# Integrating $\operatorname{sinc}(x)$ over an area

I was recently asked the following question in an exam and was unsure how to approach it:

Evaluate the integral $$\int_{0}^{\pi}\int_{y}^{\pi}\frac{\sin(x)}{x}\:\mathrm{d}x\:\mathrm{d}y$$

I recognised that this is the double integral of $\operatorname{sinc}(x)$ and that if we substitute $x=\pi\theta$, then we get:

$$\int_{0}^{\pi}\int_{\frac{y}{\pi}}^{1}\operatorname{Sinc}(\theta)\:\mathrm{d}\theta\:dy=\pi\int_{0}^{1}\int_{y}^{1}\operatorname{Sinc}(\theta)\:\mathrm{d}\theta\:\mathrm{d}y=\pi\int_{0}^{1}\operatorname{Si}(y)\:\mathrm{d}y$$

However, I'm not sure how to evaluate this, Mathematica gives the answer as $2$, but we haven't done any integrals like this before and we haven't covered special functions or $\operatorname{sinc}(x)$ in lectures.

Is there a simple way to evaluate this that I'm missing?

\begin{align} \int_0^\pi \int_y^\pi \frac{\sin x}{x}\,dx\,dy &= \int_0^\pi \int_0^x \frac{\sin x}{x}\,dy\,dx\\ &= \int_0^\pi x\frac{\sin x}{x}\,dx\\ &= \int_0^\pi \sin x\,dx\\ &= 2. \end{align}
$$I=\int_{0}^{\pi}\int_{y}^{\pi}\frac{\sin{x}}{x}\mathrm{d}x\mathrm{d}y\\ =\int_{0}^{\pi}\int_{0}^{x}\frac{\sin{x}}{x}\mathrm{d}y\mathrm{d}x\\ =\int_{0}^{\pi}\frac{\sin{x}}{x}x\mathrm{d}x$$