@Rouben Rostamian  

Yes, you are right. The changes of variables failed for the intervals not included in [0,Pi].
It is sad that we have to check each simple step that Maple does :-(

P.S. Actually I was surprised that Preben's split seemed to work because I have tried myself similar methods and failed.

Edit.
We should choose only changes of variables with "unique" inverses (mainly for trig functions), Maple not being smart enough to find the correct branch.
Change(Int(f(x), x=3*Pi/2 .. 2*Pi), cos(x)=t);
         
is wrong. We'll have to use:

Change(Int(f(x), x=3*Pi/2 .. 2*Pi), x=2*Pi-arccos(t));

@Preben Alsholm

Yes, it's correct, simple, and can be done by hand (the 4 integrals cancel in pairs). Thank you.

My workaround was

T:=IntegrationTools:-Change(J, tan(x/2)=t):
subsindets(T, `=`, u -> (t=-infinity..infinity))/2:
simplify(expand(%)):
value(%);

I did not like it because the parity and periodicity is used and the degree of the polynomial is 12.
(It can be reduced to degree 6 with another change of variables but then evala() is needed.)

@tomleslie 

It is not correct because the antiderivative obtained by int(...,x) is not continuous.

@Rohith 

Yes. But this does not imply that such a function is unbounded; e.g. f : R --> R, f(x) = arctan(x).

simplify(maximize(Expr, x=0..2*Pi))   assuming a>0,b<0;
        sqrt(a^2+b^2)

In most cases, for expressions with parameters the assumptions are necessary.

Here they should not be, but with location it's difficult otherwise.

@brian bovril 

I asked the same thing some time ago but nobody answered.

@Carl Love 

To completely understand the situation (with an .m file or %m string) it would be very nice to have an example of a simple procedure acting similar to Sol (which was generated by dsolve). Of course this procedure could be in another context.

@Carl Love 

Probably your browser is not set to recognize utf-8 (unicode) characters. I am sure that they are visible in the worksheet. I have changed the text to use ASCII characters.

I'd suggest to present a simplified version of your problem illustrating the type of the integro-differential equation/system you want to solve (which does not need several screens to be displayed).

@digerdiga 
The example was to recall what a CPV is. In our case a better example would be int(sin(sin(x)), x=-infinity..infinity, CPV) = 0.

There ar standard theorems. Note that our function has infinitely many poles in C. What formula have you used? Was it correct?
Give an example of a numerical result you have obtained.

@digerdiga 

For using residues some conditions must be verified, and the nature of the convergence of  the original integral J must be known.

Using CPV enhances the convergence. A CPV can be finite for an integrand with singularities.
E.g.  int(1/x, x=-infinity..infinity, CPV) = 0.
The table and the plot used the CPV (which equals J when this one exists).

@digerdiga 

In the CPV sense the integral converges. The problem is more maths than Maple. When I'll have time I'll post a solution at the Application Center.
In the mean-time here is a table of values. Observe some kind discontinuities at t=0 and t=±2 where the ordinary Riemann integral does not exist. I am going to investigate the case of irrational t (but this seems to be a tough task and Maple cannot help at all).

F(-21/10) = -1.34045072401727*I
F(-20/10) = -.995421082418696*I
F(-19/10) = -.562990651564561*I
F(-18/10) = -.622200895257619*I
F(-17/10) = -.687638334639360*I
F(-16/10) = -.759957889597397*I
F(-15/10) = -.839883358545194*I
F(-14/10) = -.928214662439857*I
F(-13/10) = -1.02583585065994*I
F(-12/10) = -1.13372394886877*I
F(-11/10) = -1.25295873741565*I
F(-10/10) = -1.38473355814056*I
F(-9/10) = -1.53036725774037*I
F(-8/10) = -1.69131738722984*I
F(-7/10) = -1.86919478960212*I
F(-6/10) = -2.06577972168680*I
F(-5/10) = -2.28303967155866*I
F(-4/10) = -2.52314904981962*I
F(-3/10) = -2.78851095183085*I
F(-2/10) = -3.08178120869891*I
F(-1/10) = -3.40589496772606*I
F(0) = -1.95029670411682*I
F(1/10) = -.150852890432947*I
F(2/10) = -.166718227414152*I
F(3/10) = -.184252136451252*I
F(4/10) = -.203630102799237*I
F(5/10) = -.225046067658478*I
F(6/10) = -.248714369203441*I
F(7/10) = -.274871887751179*I
F(8/10) = -.303780416539164*I
F(9/10) = -.335729281839998*I
F(10/10) = -.371038238635994*I
F(11/10) = -.410060670834519*I
F(12/10) = -.453187128052909*I
F(13/10) = -.500849234370305*I
F(14/10) = -.553524008166522*I
F(15/10) = -.611738636282314*I
F(16/10) = -.676075750282477*I
F(17/10) = -.747179257628374*I
F(18/10) = -.825760786120238*I
F(19/10) = -.912606806107378*I
F(20/10) = -.522580426869140*I
F(21/10) = -0.404209101674264e-1*I

When I said that the integral converges I had in mind the case t = 1.
For t = 0 it obviously diverges; however it converges in the sense of Cauchy Principal Value.
As a standard Riemann integral I was able to prove the convergence only for a rational t = m/n and m = odd. For t=2 the integral diverges.

@Carl Love 
This was my first idea too. But the speed gain is < 50% and the memory used is high.

@acer 

Thank you for the info. I have never used Grid:-Seq. At least for this example it's faster than Threads:-Seq.