Question

It seems that Mathematica’s handling of principal value integrals fails on some corner cases. Consider these two expressions (which should give the same result):

Integrate[UnitBox[x]/(x0 - x), {x, -Infinity, Infinity},
  PrincipalValue -> True, Assumptions -> {x0 > 0}] /. x0 -> 1 // Simplify
Integrate[UnitBox[x]/(x0 - x) /. x0 -> 1, {x, -Infinity, Infinity}, 
  PrincipalValue -> True]

In Mathematica 7.0.0 I get

I Pi+Log[3]
Log[3]

Has this been fixed in later versions? Does anybody have an idea for a (more or less) general workaround?

EDIT: The two expressions above should calculate the same result, the first by calculating a general form of the integral and evaluating it at x0=1, the second by performing the integral with x0 set to 1. Since the Cauchy principal value has a precise mathematical definition, Mathematica should give the same result or decline to answer.

EDIT 2: A perhaps simpler example of the same bug, putting a factor of -1 inside and outside the Integral give different answers (second one gives the correct answer, first one doesn’t):

-Integrate[ UnitBox[x]/(x0 - x), {x, -Infinity, Infinity}, PrincipalValue -> True, Assumptions -> {x0 > 0}]
 Integrate[-UnitBox[x]/(x0 - x), {x, -Infinity, Infinity}, PrincipalValue -> True, Assumptions -> {x0 > 0}]

Answer 1

I don’t think this is a bug in PrincipalValue. In the first line PrincipalValue does not work “correctly” because the position of the pole is not known until after Integrate is done.

EDIT:
I played around in Mathematica a bit and this is exactly what happens. You can see for yourself by using the Trace[] command. The output is a bit messy (which is why I don’t replicate it here), but you can see where the integration gets done and where the value for x0 is substituted and how that messes with PrincipalValue.

EDIT2:
So back to solving your actual problem. If you use an Assumption that specifies which side of x=1/2 x0 lies, then the two examples give the same answer.