Sub-problem, $\int^{\pi/2}_y \frac{\sin x}{x} dx $, emerging on the page 941 p4b here which asks us to find :
$\int^{\pi/2}_0 \left( \int^{\pi/2}_y \frac{\sin x}{x} dx \right) dy .$
My instructor once showed me some nice deduction for the thing inside the integral, I think it was with exponential functions or something like that. Now the upper border can be handled with the 1st fundamental rule of calculus but I cannot remember how to proceed with the integral, how to proceed here?
Ragib suggested me to draw but failure, ideas how to fix it? (some oddity with $x_{0}$)
Some observations
Let $f(x,y) := \int_{\pi/2}^y \frac{\sin(x)}{x} dx$.
if $y>\frac{\pi}{2}$, then we have no division by zero -case.
If $y<\frac{\pi}{2}$ then we may have an indefinite case a bit more problematic situation, particularly when $y<0$.
If an upper bound for $\sin(x)$ -- look they start from origin and the linear function is below $\sin(x)$ until point $p_{1}$, then there may be some theorem to use (perhaps some Cauchy-something, researching).