Finding the value of $\int_{0}^{1} \frac{\sin^2 x}{x^2}dx$

Question

I would like to find the exact value of $$\int_{0}^{1} \frac{\sin^2 x}{x^2}dx.$$

First of all we know that it exists and must be $\hspace{0.1cm}$$\leq1$$\hspace{0.1cm}$ because$\hspace{0.1cm}$ $\frac{\sin^2 x}{x^2}\leq1$.

But how would we go about finding the exact value?

Thanks.

Answer 1

Integrate by parts $$\begin{aligned}\int_0^1 \frac{\sin^2 x}{x^2} \ \mathrm{d}x&= \lim_{x\to a}\left[\frac{-\sin^2 x}{x}\right]_{x=a}^1 +\int_0^1\frac{\sin(2x)}{x} \ \mathrm{d}x\\ & \overset{t=2x}{=} -\sin^2(1) + \int_0^2 \frac{\sin t}{t}\ \mathrm{d}t\\ &= \text{Si}(2)-\sin^2(1)\approx 0.89734\end{aligned}$$ Where $\text{Si}$ is the sine integral function. I doubt the existence of a nice closed form.