From: nma@12000.org   
      
   On 10/15/2019 5:03 PM, Albert Rich wrote:   
   > On Tuesday, October 15, 2019 at 6:16:30 AM UTC-10, clicl...@freenet.de wrote:   
   >   
   >> I remember that no computer algebra system was found able to confirm   
   >> the identity:   
   >>   
   >>    2*pi*SQRT(1 - z) = SQRT(z - 1)*(2*LN(- SQRT(z - 1)) - LN(z - 1))   
   >>   
   >> for arbitrary complex z. [...]   
   >   
   > Substituting z+1 for z in this identity and rearranging terms yields the   
   identity   
   >   
   >      log(-sqrt(z)) = log(z)/2 - pi*sqrt(z)/sqrt(-z)   
   >   
   > valid for arbitrary complex z.  More generally the identity   
   >   
   >      log(-z^n) = n*log(z) - pi*sqrt(z)/sqrt(-z)   
   >   
   > appears to be valid for arbitrary complex z provided n is positive (someone   
   please confirm).   
   >   
      
   hi Albert,   
      
   A quick test shows it is valid for n>0. Only issue is when z=0.   
      
   At z=0  log(-z^n) gives -infinity while n*log(z) - pi*sqrt(z)/sqrt(-z)   
   gives indeterminant expression because of the 0/0   
      
   Btw, this the same with the original identity   
      
   2*pi*SQRT(1 - z) at z=0 is zero, but   
   SQRT(z - 1)*(2*LN(- SQRT(z - 1)) - LN(z - 1)) at z=0 is indeterminant   
   because of the 0*(-infinity - infinity) term that shows up above when z=0.   
      
   zero*infinity is not zero, Mathematica says this is indeterminant   
   and maple says undefined.   
      
   I tried Mathematica Reduce also, but it is not able to verify it,   
   it was stuck thinking for an hr so I stopped it.   
      
   > Perhaps computer algebra systems and Rubi should use this identity to   
   normalize integrands involving dependent logarithms; thereby making the second   
   integration problem in Welz_problems.txt trivial.   
   >   
   > Albert   
   >   
      
   Best,   
   --Nasser   
      
   --- SoupGate-Win32 v1.05   
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