@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.

@Axel Vogt 

The integral converges by Abel-Dirichlet test.
(Non-absolute; it diverges as a Lebesgue integral).

OK, it seems that local in add is mandatory!

f:= proc(a,b) local j; add(evalf(j), j=a..b) end;