@sgils1 

Your problem is actually an integral (so, instead of dsolve, it's possible to simply use int).

Unfortunately, Maple does not find a continuous antiderivative in the symbolic case i.e. it is valid only locally.
(This is not a bug, it is documented; for example int(1/(2+cos(x)), x) is discontinuous and it is exactly the integral obtained by hand using the substitution tan(x/2)=t; note that for this example there is a simple continuous antiderivative, but in general a CAS does not try to find it).

For example:

Here, (3) and (4) are continuous, but (2) is discontinuous.
The interesting fact is that (3) is also valid for a<0, but Maple does not care.

For your simple example, there are workarounds (looking for discontinuities and adding constants if needed; or use assuming).

@torabi 

Not quite useful. I was hoping to see an example where both input and output are Maple expressions.
I see instead a problem which must be formalized and has lots of unessential parameters such as lambda, mu, J2 etc which seem to be constants.
I don't know what the circled triple integral means. What is V? Why that change of variables? It is not clear what quantities are constans. I don't understand the assumption (4) ; Aijk are unique? Are Jijk etc given? etc, etc.
In conclusion, I cannot help here.

@torabi 

It would be useful to see:
1. A short example (just the example, no subs)
2. A description of what you want to obtain
3. The expected result.

If 1-2-3 are clear enough, then it will be probably possible to find a general method for doing this.

@tomleslie 

Actually the integral is oo (diverges).

@MacDTU 

Your f(t) := int(1/x, x = 1 .. t) does not define the desired function (Maple procedure).
[Technically it defines an "empty" procedure having an entry in the remember table].
(This is a standard mistake for a beginner.)

The correct definition (in 1D math mode) is

f := t -> int(1/x, x = 1 .. t) ;

When I saw the problem I used the substitution z = 1 - x and I found the first term of the series by hand
z ~ LambertW(4*n^2)/n.

Note the the series for z fails too.

@Gourav 

After correction, you do not have any more singularities for omega>0, but omega=0 still remains and makes the inner integral divergent.

Note that the integral cannot be computed symbolically. In general we need numerical values for parameters to compute numerically (not a very simple task for an improper double integral).
 

@student_md 

if ... then ...   is a conditional statement, `if` is a condional function (alias to ifelse), see  ?ifelse.

Do you think that the integral is convergent? I don't.

@Carl Love 

There are many such problems in number theory and most of them are unsolved.
E.g. it is conjectured that n! + 1 is not a square number for n>7.

UMG:=proc(n::posint)
local N := Normal(0,1),
      S := Statistics:-Sample(N,[n,n]) + I*Statistics:-Sample(N,[n,n]),
      j;
Matrix(LinearAlgebra:-GramSchmidt([seq(S[..,j],j=1..n)],'normalized'))
end:

@Joe Riel 

Yes, you are right.

@Mikhak 

Here symbolic solutions does not make much sense. Try:

solve(F, x, allsolutions);
F1:=convert(F, rational);
solve(F1, x, allsolutions);

It seems that you want numeric solutions.

f := (1 + a * cos(theta1*x)) * cos(theta2*x) - b:
F:=eval(f, [theta1=2, theta2=24.5, a=0.25 , b=0.1]):
RootFinding:-Analytic( F, x, re=0..4*Pi, im=-0.1 .. 0.1); sort([%]);

[0.0608404851442370, 0.195661496064193, 0.317169250177469, 0.452338844731748, 0.573321271654020, 0.709193982762240, 0.829303334433960, 0.966198716646160, 1.08517751285740, 1.22323304796064, 1.34112176470098, 1.48005365921274, 1.59739755110209, 1.73644235409529, 1.85412330733945, 1.99244553947813, 2.11113047152258, 2.24831526005946, 2.36815894926479, 2.50426092567311, 2.62505648897743, 2.76036869003974, 2.88177847976375, 3.01665365639252, 3.13832384040376, 3.27311241579994, 3.39469611285126, 3.52974541499388, 3.65089259774324, 3.78655671358430, 3.90691457307736, 4.04353049236051, 4.16280435532856, 4.30057605523926, 4.41870733647864, 4.55747629551925, 4.67487862383864, 4.81398122172356, 4.93149765193513, 5.07006063884640, 5.18846029384880, 5.32593928588975, 5.44550314841780, 5.58185358677780, 5.70244066316500, 5.83791777406930, 5.95920741126850, 6.09415862817895, 6.21579641986160, 6.35057419449755, 6.47221198618020, 6.60716320309065, 6.72845284028987, 6.86392995119415, 6.98451702758135, 7.12086746594140, 7.24043132846945, 7.37791032051040, 7.49630997551275, 7.63487296242400, 7.75238939263560, 7.89149199052055, 8.00889431883995, 8.14766327788050, 8.26579455911990, 8.40356625903060, 8.52284012199870, 8.65945604128180, 8.77981390077485, 8.91547801661595, 9.03662519936530, 9.17167450150790, 9.29325819855925, 9.42804677395540, 9.54971695796665, 9.68459213459545, 9.80600192431945, 9.94131412538175, 10.0621096886861, 10.1982116650944, 10.3180553542997, 10.4552401428366, 10.5739250748805, 10.7122473070197, 10.8299282602638, 10.9689730632571, 11.0863169551464, 11.2252488496582, 11.3431375663985, 11.4811931015018, 11.6001718977130, 11.7370672799252, 11.8571766315982, 11.9930493427051, 12.1140317696274, 12.2492013641817, 12.3707091182950, 12.5055301292148]

You have all the 98 real solutions in the interval 0 .. 4*Pi.
F is 4*Pi - periodic, so any other real solution can be obtained by adding an integer multiple of 4*Pi.

@mmcdara 

A Maple command to squeeze polynomials is compoly. Unfortunately it fails for almost any random polynomial.
My example was not presented as a challenge (and the simplified form was given).
Here is also a simple one:

F := 1238*a^9 + (6927*b + 3756)*a^8 + (40476*b^2 + 26226*b + 5334)*a^7 + (145584*b^3 + 144582*b^2 + 46947*b + 5376)*a^6 + (326904*b^4 + 436890*b^3 + 218880*b^2 + 48384*b + 4032)*a^5 + (489558*b^5 + 815670*b^4 + 543885*b^3 + 181440*b^2 + 30240*b + 2016)*a^4 + (489988*b^6 + 979926*b^5 + 816510*b^4 + 362880*b^3 + 90720*b^2 + 12096*b + 672)*a^3 + (314936*b^7 + 734882*b^6 + 734877*b^5 + 408240*b^4 + 136080*b^3 + 27216*b^2 + 3024*b + 143)*a^2 + (118084*b^8 + 314894*b^7 + 367396*b^6 + 244944*b^5 + 102060*b^4 + 27216*b^3 + 4536*b^2 + 430*b + 16)*a + 19686*b^9 + 59055*b^8 + 78735*b^7 + 61236*b^6 + 30618*b^5 + 10206*b^4 + 2268*b^3 + 323*b^2 + 25*b;

It can be written (in 1D) using 534 characters (using simplify/size), i.e.  length(convert(F,string)) = 534.
There is a simplified form using only 49 characters.
It would be very nice to have in Maple such a "squeezer".