For $|x|\leq 1$ both $1/(1+|x|)$ and $1-g(x)$ are bounded continuous functions so the statement is true, then for $x\leq -1$ use that $g(-x)=-g(x)$.