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Bug introduced in 9.0 or earlier and fixed in 10.4.0

I just discovered that Integrate with simple poles in its integrand sometimes gives different results depending on number type of its limit (Exact vs Inexact).

(*Approximate upper limit*)
Integrate[1/z, {z, -1 - I, -1 + .5 I}] // N

-0.235002 + 5.03414 I

(*Exact upper limit*)
Integrate[1/z, {z, -1 - I, -1 + 1/2 I}] // N

-0.235002 - 1.24905 I

The second result is the one that everyone would agree with as it properly accounts for crossing the logarithmic cut running along the negative axis.

Two questions:

  1. Is this the correct behavior of Integrate? I can't seem to find this behavior mentioned in the documentation.

  2. I have a huge Mathematica notebook (containing some 35 Integrate calls) that is supposed to compute integrals symbolically and/or numerically depending on the type of its limit. e.g. of the form:

    f[a_,b_] := Integrate[1/z, {z, a, b}]

    What would be some quick ways to get around this bug? I noticed NIntegrate correctly accounts for the branch cut, but it only works if there's nothing symbolic left.

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  • $\begingroup$ What happens if you do Integrate[1/z, {z, -1 - I, -1 + t I}] /. t -> {1/2, 0.5}? $\endgroup$
    – J. M.'s torpor ♦
    Dec 14 '15 at 14:01
  • $\begingroup$ @J.M. The initial Integration step takes forever and finally gives a ConditionalExpression. After replacing, you get Undefined for both cases. (btw you can't do your replace-list trick when there is a ConditionalExpression) $\endgroup$
    – QuantumDot
    Dec 14 '15 at 14:08
  • $\begingroup$ Reported as a bug. $\endgroup$
    – Daniel Lichtblau
    Dec 14 '15 at 19:37
  • $\begingroup$ @DanielLichtblau With Mathematica's branch-cut convention for Log, isn't $\int_a^b \frac{dx}{x} = \ln(b/a)$ always true for any $a, b \in \mathbb{C}$? What's the deal with all the CondintionalExpressions that it spits out? $\endgroup$
    – QuantumDot
    Jan 4 '16 at 17:29
  • $\begingroup$ Maybe. But Integrate is not going to figure out that that always corrects for Log[b]-Log[a]. $\endgroup$
    – Daniel Lichtblau
    Jan 4 '16 at 20:05
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These integrals in the complex plane are ambiguous in that the paths are not specified. Perhaps, the first Integrate to the long way around z = 0, as in

Integrate[1/z, {z, -1 - I, +1 - I, +1 + 1/2 I, -1 + 1/2 I}] // Simplify
(* 1/4 I (7  π - 4 ArcTan[1/2] + I Log[64/25]) *)
% // N
(* -0.235002 + 5.03414 I *)

The same result is obtained from

Integrate[1/z, {z, -1 - I, +1 - I, +1 + .5 I, -1 + .5 I}]

The only way I know to be sure that the desired path is taken is to express the integral as

I Integrate[1/(-1 + I y), {y, -1, .5 }]
(* -0.235002 - 1.24905 I *)
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  • $\begingroup$ Well, I thought that in a computer program like Mathematica, one would have a consistent convention which would most naturally be straight path. If Mathematica just picks whatever contour it wants, I would call this a catastrophic design flaw. $\endgroup$
    – QuantumDot
    Dec 14 '15 at 19:22
  • $\begingroup$ @QuantumDot I would think so too, but I am unable to find a discussion of this point in the documentation. Contour integrals are so common that they should be covered explicitly. $\endgroup$
    – bbgodfrey
    Dec 14 '15 at 19:40
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This bug has been fixed as of Mathematica 10.4.0.

Integrate[1/z, {z, -1 - I, -1 + .5 I}]

(* -0.235002 - 1.24905 I *)
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