Integral $\int_0^1 \frac{\sin^2 \pi x}{x} dx.$

Question

I would like to calculate the following definite integral with trigonometric function.

$$\int_0^1 \frac{\sin^2 \pi x}{x} dx.$$

I am seeking analytical solution.

Answer 1

If you do not want to face the special function (sine and cosine integrals), a quite good approximation could already have been obtained $1,400$ years ago using $$\sin(t) \simeq \frac{16 (\pi -t) t}{5 \pi ^2-4 (\pi -t) t}\qquad (0\leq t\leq\pi)$$ proposed by Mahabhaskariya of Bhaskara I, a seventh-century Indian mathematician.

It would give $$\int_0^1 \frac{\sin^2 (\pi x)}{x} dx\simeq\int_0^1 \frac{256 (x-1)^2 x}{\left(4 x^2-4 x+5\right)^2}\,dx$$

Using partail fraction decomposition, this would give $$\int\frac{256 (x-1)^2 x}{\left(4 x^2-4 x+5\right)^2}\,dx=\frac{20 x+30}{4 x^2-4 x+5}+8 \log \left(4 x^2-4 x+5\right)+3 \tan ^{-1}\left(\frac{1}{2}-x\right)$$ and, using the bounds, the nice $$\int_0^1 \frac{\sin^2 (\pi x)}{x} dx\simeq 4-6 \tan ^{-1}\left(\frac{1}{2}\right)\approx 1.21811$$ while, as given by Wolfram Alpha $$\int_0^1 \frac{\sin^2 (\pi x)}{x} dx=\frac{1}{2} (-\text{Ci}(2 \pi )+\gamma +\log (2 \pi ))\approx 1.21883$$ So, this so old and beautiful approximation leads to a result which is in relative error of $0.058$%.

A few years ago (have a look here), I worked an approximation $$\sin(t)=\sum_{i=1}^n a_i \Big((\pi-t)t\Big)^i$$ Using the coefficients given in the linked page, the result would be the monstrous $$\frac{80141757696000}{\pi ^{14}}-\frac{18083268403200}{\pi ^{12}}+\frac{1207036131840}{\pi ^{10}}-\frac{21105342720}{\pi ^8}+\frac{110355840}{\pi ^6}$$ which is $1.21882860$ while the exact value is $1.21882670$.