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I want to calculate an integral, but Mathematica gives me different results if I am more specific with the Assumptions option of Integrate.

I have :

Integrate[(1 - Cos[x])/x^2 Exp[I t x], {x, -∞, ∞}]

that returns :

ConditionalExpression[π - π Abs[t], -1 < Re[t] < 1 && Im[t] == 0]

and :

Integrate[(1 - Cos[x])/x^2 Exp[I t x], {x, -∞, ∞}, Assumptions :> t ∈ Reals]

that returns :

1/2 π (Abs[-1 + t] - 2 Abs[t] + Abs[1 + t])

How to explain the difference between the two results ?

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  • $\begingroup$ I would say it's a bug that there is not every conditional in the first result. But the results are the same if you assume this condition. $\endgroup$ – swish Jan 4 '14 at 15:54
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The second one is more general:

Simplify[1/2 π (Abs[-1 + t] - 2 Abs[t] + Abs[1 + t]), -1 < t < 1]
π - π Abs[t]
Plot[{π - π Abs[t], 1/2 π (Abs[-1 + t] - 2 Abs[t] + Abs[1 + t])}, {t, -3, 3}, 
     BaseStyle -> Thick]

enter image description here

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  • $\begingroup$ Try Integrate[(1 - Cos[x])/x^2 Exp[I 2 x], {x, -∞, ∞}] $\endgroup$ – Dr. belisarius Jan 4 '14 at 19:12
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    $\begingroup$ @belisarius It returns 0 as expected. See red line in the plot. $\endgroup$ – ybeltukov Jan 4 '14 at 19:21
  • $\begingroup$ Sorry, I misread your answer $\endgroup$ – Dr. belisarius Jan 4 '14 at 20:25
  • $\begingroup$ Thank you for your answer, but I still don't understand why the first one (which should be more general, as there is no Assumptions) does not give a complete result. Is it a bug ? $\endgroup$ – deltux Jan 5 '14 at 9:17

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