calculus - 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?

Answer

Change the order of integration:

$$\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}$$

Since the integrand is continuous, there is no problem changing the order of integration.