Here is my attempt: As $[0,\pi/2] \times [0,\pi/2]$ is a finite measure space(hence $\sigma$ finite) and also its $x \times y$ integrable therefore hypothesis of the fubini's theorem are satisfied and
By change of limit we get:
$\int_{0}^{\pi/2}\int_{x}^{\pi/2}y\frac{\sin x}{x}dx dy$
$\implies$ $\frac{\pi^2}{8}\int_{0}^{\pi/2}\frac{\sin x}{x}-\int_{0}^{\pi/2}\frac{x\sin x}{2}dx$
and then I don't know how to integrate the first integral $\int_0^{\pi/2}\frac{\sin x}{x}dx$.
Thanks for any help!!
I think you messed up the change of order. In the first integral, we have $0\leq y \leq \frac \pi 2$. Then, in the second integral, we have $y \leq x \leq \frac\pi 2$. In total, this gives us:
$$0 \leq y \leq x \leq \frac \pi 2$$
Now, if we want to switch the order, we need to remember abide by this. First, $0\leq x\leq \frac\pi 2$, which determines the limits of the new outer integral. Then, we have $0\leq y \leq x$, which determines the limits of the new inner integral. Thus, the new integral becomes:
$$\int_0^{\frac \pi 2}\int_0^x y\frac{\sin x}{x}dydx$$
Now, we can take $\frac{\sin x}{x}$ out of the inner integral since it has nothing to do with $y$, so we get:
$$\int_0^{\frac \pi 2}\frac{\sin x}{x}\int_0^x ydydx$$
$\int_0^x ydy=\frac{x^2}{2}$, so when we substitute back in and multiply it with $\frac{\sin x}{x}$, we get:
$$\int_0^{\frac \pi 2}\frac{x\sin x}{2}dx$$
I will leave the final integration by parts for you to do.