Suspected bug in Integrate

Question

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.

Indeed, it is related. It seems surprising that this problem has not been fixed after so many months. — bbgodfrey, Dec 22 '14 at 14:01
If the imaginary part of t0 is greater than 2 ArcTanh[(a - b)/(a + b)], then the integral is zero. However, I still get zero (erroneously) if the assumption Element[t0, Reals] is added. — Michael E2, Dec 22 '14 at 16:17
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. — Michael E2, Dec 22 '14 at 17:24
In version 10.2, I get the following solution for the second integral:$$\text{ConditionalExpression}\left[\frac{2 \pi }{a^2-b^2},-3 \pi \leq \Re(\text{t0})\leq -\pi \land \Im(\text{t0})=0\right]$$ — Sepideh Abadpour, Oct 14 '15 at 19:11

score 3 · Answer 1 · answered Jul 19 '15 at 09:41

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] *)

Suspected bug in Integrate

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