Proper way to simplify integral result in Mathematica given integer constraints

Question

Evaluating the following integral should be non-zero, and mathematica correctly gives a non-zero result

Integrate[ Cos[ (Pi * x)/2 ]^2 * Cos[ (3*Pi*x)/2 ]^2, {x,         


        

Answer 1

Lets use some conclusive conditions about the two integers m=n||m!=n.

Assuming[{(n \[Element] Integers && m \[Element] Integers && m == n)},
Integrate[Cos[(Pi x)/2]^2 Cos[((2 n + 1) Pi x)/2] Cos[((2 m + 1) Pi x)/2],
{x, -1, 1}]]

The answer for this case is 1/2. For the other case it is

Assuming[{(n \[Element] Integers && m \[Element] Integers && m != n)},
Integrate[
Cos[(Pi x)/2]^2 Cos[((2 n + 1) Pi x)/2] Cos[((2 m + 1) Pi x)/
 2], {x, -1, 1}]]

and the answer is 0.

However I am amazed to see that if we add this two conditions as an "either or stuff", Mathematica returns one zero after integration. I mean in case of the following I am getting only zero but not ``1/2||0`.

Assuming[{(n \[Element] Integers && m \[Element] Integers && 
 m == n) || (n \[Element] Integers && m \[Element] Integers && 
 m != n)}, 
Integrate[
Cos[(Pi x)/2]^2 Cos[((2 n + 1) Pi x)/2] Cos[((2 m + 1) Pi x)/
 2], {x, -1, 1}]]

By the way we can see the conditions exclusively where this integral becomes Indeterminate.

res = Integrate[
Cos[(Pi x)/2]^2 Cos[((2 n + 1) Pi x)/2] Cos[((2 m + 1) Pi x)/
  2], {x, -1, 1}] // Simplify

The output is here.

Now lets see all the relations m and n can have to make the Integral bad!

BadPart = (res*4 Pi);
Flatten@(Solve[(Denominator[#] == 0), m] & /@ 
 Table[BadPart[[i]], {i, 1, Length@BadPart}] /. 
Rule -> Equal) // TableForm

So these are the special cases which as Sjoerd mentioned are having infinite instances.

BR