Hello Henrik, so apparently I misunderstood your post, thank you for pointing this out to me. In particular, I was mislead by ...can only be applied on the integrand....

Since we are dealing with a vanishing denominator, it makes sense to use Limit[exprInt, qt -> -2 q] instead of exprInt /. qt -> 2 q Subsequent simplification with the condition that q should be integer produces the correct result for this condition. But as I mentioned there are two other conditions ( qt == 0, qt == -2 q) which will also lead to non-zero value of the the integral expression.

Best regards, Michael

So I do not agree to Hendrik's statement that the conditions should be applied to the integrand ...

I did not make that statement. I was just stating that the respective condition cannot be applied to the calculated general integral:

But FullSimplify does not see anything else then exprInt. So - how could this function ever come up with this special condition?!

Dimitrios,

I suspect that the point here is that the special case (qt -> 2*q) can only be applied on the integrand, not on the calculated integral (because (-4 q^2 + qt^2) is part of its denominator). Therefore for any simplification afterwards there is no chance for a conditional outcome like

$$1/2\cdot Sin(\phi1 + \phi2 - \phi3)\quad \mbox{if} \quad qt == 2*q$$

Does that make sense to you?