Double integral problem: $\int_0^\pi\int_x^\pi \frac{\sin y}{y} dy\, dx$ Calculate:
$$\int_0^\pi \int_x^\pi \frac{\sin y}{y} dydx$$
How to calculate that? This x is terribly confusing for me. I do not know how to deal with it properly.
 A: You're looking to integrate $\dfrac{\sin y}{y}$ over this area: 

where $x\le y\le \pi$ and then $0 \le x \le \pi$.
But notice that if you swap the axes:

You'll have the same area but seen from a different perspective where $0 \le x\le y$ and $0\le y \le \pi$, which graphically justifies the following application of Fubini's Theorem:
$$\int_0^\pi \int_x^\pi \frac{\sin y}{y} dydx=\int_0^\pi \int _0^y \frac{\sin y}{y} dxdy=\int_0^\pi \sin y dy=2 $$
A: Let $f: (x,y) \mapsto y^{-1}\sin y$ for all $(x,y) \in \mathbb{R}^{2}$ such that $y \neq 0$; let $S := \{ (x,y) \in \mathbb{R}^{2} \mid 0 \leq x \leq \pi, x \leq y \leq \pi \}$; and let $\int_{S} f$ exist. Then by Fubini's theorem we have
$$
\int_{S}f = \int_{0}^{\pi}\int_{x}^{\pi} \frac{\sin y}{y} dy dx = \int_{0}^{\pi}\int_{0}^{y} \frac{\sin y}{y} dx dy = \int_{0}^{\pi}\sin y dy = -\cos y \big|_{0}^{\pi} = 2.
$$ 
A: You need to invert both integrals with Fubini theorem (the function is positive, so no problem here):
$$I = \int_0^\pi \int_x^\pi \frac{\sin(y)}{y} dy dx = \int_D \frac{\sin(y)}{y} dx dy$$
Where $D = \{ (x,y) : 0\leq x \leq \pi\text{ and } x \leq y \leq \pi \}$ 
But you can also rewrite $D$ as $D = \{ (x,y) : 0\leq x \leq y \leq \pi\}$
Or, another form, $D = \{ (x,y) : 0\leq y \leq \pi \text{ and } 0 \leq x \leq y\}$
So we have  
$$I = \int_0^\pi \int_0^y \frac{\sin(y)}{y} dx dy$$
$$ = \int_0^\pi  \frac{\sin(y)}{y}  \left( \int_0^ydx \right) dy$$
$$ = \int_0^\pi  \frac{\sin(y)}{y}  y dy$$
$$ = \int_0^\pi  \sin(y) dy$$
