10
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Bug introduced in 9.0.1 or earlier and fixed in 10.1.0


In version 10.0:

Integrate[1/(a^2 + b^2 - 2 a b Cos[t]), {t, 0, 2 Pi},  Assumptions -> {a > b > 0}]
(* (2 Pi)/(a^2 - b^2) *)
Integrate[1/(a^2 + b^2 - 2 a b Cos[t + t0]), {t, 0, 2 Pi},  Assumptions -> {a > b > 0}]
(* 0 *)

However, a phase shifting on t should not affect the integral.

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  • 2
    $\begingroup$ Maybe related to 48437 ? $\endgroup$ – b.gates.you.know.what Dec 22 '14 at 13:38
  • $\begingroup$ @b.gatessucks Quite similar. $\endgroup$ – njpipeorgan Dec 22 '14 at 13:58
  • $\begingroup$ Indeed, it is related. It seems surprising that this problem has not been fixed after so many months. $\endgroup$ – bbgodfrey Dec 22 '14 at 14:01
  • $\begingroup$ @bbgodfrey, yes, same thing under 10.0.2. $\endgroup$ – Alexey Bobrick Dec 22 '14 at 14:47
  • 1
    $\begingroup$ This will give the right answer: Integrate[1/(a^2 + b^2 - 2 a b Cos[t + t0]), {t, 0, 2 Pi}, Assumptions -> a > b > 0 && -Pi < t0 < Pi]. The problem I suspect is that the integral cannot really be done by substitution t = 2 ArcTan[u/2], since you have to count how many times the angle winds around the origin. $\endgroup$ – Michael E2 Dec 22 '14 at 17:24
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This appears to be fixed in 10.1.0:

Integrate[1/(a^2 + b^2 - 2 a b Cos[t]), {t, 0, 2 Pi}, 
 Assumptions -> {a > b > 0}]
(* (2 π)/(a^2 - b^2) *)

Integrate[1/(a^2 + b^2 - 2 a b Cos[t + t0]), {t, 0, 2 Pi}, 
 Assumptions -> {a > b > 0}]
(* ConditionalExpression[(2 π)/(a^2 - b^2), -3 π <= Re[t0] <= -π && Im[t0] == 0] *)
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