A calculus problem by Aly Ahmed

$\begin{aligned} I & = \int_{-\infty}^\infty \frac {\sin 2020x}x dx & \small \blue{\text{Since the integrand is even,}} \\ & = 2\int_0^\infty \frac {\sin (2020x)} x dx & \small \blue{\text{Let }u = 2020 x \implies du = 2020 \ dx} \\ & = 2 \int_0^\infty \frac {\sin u}u du \\ & = 2 \text{ Si }(\infty) & \small \blue{\text{where Si }(\cdot) \text{ denotes the sine integral.}} \\ & = 2 \cdot \frac \pi 2 = \pi \approx \boxed{3.14} \end{aligned}$

Since, $\int_{-\infty}^\infty \frac{\sin(2020x)}{x}dx=\int_{-\infty}^{\infty} \frac{\sin(x)}{x}dx$ Then splitting the integral, $\int_{-\infty}^0 \frac{\sin(x)}{x}dx+\int_0^\infty\frac{\sin(x)}{x}dx$ The positive integral is simply $\frac{\pi}{2}$ and the negative one is also $\frac{\pi}{2}$ ( you can prove that by substituting $x=-t$ which would flip the integral), $\frac{\pi}{2}+\frac{\pi}{2}=\pi$ As our result.