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Unexpected result from trigonometric integral simplification?

In the notebook below, I calculate a general expression and get zero. This is wrong in general because of the following special case: for qt = 2q, I get the sine of some phases. I have confirmed the special case with pen and paper. Can somebody please tell me why Mathematica screws up like this?

P.S. I know that the attached notebook was created in Mathematica 11. I assure you that Mathematica 13 makes the exact same mistake.

POSTED BY: Dimitrios Tsevas
5 Replies
Posted 3 years ago
POSTED BY: Michael Helmle
Posted 3 years ago
POSTED BY: Michael Helmle
POSTED BY: Henrik Schachner

It seems that Integrate in such cases with parameters will give the "generic" answer and it is up to you to work out the special cases. A simpler example:

Integrate[Cos[a x], {x, 0, 1}]
Limit[%, a -> 0]

One may wish the following to give a conditional expression:

Integrate[Cos[a x], {x, 0, 1}, GenerateConditions -> True]

but, sadly, it does not.

POSTED BY: Gianluca Gorni

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?

POSTED BY: Henrik Schachner
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