advanced improper integral?

Note that
∫(x = 0 to ∞) [arctan(πx) - arctan x] dx/x
= ∫(x = 0 to ∞) (1/x) arctan y {for y = x to πx} dx
= ∫(x = 0 to ∞) ∫(y = x to πx} (1/x) * (1/(1 + y^2)) dy dx.

Now, we interchange the order of integration.
We rewrite the region bounded between y = x and y = πx for x ≥ 0
as x = y/π to x = y with y ≥ 0.

So, we obtain
∫(y = 0 to ∞) ∫(x = y/π to y} (1/x) * (1/(1 + y^2)) dx dy.

Evaluating this:
∫(y = 0 to ∞) ln x * (1/(1 + y^2)) {for x = y/π to y} dy
= ∫(y = 0 to ∞) [ln y - ln(y/π)] * (1/(1 + y^2)) dy
= ∫(y = 0 to ∞) ln [y / (y/π)] * (1/(1 + y^2)) dy
= ∫(y = 0 to ∞) ln π * (1/(1 + y^2)) dy
= ln π * arctan y {for y = 0 to ∞}
= (π/2) ln π.

I hope this helps!