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MathGroup Archive 1998

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RE: Unexpected behavior with S

Adrian Cable  wrote:
|In fact, it turns out that for any integer k,
|(Series[x^x,{x,0,k}]-1)/(Log[x]) = Integrate[Series[x^x,{x,0,k}]].
Now, |this result doesn't make sense, because it would imply that
|Integrate[x^x,x] = (x^x - 1)/Log[x], whereas in fact Integrate[x^x,x]
|cannot be expressed in closed form. What have I (or Mathematica, for
|that matter) done wrong here to get this obviously incorrect result? |
   I looked into it, and it may be that Mathematica incorrectly does the
integral for certain cases of SeriesData (a series expansion with an
O[x]^n  term).  See below.

I may have missed something, but I would expect zero for Out[2] and
Out[3]  below.

poly=Series[x^x, {x,0,2}];
int=Integrate[poly, x];

Normal[ D[int, x] - poly ]


D[ Normal[int], x ] - Normal[poly]

x/2 + 1/3*x^2*Log[x]

Ted Ersek

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