Re: v7.0.0 bug with PrincipalValue -> True

*To*: mathgroup at smc.vnet.net*Subject*: [mg110111] Re: v7.0.0 bug with PrincipalValue -> True*From*: Bill Rowe <readnews at sbcglobal.net>*Date*: Wed, 2 Jun 2010 02:05:31 -0400 (EDT)

On 6/1/10 at 7:40 AM, januswesenberg at gmail.com (janus) wrote: >Hi, >as a physicist I'm allowed to use Cauchy principal values (tongue in >cheek), but it seems that my currently installed version of >Mathematica (7.0.0 for Mac) is having some difficulties. Consider >the following integral (same result with PrincipalValue- >>False as expected): >Integrate[UnitBox[x]/(x0-x)/. x0->1,{x,- >Infinity,Infinity},PrincipalValue->True] Log[3] >Now, do the same integral with an unknown position of the pole, and >an I sneaks in: >Integrate[UnitBox[x]/(x0-x),{x,-Infinity,Infinity},PrincipalValue- >>True,Assumptions-> {x0>0}]/.x0->1//Simplify >I \[Pi]+Log[3] >Would somebody with a newer kernel mind checking whether this is a >fixed bug, or is there perhaps a workaround? >Take[SystemInformation["Kernel"],2] {Version->7.0 for Mac OS X x86 >(64-bit) (November 11, 2008),ReleaseID- >>7.0.0 (1148398, 1148273)} I get the same results using In[1]:= Take[SystemInformation["Kernel"], 2] Out[1]= {Version->7.0 for Mac OS X x86 (64-bit) (February 19, 2009),ReleaseID->7.0.1 (1214020, 1213835)} But if I replace x0 with 1, things are different. That is: In[18]:= Integrate[UnitBox[x]/(1 - x), {x, -Infinity, Infinity}, PrincipalValue -> True] // Simplify Out[18]= log(3)