how to calculate $\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\sin{|x-y|}dxdy$? Can you provide your solution step by step? particularly how to address the absolute.
I do know how to address some similar integral, like
$\int_{-\pi/2}^{\pi/2}\sin|x-\pi/4|dx$ $=\int_{-\pi/2}^{\pi/4}\sin{(\pi/4-x)dx + \int_{\pi/4}^{\pi/2}}\sin(x-\pi/4)dx$.
But, when it comes to double integral, I am not sure what the $y$ exactly is, and I do not know how to divide the intergal containing an absolute calculation.
If you take values of $(x - y)$ below the line $y = x$, then $y \lt x$ and hence $(x - y)$ will be positive. So, from symmetry of the region of integration, the integral is, ($a = \dfrac{\pi}{2}$ )
$\begin{equation} \begin{split} I &= \displaystyle 2 \int_{x=-a}^{a} \int_{-a}^{x} \sin(x - y) dy dx \\ &= \displaystyle 2 \int_{x = -a}^{a} \cos(x - y) \bigg|_{-a}^{x} dx \\ &= \displaystyle 2 \int_{x = -a}^{a} (1 - \cos(x + a) ) dx \\ &= \displaystyle 2 ( x - \sin(x + a) ) \bigg|_{-a}^{a} \\ &= 2 ( 2 a - \sin(2a) + \sin(0) ) \\ &= 2 (\pi - 0 + 0 ) \\ &= 2 \pi \end{split} \end{equation}$