Question: How to calculate this integral with Maple?

Let us consider the improper integral

int((abs(sin(2*x))-abs(sin(x)))/x, x = 0 .. infinity);

Si(Pi)-Si((1/2)*Pi)+sum(-(-1)^_k*Si(Pi*_k)+signum(sin((1/2)*Pi*_k))*Si((1/2)*Pi*_k)+Si(Pi*_k+Pi)*(-1)^_k-signum(cos((1/2)*Pi*_k))*Si((1/2)*Pi*_k+(1/2)*Pi), _k = 1 .. infinity)
                    

Mathematica 11 produces a similar expression and a warning

Integrate::isub: Warning: infinite subdivision of the integration domain has been used in computation of the definite integral \!\(\*SubsuperscriptBox[\(\[Integral]\), \(0\), \(\[Infinity]\)]\(\*FractionBox[\(\(-Abs[Sin[x]]\) + Abs[Sin[2\ x]]\), \(x\)] \[DifferentialD]x\)\). If the integral is not absolutely convergent, the result may be incorrect.

Up to Pedro Tamaroff http://math.stackexchange.com/questions/61828/proof-of-frullanis-theorem , the answer is 2/Pi*ln(2) because of 

J := int(abs(sin(2*x))-abs(sin(x)), x = 0 .. T) assuming T>2;
-1/2-signum(sin(T))*signum(cos(T))*cos(T)^2+(1/2)*signum(sin(T))*signum(cos(T))+cos(T)*signum(sin(T))+floor(2*T/Pi)

B := limit(J/T, T = infinity);
                               2 /Pi

K := x*(int((abs(sin(2*t))-abs(sin(t)))/t^2, t = x .. 1)) assuming x>0,x<1;

     2*sin(x)*cos(x)-2*Ci(2*x)*x+Ci(x)*x+sin(1)*x-sin(2)*x+2*Ci(2)*x-Ci(1)*x-sin(x)

                         
A := limit(K, x = 0, right);
                               0

Its numeric calculation results 

evalf(Int((abs(sin(2*x))-abs(sin(x)))/x, x = 0 .. infinity));
                        Float(undefined)

which seems not to be true.

The question is: how to obtain the reliable results for it with Maple, both symbolic and numeric?